Math 365, Elementary Statistics

Lesson 9 :Testing Hypotheses

Satya Mandal

The Philosophy of Testing Hypotheses	9.1 Significance Test for mean μ, when σ is known (Homework 29)	9.2 Significance Test for μ, Case of σ Unknown (Homework 30-31)
9.3 Population Proportion(Homework 32)	9.4 Testing Hypotheses on Variance σ² (Homework 33)	9.5 A Significance Test to Compare Two Populations (Homework 34)
9.6 Compare Means of Two Populations: σ₁, σ₂ Unknown (Homework 35)	9.7 Comparing Proportions p₁, p₂ of Two Populations (Homework 36)	Homework 29 - 36

Due Date: Visit the homework site.

The Philosophy of Testing Hypotheses

The Testing of hypotheses is another approach to estimation of parameters. A hypothesis H₀, called the Null hypothesis, is tested against another hypothesis H_A, called the alternative hypothesis. Only one of these two hypotheses is true. Based on the collected sample and established testing criterion, one of them is accepted and the other one rejected. The following two examples would provide further insight.

Example 1. An assertion is made that the disparity between the wages (annual income) of working men and women does not exist any more. To test this assertion, the mean annual incomes μ₁, μ₂, respectively, of the working male and female populations were compared. Our Null hypothesis H₀ would be that the mean annual income μ₁ of the working male population would be higher than the mean annual income μ₂ of the working female population. The Alternative Hypothesis H_A would be, as the assertion suggests, that these two means would be equal. We write them formally as:

H₀ : μ₁- μ₂ > 0
H_A : μ₁- μ₂ = 0

Example 2. A TV commentator mentioned that, during the last decade, the life expectancy of human being has increased substantially from 75 years. To test this assertion, the mean life expectancy μ was compared with 75. The Null hypothesis H₀ would be that the mean life expectancy μ remains equal to 75, as it was before. The Alternative Hypothesis H_A would be that, as the assertion suggests, the mean μ rose above 75 year by now. We write them formally as:

H₀ : μ =75
H_A : μ >75

Definitions and Terminologies.

Following are some definitions and terminologies.

Definition. A statistical hypothesis is defined to be a statement, claim, or proposition regarding a population. Usually, it would be about the values of the population parameters. The hypotheses H₀ and H_A in the above two examples would examples of statistical hypotheses.
It would be important to distinguish which one would be the Null hypothesis and which one would be the alternative hypothesis in a given context. One of them would, essentially, be the negation of the other.
The Null hypothesis H₀ represents the status quo. It would be the conventional wisdom. It represents something that was accepted for a long time, or some assumption or method that has been working reliably for a long time. Null hypothesis would remain as the default, unless the collected data provides very strong evidence against it, in favor of the alternative. There is a clear bias in favor of the Null Hypothesis.

The alternative hypothesis represents a new claim or something out of the ordinary. It could be a researcher's new technology or some sales person's claim. The bar for acceptance of the Alternative Hypothesis is very high. The burden of proof of its validity belongs to those who assert the same. There may even be resistance or skepticism about its validity. It would be accepted only if there is very strong evidence, in the collected data, in its support.

There are reasons for such favoritism in favor of Null Hypothesis. This is because an incorrect decision to reject the Null may have more serious consequences than rejecting the Alternative incorrectly. For example, in any medical test, erroneously concluding that the patient does not have an ailment would have more grievous consequences than erroneously concluding that the patient has the same ailment. In particular, when one designs a pregnancy test, the priority would be to minimize chances (probability) of erroneously concluding that one is not pregnant when one is indeed pregnant, than the converse. Common sense dictates that such a test could only allow a maximum of five percent of such erroneous conclusions. Such erroneous conclusions are also known as false negative and false positive.
Given a Null hypothesis H₀ and an alternative hypothesis H_A, a test of hypothesis is a rule or a procedure to decide, based on the collected sample, whether to accept H₀ or H_A. The test will be based on the value of a test statistic. The rule is also called the decision rule.
A test of hypothesis is also known as a Significance Test. The test will be based on the value of a test statistic.
Two Types of errors. In such testing of hypotheses, two types of mistaken conclusions (errors) are possible as follows.
1. Rejecting the Null H₀ when it is in fact true would be called the type one error. The analogy would be a false negative.
2. Accepting the Null H₀ when it is in fact false would be called the type two error. The corresponding analogy would be a false positive.
3. The probability of type one error would be called the level of significance. It would be denoted by α. Since the priority would be to minimize the frequency of false negative, α would be a small number. Most often, α will be a .1, .05, .01 or a small number.

The rest of this chapter would be analogous to Lesson 7 and 8. Corresponding to each interval estimation we considered, there would be one Significance Test.

9.1 A Significance Test for mean μ, when σ is known

Let X be a random variable with mean μ and standard deviation σ. Some of our hypotheses testing would look like the following.

Two Tail Test	Left Tail Test	Right Tail Test
H₀ : μ = 75 H_A : μ ≠ 75	H₀ : μ = 75 H_A : μ < 75	H₀: μ = 75 H_A: μ > 75

More generally, they would look like one of the following.

Two Tail Test	Left Tail Test	Right Tail Test
H₀ : μ = μ ₀ H_A : μ ≠ μ ₀	H₀ : μ = μ ₀ H_A : μ < μ ₀	H₀: μ = μ ₀ H_A: μ > μ ₀

In this course, all the Null Hypotheses H₀ would be an equality. The alternative Hypotheses H_A would be one of the three inequalities as above.

Develop a Significance Test

A Significance Test for the mean μ would be developed for the following Null and Alternative hypotheses:

H₀ : μ = μ ₀
H_A: μ ≠ μ ₀

Take a sample X₁,X₂, …, X_m of size m from the X population and let X be the sample mean.

The sample size m is assumed to be large. Therefore, by CLT X has

N(μ, σ_X) distribution, where σ _X = σ/√m.
By increasing the sample size m, both type one and type two errors can be controlled. Once the sample size is fixed, it is not possible to control both simultaneously. As one of them is minimized the other one goes up. As was mentioned above, priority would be to control the probability of type one error, that is the level of significance α. Therefore, a Test of Significance at the level of significance α will be developed.
As usual, use X as an estimator of μ. As the alternative hypothesis is H_A : μ ≠ μ₀, the null hypothesis H₀ would be rejected, only if X and μ₀ are far apart, that is, if

| X - μ₀| is large.
If H₀ is true, then μ = μ₀ and

Z=(X-μ₀) /σ_X has N(0,1) distribution, where σ_X = σ/√m.

Expression Z above will be called a test statistic and we will accept H₀ if the observed (absolute) value |z| of |Z| is small and reject H₀ if the observed value |z| of |Z| is large.
If H₀ is true, then
P(Z is not within [ -z_{α /2,} z_α/2 ]) = α
So, at the level of significance α, the decision rule is set as:

Reject H₀ if z is not within [ -z_α/2, z_α/2 ] where z = (x-μ₀) /σ_X = (x-μ ₀)√m/σ.

Accept H₀ otherwise.

Obviously, rejection of H₀ is synonymous to acceptance of H_A.
The above decision rule works only if we know the value of σ.

The Hypothesis Test for the mean μ, when σ is known

Arguing similary, set the Decision Rules for all three tests for mean μ. It is assumes that the value of σ is known.

Two-tail test: The decision rule for testing the hypotheses

H₀: μ = μ₀
H_A: μ ≠ μ₀

is set as, at the level of significance α,

Reject H₀if z is not within [ -z_α/2, z_α/2 ] where z = (x-μ ₀)√m/σ.

Accept H₀ otherwise.
Left-tail test: The decision rule for testing the hypotheses

H₀ : μ = μ₀
H_A : μ < μ₀

is set as, at the level of significance α

Reject H₀if z < -z_α where z = (x-μ₀)√m/σ.

Accept H₀ otherwise.
Right-tail test: The decision rule for testing the hypotheses

H₀ : μ = μ₀
H_A : μ > μ₀

is set as, at the level of significance α,

Reject H₀ if z > z_α where z = (x-μ₀)√m/σ.

Accept H₀ otherwise.
Informally, these will be called Z-Tests.

Definition. The set of values (that is, the intervals) that leads to the rejection of the Null hypothesis H₀ is called the rejection region or the critical region.

p-Value based Decision Rules

There is an alternative and equivalent way to describe the decision rules, stated above. This is based on some probability computations.

Definition. Let T be a test statistic to test H₀ against H_A. Let the observed value of T = t. The p-value, for this test, is defined as the probability, assuming H₀ is true, that T will take a value at least as extreme as t or worse. In the above decision rules, the test statistic is

Z = (X-μ₀) /σ_X = (x-μ₀)√m/ σ

In particular for the Z-test, if Z = z is the observed value of Z, then p-value is define as follow. The normalcdf function of TI-84 can be used to compute the same.

For the two-tail test, the p-value is given by
p=P(Z ∉[-|z|,|z|]) = 1 - P(-|z| < Z < |z|) = 1 - normalcdf(-|z|, |z|).
For the left-tail test, the p-value is given by
p=P(Z < z) = normalcdf(-5, z).
For the right-tail test, the p-value is given by
p=P(Z > z) = normalcdf(z, 5).

p-value based Decision Rules:

For all three Z-Tests, the p-values can be computed as above. Then, the above decision rules could, equivalently, be written as, at the level of significance α,

Reject H₀if p < α
Accept H₀ otherwise.

Remark. For the rest of this chapter, the decision rules for various significance tests will be described in two ways: (1) By checking whether the value of the test statistics T falls within the critical region or not. (2) By checking whether the p-value < α or not?

Problems on 9.1: On Z-Tests

Exercise 9.1.1. The standard deviation of life expectancy of a population is σ = 15 years. A a sample of size 25 had mean life expectancy X = = 81 years. Perform a significence test for the null and alternative hypothesis, regarding the mean life expectancy μ:

H₀ : μ = 75
H_A : μ ≠ 75.

Compute the value of the test statistic Z.
Compute the p-value.
At the 5 percent level of significance will you reject or accept the null hypothesis?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would REJECT the null hypothesis?

Solution:
Here the population standard deviation σ = 15,
the sample size n = 25,
the sample mean X = 81,
Also, μ₀ = 75

The Test Statistics z = (x- μ₀)√n/σ =(81-75)√25/15 = 2.
This is a Two Tail Test. So,
p-value = 1 - P(-|z| < Z < |z|) = 1 - normalcdf(-|z|, |z|) = 1 - normalcdf(-2, 2) = 1 - .9545 = .0455.
Five percent level of significance means α = .05. Since,
p-value = .0455 < α = .05, we REJECT the null hypothesis at 5 percent level of significance.
That means, at five percent level of significance, we accept that the mean life expectancy μ ≠ 75.
Since p-value = .0455, from this possibilities of .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent; 5 percent would be the lowest level at which we would reject the null hypothesis.

p-value demo. The problem may have changed.

Exercise 9.1.2. (Change the level of significance.) Assume the same situation as in exercise 9.1.1. At the 1 percent level of significance will you reject or accept the null hypothesis?

Solution:
This is also a two tail test. From Exercise 9.1.1, p-value = .0455.

One percent level of significance means α = .01.Since,
p-value = .0455 is not less than α = .01. We ACCEPT the null hypothesis at one percent level of significance.
That means, at one percent level of significance, we do not accept that the mean life expectancy μ ≠ 75.

Exercise 9.1.3. (Change the alternative hypothesis) Assume the same situation as in exercise 9.1.1 and change the hypotheses as follows:

H₀ : μ = 75
H_A : μ > 75

Answer all the four questions as in exercise 9.1.1.

Solution:

From exercise 9.1.1 the test statistics z = = 2.
This is a Right Tail Test. So,
p-value = =P(Z > z) = normalcdf(2, 5) = .02275
Five percent level of significance means α = .05. Since,
p-value = .02275 < α = .05, we REJECT the null hypothesis at 5 percent level of significance.
That means, at five percent level of significance, we accept that the mean life expectancy μ is higher than 75 years.
Since p-value = .02275, from this possibilities of .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent; 3 percent would be the lowest level at which we would reject the null hypothesis (i.e. accept the alternative).

p-value demo. The problem may have changed.

Exercise 9.1.4. The time taken by an athlete to run an event is normally distributed with mean μ and known standard deviation σ = 3.5 seconds. The coach believes that his/her mean time μ has improved from last year's mean 34 seconds. To test, the athlete ran 16 times and the sample mean was found to be X = 31 seconds.

Formulate the null and alternative hypotheses to perform a significance test for coach's belief.
Compute the value of the test statistic Z.
Compute the p-value.
At the 5 percent level of significance will you reject or accept the null hypothesis (or that his/her time has improved or not)?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would REJECT the null hypothesis?

Solution:

The alternative hypothesis is the coach's belief: H_A: μ < 34. The null and alternative hypotheses are:

H₀ : μ = 34
H_A : μ < 34.

We summarize the given data:

The population standard deviation σ = 3.5,
the sample size n = 16,
the sample mean X = 31,
Also, μ₀ = 34
The Test Statistics z = (x- μ₀)√n/σ =(31-34)√16/3.5 = -3.4286
This is a Left Tail Test. So,
p-value = P(Z < z) = normalcdf(-5, z) = normalcdf(-5, -3.4286) = 3.0311*10^-4.
Five percent level of significance means α = .05. Since,
p-value = 3.0311*10^-4 < α = .05, we REJECT the null hypothesis at 5 percent level of significance.
That means, at five percent level of significance, we accept that the mean time μ has improved from last year's mean 34 seconds.
Since p-value = 3.0311*10^-4, from this possibilities of .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent; .1 percent (because 3.0311*10^-4 < .001) would be the lowest level at which we would reject the null hypothesis.

p-value demo .The problem may have changed.

Exercise 9.1.5. The effectiveness of a weight loss program is to be tested on a group of 83 participants. At the beginning of the program, the mean weight of group is 210 pounds. At the end of the program the mean weight of the group is 199 pounds. The standard deviation of weight is known to be σ = 53.1 pounds. In terms of mean weight μ, perform a significance test that the program is effective.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic Z.
Decide if it is a Two Tail, Left Tail or Right Tail Test and compute the p-value.
At the 2 percent level of significance will you reject or accept the null hypothesis (or that the program is effective or not)?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would REJECT the null hypothesis?

Solution:
The population mean weight (after completing such a program) will be denoted by μ.

The alternative hypothesis is program was effective to reduce the mean weight feom 210 pounds: H_A: μ < 210. The null and alternative hypotheses are:

H₀ : μ = 210
H_A : μ < 210.

We summarize the given data:

The population standard deviation σ = 53.1,
the sample size n = 83,
the sample mean X = 199,
Also, μ₀ = 210
The Test Statistics z = (x- μ₀)√n/σ =(199-210)√83/53.1 = -1.8872
This is a Left Tail Test. So,
p-value = P(Z < z) = normalcdf(-5, z) = normalcdf(-5, -1.8872) = .0296.
Two percent level of significance means α = .02. Since,
p-value = .0296 is NOT less than α = .02, we accept the null hypothesis at 2 percent level of significance.
That means, at two percent level of significance, we do not accept that the weight loss program is effective.
Since p-value = .0296, from this possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 3 percent would be the lowest level at which we would reject the null hypothesis.

Exercise 9.1.6. A manufacturer of heating furnace is marketing a new model of energy efficient furnace. The mean gas consumption in January by ordinary furnaces is 153 CCF. A sample of 93 new model furnace had a mean consumption of 142 CCF in January. The standard deviation of consumption in January is known to be σ = 46 CCF. In terms of mean consumption μ, perform a significance test that the new model is really energy efficient.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic Z.
Compute the p-value.
At the 1 percent level of significance will you reject or accept the null hypothesis (or that the new model is energy efficient or not)?
What would be the lowest level of significance, percent among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would REJECT the null hypothesis?

Solution:
The population mean consumption in January will be denoted by μ.

The alternative hypothesis is the claim of the manufacturer: H_A: μ < 153. The null and alternative hypotheses are:

H₀ : μ = 153
H_A : μ < 153.

We summarize the given data:

The population standard deviation σ = 46,
the sample size n = 93,
the sample mean X = 142,
Also, μ₀ = 153
The Test Statistics z = (x- μ₀)√n/σ =(142 - 153)√83/46 = -2.3061
This is a Left Tail Test. So,
p-value = P(Z < z) = normalcdf(-5, z) = normalcdf(-5, -2.3061) = .01055.
One percent level of significance means α = .01. Since,
p-value = .01055 is NOT less than α = .01, we accept the null hypothesis at 1 percent level of significance.
That means, at one percent level of significance, we do not accept that the this model is energy efficient.
Since p-value = .01055, from this possibilities of .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent; 2 percent would be the lowest level at which we would reject the null hypothesis.

Exercise 9.1.7. It is believed that due to favorable weather conditions the mean weight μ of King salmon in Anchor River would be higher than the last year's mean of 33 pounds . The standard deviation of the weight is known to be σ = 16 pounds. A catch of 53 King had a mean of 39 pounds. In terms of mean weight μ, perform a significance test that the weight would be higher.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic Z.
Compute the p-value.
At the 2 percent level of significance will you reject or accept the null hypothesis (or that the mean weight hasincreased or not)?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would REJECT the null hypothesis?

Solution:
The population mean weight will be denoted by μ.

The alternative hypothesis is that the mean weight has increased: H_A: μ > 33. The null and alternative hypotheses are:

H₀ : μ = 33
H_A : μ > 33.

We summarize the given data:

The population standard deviation σ = 16,
the sample size n = 53,
the sample mean X = 39,
Also, μ₀ = 33
The Test Statistics z = (x- μ₀)√n/σ =(39 - 33)√53/16 = 2.7300
This is a Right Tail Test. So,
p-value = P(z < Z) = normalcdf(2.7300, 5) = .0032
Two percent level of significance means α = .02. Since,
p-value = .0032 < α = .02, we REJECT the null hypothesis at 2 percent level of significance.
That means, at two percent level of significance, we accept that the mean weight has increased.
Since p-value = .0032, from this possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; .5 percent (because .0032 < .005) would be the lowest level at which we would reject the null hypothesis.

Exercise 9.1.8. The instructor of Math 365 claims that due to his updated method of teaching, the student's learning has improved. The mean percent score of all his Math 365 courses before this semester was 68 percent. This semester in his call of 79 students, the mean percent score is 74 percent. The standard deviation of the percent score is known to be σ = 22 percent. In terms of mean consumption μ, perform a significance test that the percent score is higher.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic Z.
Compute the p-value.
At the 2 percent level of significance will you reject or accept the null hypothesis (or that the student's learning has improved or not)?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would REJECT the null hypothesis?

Solution:
The population mean percent score will be denoted by μ.

The alternative hypothesis is that the mean percent score has increased from 68 percent: H_A: μ > 68. The null and alternative hypotheses are:

H₀ : μ = 68
H_A : μ > 68.

We summarize the given data:

The population standard deviation σ = 22,
the sample size n = 79,
the sample mean X = 74,
Also, μ₀ = 68
The Test Statistics z = (x- μ₀)√n/σ =(74 - 68)√79 /22 = 2.4241
This is a Right Tail Test. So,
p-value = P(z < Z) = normalcdf(z, 5) = normalcdf(2.4241, 5)= .0077
Two percent level of significance means α = .02. Since,
p-value = .0077 < α = .02, we REJECT the null hypothesis at 2 percent level of significance.
That means, at two percent level of significance, we accept that the mean percent score has increased.
Since p-value = .0077, from this possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 1 percent (because .0077 < .001) would be the lowest level at which we would reject the null hypothesis.

Exercise 9.1.9. It is believed that the annual mean expenditure, including tuition, for students has increased from the corresponding mean in year 2000. In year 2000, the mean annual expenditure was $17,000. A sample of 87 students had annual mean expenditure of $19,500. The standard deviation annual expenditure is known to be σ = $7,500 percent. In terms of mean expenditure μ, perform a significance test that the mean annual expenditure μ has increased.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic Z.
Compute the p-value.
At the 2 percent level of significance will you reject or accept the null hypothesis (or that the expenditure has or not)?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would REJECT the null hypothesis?

9.2 Significance Test for μ, Case of σ Unknown

Let X be a random variable with mean μ and standard deviation σ. In this section also, another confidence interval of the mean μ. In contrast to Z-Test, this section deals with the situation when σ is unknown. As in the case of T-itervals (section 7.2), X would be assumed to have a normal distribution. Two Tail, Left Tail and Right Tail Tests would be developed to test the null hypothesis H₀: μ = μ₀, in the case when the value of σ is not known.

A sample X₁,X₂,…,X_m of size m is drawn from the X population. Let X and S² denote the sample mean and variance, respectively. The statistic

T=((X-μ₀) √m) /S would be the Test Statistic.

Similar to the situation of T-intervals (section 7.2), when the null hypthesis H₀: μ = μ₀ is true, T has t-distribution with degrees of freedom m-1. Using the same kind of arguments as in section 9.1, the decision rules are set as follows:

Two-tail test: The decision rule for testing the hypotheses

H₀ : μ= μ₀
H_A : μ ≠ μ₀

is set as, at the level of significance α,

Reject H₀ if t is not within [ -t_{m-1, α/2,} t_{m-1, α/2} ] where t = ((x-μ₀) √m) /s.

Accept H₀ otherwise.
Left-tail test: The decision rule for testing the hypotheses
H₀ : μ = μ₀
H_A : μ < μ₀

is set as, at the level of significance α,

Reject H₀ if t < -t_{m-1, α} where t = ((x-μ₀) √m) /s.

Accept H₀ otherwise.
Right-tail test: The decision rule for testing the hypotheses

H₀ : μ = μ₀
H_A : μ > μ₀

is set as, at the level of significance α,

Reject H₀ if t > t_m-1,_α where t = ((x-μ₀) √m) /s

Accept H₀ otherwise.

These are known as T-Tests.

p-Value based Decision Rules

The test statistic

T = (x-μ₀) √m/S has a t-distribution, with degrees of freedom m-1.

For the T-Tests, if T = t is the observed value of T, then p-value is define as follow. The tcdf function of TI-84 can be used to compute the same.

For the two-tail test, the p-value is given by
p=P(T not within [-|t|,|t|]) = 1 - P(-|t| < Z < |t|) = 1 - tcdf(-|t|, |t|, m-1).
For the left-tail test, the p-value is given by
p=P(T < t) =tcdf(-5, t, m-1) + [1 - tcdf(-5, 5, m-1)]/2 ≈ tcdf(-5, t, m-1) whenever m is large.
For the right-tail test, the p-value is given by
p=P(T > t) = tcdf(t, 5, m-1) + [1 - tcdf(-5, 5, m-1)]/2 ≈ tcdf(t, 5, m-1) whenever m is large.

p-value based Decision Rules:

For all three T-Tests, the p-values can be computed as above. Then, the above decision rules could, equivalently, be written as, at the level of significance α,

Reject H₀if p < α
Accept H₀ otherwise.

Problems on 9.2: on T-Tests

Exercise 9.2.1. A supplier of lamps claims that the mean lifetime of his lamps is longer than that of the lamps in the market. The mean lifetime of the bulbs on the market is 3456 hours. To test the claim of the supplier, a sample of 26 bulbs were examined. The sample mean was found to be 3720 hours and the sample standard deviation was s = 552 hours. In terms of mean lifetime μ, perform a significance test that the supplier's lamps last longer.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic T.
Compute the p-value.
At the 5 percent level of significance will you reject or accept the null hypothesis (or that the mean lifetime of the lamps is higher or not)?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would REJECT the null hypothesis?

Solution:

Here the population standard deviation is not known. So, the T-test will be used.

The alternative hypothesis is that mean lifetime μ of the supplier's lamps is higher than 3456 hours: H_A: μ > 3456. The null and alternative hypotheses are:

H₀ : μ = 3456
H_A : μ > 3456.

We summarize the given data:

The sample size n = 26,
The sample mean X = 3720,
The sample standard deviation S = 552,
Also, μ₀ = 3456
The Test Statistics t = (x- μ₀)√n/S =(3720 - 3456)√ 26/552 = 2.4387
This is a Right Tail T-Test.
The degrees for freedom df = m-1 26-1 =25.

p-value = P(t < T) = tcdf(2.4387, 5, 25) = .0111
Five percent level of significance means α = .05. Since,
p-value = .0111 < α = .05, we REJECT the null hypothesis at 5 percent level of significance.
That means, at five percent level of significance, we accept that the supplier's claim.
Since p-value = .0111, from this possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 2 percent (because .0111 < .02) would be the lowest level at which we would reject the null hypothesis.

p-Value demo . The problem may have changed.

Exercise 9.2.2. It is believed that the mean length of babies at birth in the United States is higher than the mean of 16.7 inches in some other nation. A sample of 33 babies in the United States was collected, and the sample mean and standard deviation was found to be X = 19 inches, S = 5.5 inches. Perform a of significance test for this beleif as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic T.
Compute the p-value.
At the 1 percent level of significance will you reject or accept the null hypothesis (or that the birth length is higher or not)?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would REJECT the null hypothesis?

Solution:

Here the population standard deviation is not known. So, the T-test will be used.

The alternative hypothesis is that mean birth length μ in US is higher than 16.7 inches: H_A: μ > 16.7 . The null and alternative hypotheses are:

H₀ : μ = 16.7
H_A : μ > 16.7 .

We summarize the given data:

The sample size n = 33,
The sample mean X = 19,
The sample standard deviation S = 5.5,
Also, μ₀ = 16.7
The Test Statistics t = (x- μ₀)√n/S =(19 - 16.7)√ 33/5.5 = 2.4023
This is a Right Tail T-Test.
The degrees for freedom df = m-1 33-1 = 32.

p-value = P(t < T) = tcdf(2.4023, 5, 32) = .0111
One percent level of significance means α = .01. Since,
p-value = .0111 is not less than α = .01, we ACCEPT the null hypothesis at 1 percent level of significance.
That means, at one percent level of significance, we do NOT accept that the mean birth length μ is longer than 16.7 .
Since p-value = .0111, from this possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 2 percent (because .0111 < .02) would be the lowest level at which we would reject the null hypothesis.

p-Value demo . The problem may have changed.

Exercise 9.2.3. A car manufacturer claims that a new model of car will get more mileage per gallon than the old model. The old model gets a mean mileage of 33 miles per gallon. To test the claim, 19 cars from the new model were tested and the sample mean was found to be x = 35 miles and standard deviation s = 3.3 miles. Perform a significance test for this manufacturer's claim as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic T.
Compute the p-value.
At the 1 percent level of significance will you reject or accept the null hypothesis (or that the milage is higher or not)?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent,at which you would REJECT the null hypothesis?

Solution:

Here the population standard deviation is not known. So, the T-test will be used.

The alternative hypothesis is that mean milage μ per gallon for the new model is higher than 33 miles: H_A: μ > 33 . The null and alternative hypotheses are:

H₀ : μ = 33
H_A : μ > 33 .

We summarize the given data:

The sample size n = 19,
The sample mean X = 35,
The sample standard deviation S = 3.3,
Also, μ₀ = 33
The Test Statistics t = (x- μ₀)√n/S =(35 - 33)√ 19/3.3 = 2.6218
This is a Right Tail T-Test.
The degrees for freedom df = m-1 = 19 - 1 = 18.

p-value = P(t < T) = tcdf(2.6218, 5, 18) = .0086
One percent level of significance means α = .01. Since,
p-value = .0086 is less than α = .01, we REJECT the null hypothesis at 1 percent level of significance.
That means, at one percent level of significance, we accept that the mean milage μ is higher than 33 miles .
Since p-value = .0086, from this possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 1 percent (because .0086 < .01) would be the lowest level at which we would reject the null hypothesis.

p-Value demo . The problem may have changed.

Exercise 9.2.4. It is assumed that the lifetime (in hours) of light bulbs produced in a factory is normally distributed with mean μ and standard deviation σ. The mean lifetime for an average light bulb on the market is 4500 hours. To estimate μ, the following data was collected on the lifetime of light bulbs.

5110	4671	6441	3331	5055	5270	5335	4973	1837	5487
7783	4560	6074	4777	4707	5263	4978	5418	5123	5017

The producer claims that the mean lifetime of the bulbs is more than the average bulbs on the market. Perform a significance test for this claim.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic T.
Compute the p-value.
At the 2 percent level of significance will you reject or accept the null hypothesis (or that the lifetime is higher or not)?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent,at which you would REJECT the null hypothesis?

Solution:

Here the population standard deviation is not known. So, the T-test will be used.

The alternative hypothesis is that the mean lifetime μ of the lamps is higher than 4500 hours: H_A: μ > 4500 . The null and alternative hypotheses are:

H₀ : μ = 4500
H_A : μ > 4500.

Use TI-84, as in Lesson 2, summarize the raw data:

The sample size n = 20,
The sample mean X = 5060.5,
The sample standard deviation S = 1143.1106,
Also, μ₀ = 4500
The Test Statistics t = (x- μ₀)√n/S =(5060.5 - 4500)√ 20/1143.1106 = 2.1928
This is a Right Tail T-Test.
The degrees for freedom df = m-1 = 20 - 1 = 19.

p-value = P(t < T) = tcdf(2.1928, 5, 19) = .0204
Two percent level of significance means α = .02. Since,
p-value = .0204 is not less than α = .02, we ACCEPT the null hypothesis at 2 percent level of significance.
That means, at two percent level of significance, we do NOT accept that the mean lifetime μ is higher than 4500 hours.
Since p-value = .0204, from this possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 3 percent (because .0204 < .03) would be the lowest level at which we would reject the null hypothesis.

p-Value demo . The problem may have changed.

Exercise 9.2.5. To estimate the mean weight (in pounds) of salmon in a river, the following sample was collected.

34.7	33.8	38.2	20.3	27.8
45.3	43.1	37.3	32.5	32.3
31.8	41.5	44.5	29.2	25.3
29.6	39.5	29.1	37.3

It is suspected that, due to polution, the mean weight has reduced from last year's the mean weight 37 pounds. Perform a significance test as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic T.
Compute the p-value.
At the 10 percent level of significance will you reject or accept the null hypothesis (or that the mean weight has reduced or not)?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent,at which you would REJECT the null hypothesis?

Solution:

Here the population standard deviation is not known. So, the T-test will be used.

The alternative hypothesis is that the mean weight μ has reduced from 37 pounds: H_A: μ < 37 . The null and alternative hypotheses are:

H₀ : μ = 37
H_A : μ < 37.

Use TI-84, as in Lesson 2, summarize the raw data:

The sample size n = 19,
The sample mean X = 34.3737,
The sample standard deviation S = 6.7608,
Also, μ₀ = 37
The Test Statistics t = (x- μ₀)√n/S =(34.3737 - 37)√ 19/6.7608 = -1.6933
This is a Left Tail T-Test.
The degrees for freedom df = m-1 = 19 - 1 = 18.

p-value = P(T < t) = tcdf(-5, -1.6933, 18) = .0538
Ten percent level of significance means α = .10. Since,
p-value = .0538 < α = .10, we REJECT the null hypothesis at 10 percent level of significance.
That means, at ten percent level of significance, we accept that the mean weight μ has reduced from 37 pounds.
Since p-value = .0538, from this possibilities of .1, .5, 1 ,2, 3, 4, 5, 6, 7,8, 9,10 percent; 6 percent (because .0538 < .06) would be the lowest level at which we would reject the null hypothesis.

Exercise 9.2.6. It is speculated that the teenage boys in a certain community are under weight. Under normal circumstances the mean weigh of this age group should be 155 pounds. A sample of 27 teenage boys had a mean weight 135 pound and sample standard deviation 32 pounds. Perform a significance test whether the mean weigh μ of this group is below 155 pounds, as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic T.
Compute the p-value.
At the 3 percent level of significance will you reject or accept the null hypothesis (or that the mean weight is lower or not)?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent,at which you would REJECT the null hypothesis?

Exercise 9.2.7. A guess is that the mean time μ needed for a student to arrive at the class from his/her residence would be less than 30 minutes. To test his guess, a sample 37 was collected. The sample mean time needed was 27 minutes and the sample standard deviation was 9 minutes. Perform a significance test whether the mean time μ needed would be below 30 minutes, as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic T.
Compute the p-value.
At the 3 percent level of significance will you reject or accept the null hypothesis (or that the mean weight is lower or not)?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent,at which you would REJECT the null hypothesis?

9.3 Population Proportion

Let p be the population proportion that has a particular attribute A. In this section, decision rules would be formulated to to test the Null hypothesis

H₀ : p = p₀.

As in section 7.3, a sample of size m is drawn. Let X be the number of sample members that has this attribute and X = X/m be the sample proportion. (In other words, X is the sample proportion of "success.") The test statistic to be used is

Z=(X-p₀) /σ_X where σ_X = √[(p₀(1-p₀)) /m].

If H₀ : p = p₀ is true, then Z has approximately N(0,1) distribution. As before, the decision rules are set as follows:

Two-tail test: The decision rule for testing the hypotheses

H₀ : p = p₀
H_A : p ≠ p₀

is set as, at the level of significance α,

Reject H₀ if z is not within [ -z_α/2, z_α/2 ] where z = (x-p₀) /σ_X.

Accept H₀ otherwise.
Left-tail test: The decision rule for testing the hypotheses

H₀ : p = p₀
H_A : p < p₀

is set as, at the level of significance α,

Reject H₀if z < -z_α where z = (x-p₀) /σ_X

Accept H₀ otherwise.
Right-tail test: The decision rule for testing the hypotheses

H₀ : p = p₀
H_A : p > p₀

is set as, at the level of significance α,

Reject H₀if z > z_α where z = (x-p₀) /σ_X

Accept H₀ otherwise.

These are known as 1-Proportion Z-Test

p-Value based Decision Rules

With thhe test statistics

Z = (X - p₀) /σ_X = (x- p₀)√m/ √[p₀(1-p₀)]

p-Values are defined as

For the two-tail test,
p=P(Z ∉[-|z|,|z|]) = 1 - P(-|z| < Z < |z|) = 1 - normalcdf(-|z|, |z|).
For the left-tail test,
p=P(Z < z) = normalcdf(-5, z).
For the right-tail test,
p=P(Z > z) = normalcdf(z, 5).

p-value based Decision Rules:

For all three 1-Proportion Z-Tests, the p-values can be computed as above. Then, the above decision rules could, equivalently, be written as, at the level of significance α,

Reject H₀if p < α
Accept H₀ otherwise.

Problems on 9.3: 1-Proportion Z-Test

Exercise 9.3.1. In a sample of 197 apples from a lot, 26 were found to be sour. The lot will be rejected if more than 10 percent is sour. Perform a significance test for the acceptability of this lot.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic Z.
Compute the p-value.
At the 3 percent level of significance will you reject or accept the null hypothesis (or whether the lot is acceptable or not)?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent,at which you would REJECT the null hypothesis?

Solution:
The proportion of sour apples will be denoted by p. A 1-Proportion Z-Test will be performed.

The alternate hypothesis is that p is more that .1 (ten percent).
The null and alternative hypotheses are:

H₀ : p = .1
H_A : p > .1.

We summarize the given data:

The sample size n = 197,
The number of success X = 26,
The sample proportion of success X = X/n = 26/197 = .1320
Also, p₀ = .1
The Test Statistics
Z = (X - p₀)√n/ √[p₀(1-p₀)] = (.1320 - .1)√ 197/ √[.1(1 - .1)] = 1.4971
This is a Right Tail Test. So,
p-value = P(z < Z) = normalcdf(z, 5) = normalcdf(1.4971, 5) = .0672
Three percent level of significance means α = .03. Since,
p-value = .0672 is NOT less than α = .03, we accept the null hypothesis at 3 percent level of significance.
That means, at three percent level of significance, we conclude that the lot is acceptable.
In fact, .06 < p-value = .0296 < .07. Therefore, from these possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 7 percent would be the lowest level at which we would reject the null hypothesis.

a p-Value demo . The problem may have changed.

Exercise 9.3.2. A new vaccine was tried on 147 randomly selected individuals, and it was determined that 61 of them got the virus. It is known that usually fifty percent of the population get the virus. Perform a significance test to decide if this vaccine is indeed effective or not.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic Z.
Compute the p-value.
At the 3 percent level of significance will you reject or accept the null hypothesis (or that the vaccine is effective or not)?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would REJECT the null hypothesis?

Solution:
The proportion of the vaccinated population who benefit from it would be denoted by p.
A 1-Proportion Z-Test will be performed.

The alternate hypothesis is that p is less that .5 (better than 50 percent).
The null and alternative hypotheses are:

H₀ : p = .5
H_A : p < .5.

We summarize the given data:

The sample size n = 147,
The number of success X = 61,
The sample proportion of success X = X/n = 61/147 = .4150
Also, p₀ = .5
The Test Statistics
Z = (X - p₀)√n/ √[p₀(1-p₀)] = (.4150 - .5)√ 147/ √[.5(1 - .5)] = -2.0611
This is a Left Tail Test. So,
p-value = P(Z < z) = normalcdf(-5, z) = normalcdf(-5, -2.0611) = .0196
Three percent level of significance means α = .03. Since,
p-value = .0196 < α = .03, we REJECT the null hypothesis at 3 percent level of significance.
That means, at three percent level of significance, we conclude that the vaccine in EFFECTIVE.
In fact, .01 < p-value = .0196 < .02.
Therefore, from these possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 2 percent would be the lowest level at which we would reject the null hypothesis.

a p-Value Demo . The problem may have changed.

Exercise 9.3.3. Before an election for a congressional seat, a poll was conducted. Out of 887 randomly selected voters interviewed, 389 said that they would vote for Candidate A. The election strategists have decided that, to win Candidate A needs to get more than 40 percent votes. Perform a significance test whether he/she will get more than 40 percent or not.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic Z.
Compute the p-value.
At the 3 percent level of significance will you reject or accept the null hypothesis (or whether he/she will get more than 40 percent or not.)?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent,at which you would REJECT the null hypothesis?

Solution:
The proportion of the voter population who would vote for the candidate would be denoted by p.
A 1-Proportion Z-Test will be performed.

The alternate hypothesis is that p is more that .4 (higher than 40 percent).
The null and alternative hypotheses are:

H₀ : p = .4
H_A : p > .4.

We summarize the given data:

The sample size n = 887,
The number of success X = 389,
The sample proportion of success X = X/n = 389/887 = .4386
Also, p₀ = .4
The Test Statistics
Z = (X - p₀)√n/ √[p₀(1-p₀)] = (.4386 - .4)√ 887/ √[.4(1 - .4)] = 2.3466
This is a Right Tail Test. So,
p-value = P(z < Z) = normalcdf(z, 5) = normalcdf(2.3466, 5) = .0095
Three percent level of significance means α = .03. Since,
p-value = .0095 < α = .03, we REJECT the null hypothesis at 3 percent level of significance.
That means, at three percent level of significance, we conclude that the Candidate A will win.
In fact, .005 < p-value = .0095 < .01.
Therefore, from this possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 1 percent would be the lowest level at which we would reject the null hypothesis.

a p-Value demo. The problem may have changed.

Exercise 9.3.4. A pollster was asked to make decision whether the proportion p of the US population who would support Government shutdown due to budget dispute, would be above 55 percent or not? A sample 898 were polled and 522 of them said they would support government shutdown. Perform a significance test whether p would be above 55 percent?

Formulate the null and alternative hypotheses.
Compute the value of the test statistic Z.
Compute the p-value.
At the 3 percent level of significance will you reject or accept the null hypothesis.?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent,at which you would REJECT the null hypothesis?

Exercise 7.3.5. A telephone company wants to know whether the proportion p of calls that are longer than 20 minutes, in a town, would exceed 65 percent. A sample of 1123 class, 761 were longer than 20 minute. Perform a significance test whether p would be above 65 percent?

Formulate the null and alternative hypotheses.
Compute the value of the test statistic Z.
Compute the p-value.
At the 3 percent level of significance will you reject or accept the null hypothesis.)?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent,at which you would REJECT the null hypothesis?

Exercise 7.3.6. It is believed that, this year, proportion p of infected oranges will remain below 15 percent. A sample of 1333 class, 175 were were infected. Perform a significance test whether p would remain below 15 percent?

Formulate the null and alternative hypotheses.
Compute the value of the test statistic Z.
Compute the p-value.
At the 3 percent level of significance will you reject or accept the null hypothesis?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent,at which you would REJECT the null hypothesis?

9.4 Testing Hypotheses on Variance σ²

in this section, we formulate a significance test for the population variance σ² (equivalently, of the standard deviation σ). It is assumed that X has a N(μ, σ) distribution. The Null hypothesis would be

H₀ : σ² = σ²₀.

As usual, a sample X₁,X₂, …,X_m of size m is drawn from the X population. Let S² be the sample variance. The following test statistic

Y = (m-1)S²/σ₀². will be used.

If H₀ : σ² = σ₀² is true, then Y has χ²-distribution with degrees of freedom m-1. Using the arguments as in the above sections, the decision rules are formuated as follows.

Two-tail test: The decision rule for testing the hypotheses

H₀ : σ²= σ₀²
H_A : σ² ≠ σ₀²

is set as, at the level of significance α,

Reject H₀, if y is not within [ χ² _m-1,1-α/2, χ² _{m-1, α/2} ] where y = (m-1)s²/σ₀²

Accept H₀ otherwise.
Left-tail test: The decision rule for testing the hypotheses

H₀ : σ²= σ₀²
H_A : σ² < σ₀²

is set as, at the level of significance α,

Reject H₀, if y < χ² _m-1,1-α where y = (m-1)s²/σ₀²

Accept H₀ otherwise.
Right-tail test: The decision rule for testing the hypotheses

H₀ : σ²= σ₀²
H_A : σ² > σ₀²

is set as, at the level of significance α,

Reject H₀, if y > χ² _m-1,α where y = (m-1)s²/σ₀²

Accept H₀ otherwise.

The significance tests for σ² are called Chi Square Tests, also written as Χ² Tests.

p-Value based Decision Rules

The test statistic

Y = (m-1)S²/σ₀² has a chi Square Distribution with degrees of freedon df = m-1.

For the Chi Square Tests, if Y = y is the observed value of Y, then p-value and the decision rules are define as follows. The Χ²cdf function of TI-84 can be used to compute the same.

For the two-tail test, two p-Values (left and right) will be defined:
p₁ = P(Y < y) =Χ²cdf(0, y, m-1) p₂ = P(y < Y) =1 - Χ²cdf(0, y, m-1)

Decision Rule: Reject H₀, if p₁ < α/2 OR if p₂ < α/2. Accept H₀ otherwise.
For the left-tail test, the p-value is given by
p=P(Y < y) =Χ²cdf(0, y, m-1).

Decision Rule: Reject H₀, if p < α. Accept H₀ otherwise.
For the right-tail test, the p-value is given by
p=P(y < Y) =1 - Χ²cdf(0, y, m-1).

Decision Rule: Reject H₀, , if p < α. Accept H₀ otherwise.

Problems on 9.4: Χ²-Test

Exercise 9.4.1 Suppose that we have collected a sample of size n = 23 from a normal population with mean μ and variance σ². The sample variance was found to be s² = 46.7. It is believed that the variance σ² is higher than 25. Perform a significance test as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Compute the p-value.
At the 3 percent level of significance will you reject or accept the null hypothesis?
What would be the lowest level of significance, percent among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent,at which you would REJECT the null hypothesis?

Solution:
A Chi SquareTest will be performed.

The null and alternative hypotheses are:

H₀ : σ² = 25
H_A : σ² > 25.

We summarize the given data:

The sample size n = 23,
The Sample Variance S² = 46.7
Also, σ²₀ = 25
The Test Statistics
y = (n-1)S²/σ₀² = (23 - 1)46.7/25 = 41.096
This is a Right Tail Test.
The degrees of freedom = df = n - 1 = 23 - 1 =22.
p-value = P(y < Y) = 1 - P(Y < y) = 1 - Χ²cdf(0, y, n - 1) = 1 - Χ²cdf(0, 41.096, 22) = 1 - .9920 = .008
Three percent level of significance means α = .03. Since,
p-value = .008 < α = .03, we REJECT the null hypothesis at 3 percent level of significance.
That means, we conclude that the σ² is higher than 25.
In fact, .005 < p-value = .008 < .01. Therefore, from these possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 1 percent would be the lowest level at which we would reject the null hypothesis.

a p-Value demo

Exercise 9.4.2 Following is data on the life expectancies of a group of people older than 75.

It is believed that the variance σ² life expectancy of this group is higher than 16. Perform a significance test as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Compute the p-value.
At the 3 percent level of significance will you reject or accept the null hypothesis?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent,at which you would REJECT the null hypothesis?

Solution:
A Chi SquareTest will be performed.

The null and alternative hypotheses are:

H₀ : σ² = 16
H_A : σ² > 16.

We summarize the given data:

The sample size n = 15,
Use TI-84 to compute the sample standard deviation S.
The Sample Variance S² = (5.8578)² = 34.3138
Also, σ²₀ = 16
The Test Statistics
y = (n-1)S²/σ₀² = (15 - 1)34.3138/16 = 30.0246
This is a Right Tail Test.
The degrees of freedom = df = n - 1 = 15 - 1 = 14.
p-value = P(y < Y) = 1 - P(Y < y) = 1 - Χ²cdf(0, y, n - 1) = 1 - Χ²cdf(0, 30.0246, 14) = 1 - .9924 = .0076
Three percent level of significance means α = .03. Since,
p-value = .0076 < α = .03, we REJECT the null hypothesis at 3 percent level of significance.
That means, at the level of significance 3 percent, we conclude that the σ² is higher than 16.
In fact, .005 < p-value = .0076 < .01. Therefore, from these possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 1 percent would be the lowest level at which we would reject the null hypothesis.

p-Value demo

Exercise 9.4.3 The following is data on monthly gas consumption (in ccf) by the households in a town during the winter months.

154

222

264

257

127

228

240

393

278

140

It believed that variance σ² is less that 15600 ccf². Perform a significance test as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Decide if the is a Left tail or Right tail test and compute the p-value.
At the 3 percent level of significance will you reject or accept the null hypothesis?
What would be the lowest level of significance, percent among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent,at which you would REJECT the null hypothesis?

Solution:
A Chi SquareTest will be performed.

The null and alternative hypotheses are:

H₀ : σ² = 15600
H_A : σ² < 15600.

We summarize the given data:

The sample size n = 10,
Use TI-84 to compute the sample standard deviation S.
The Sample Variance S² = (78.4645)² = 6156.6778
Also, σ²₀ = 15600
The Test Statistics
y = (n-1)S²/σ₀² = (10 - 1)6156.6778/15600 = 3.5519
This is a Left Tail Test.
The degrees of freedom = df = n - 1 = 10 - 1 = 9.
p-value = P(Y < y) = Χ²cdf(0, y, n - 1) = Χ²cdf(0, 3.5519, 9) = = .0616
Three percent level of significance means α = .03. Since,
p-value = .0616 is not less than α = .03, we ACCEPT the null hypothesis at 3 percent level of significance.
That means, at the level of significance 3 percent, we do NOT accept that the σ² is lower than 15600.
In fact, .006 < p-value = .0616 < .07. Therefore, from these possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 7 percent would be the lowest level at which we would reject the null hypothesis.

Solution

Exercise 9.4.4. The birth weight of babies has a normal distribution, with variance σ². Because of the economic and social diversity of the community, there are concerns about variability of the birth weight. It is believed that the variance σ² may be higher than 17 pounds-square. A sample of 26 birth-weight was collected and the sample variance was found to be s² = 26.7 pounds-square. Perform a significance test as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Decide if the is a Left tail or Right tail test and compute the p-value.
At the 3 percent level of significance will you reject or accept the null hypothesis?
What would be the lowest level of significance, percent among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent,at which you would REJECT the null hypothesis?

Solution:
A Chi SquareTest will be performed.

The null and alternative hypotheses are:

H₀ : σ² = 17
H_A : σ² > 17.

We summarize the given data:

The sample size n = 26,
The Sample Variance S² = 26.7
Also, σ²₀ = 17
The Test Statistics
y = (n-1)S²/σ₀² = (26 - 1)26.7/17 = 39.2647
This is a Right Tail Test.
The degrees of freedom = df = n - 1 = 26 - 1 = 25.
p-value = P(Y < y) = 1 - Χ²cdf(0, y, n - 1) = 1 - Χ²cdf(0, 39.2647, 25) = = .0346
Three percent level of significance means α = .03. Since,
p-value = .0346 is not less than α = .03, we ACCEPT the null hypothesis at 3 percent level of significance.
That means, at the level of significance 3 percent, we do NOT accept that the σ² is higher than 17.
In fact, .003 < p-value = .0346 < .04. Therefore, from these possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 4 percent would be the lowest level at which we would reject the null hypothesis.

Exercise 9.4.5. Because of homogeneity of a social group it is speculated that the variability of length of babies at birth may be small. It is speculated that variance length σ² of babies may be lower than 25 square-inches. A sample of size n = 16 on birth-lengths was collected. The sample variance was found to be s² = 13 square-inches. Perform a significance test as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Decide if the is a Left tail or Right tail test and compute the p-value.
At the 3 percent level of significance will you reject or accept the null hypothesis?
What would be the lowest level of significance, percent among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would REJECT the null hypothesis?

Solution:
A Chi SquareTest will be performed.

The null and alternative hypotheses are:

H₀ : σ² = 25
H_A : σ² < 25.

We summarize the given data:

The sample size n = 16,
The Sample Variance S² = 13
Also, σ²₀ = 25
The Test Statistics
y = (n-1)S²/σ₀² = (16 - 1)13/25 = 7.8
This is a Left Tail Test.
The degrees of freedom = df = n - 1 = 16 - 1 = 15.
p-value = P(Y < y) = Χ²cdf(0, y, n - 1) = Χ²cdf(0, 7.8, 15) = = .0684
Three percent level of significance means α = .03. Since,
p-value = .0684 is not less than α = .03, we ACCEPT the null hypothesis at 3 percent level of significance.
In fact, .06 < p-value = .0684 < .07. Therefore, from these possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 7 percent would be the lowest level at which we would reject the null hypothesis.

Exercise 9.4.6. The variability of the length of the telephone calls is a concern of the telephone company. It is speculated that the variance may be higher that 64 minutes-square. A sample of 14 calls had a sample variance s² = 99 minutes-square. Perform a significance test as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Decide if the is a Left tail or Right tail test and compute the p-value.
At the 3 percent level of significance will you reject or accept the null hypothesis?
What would be the lowest level of significance, percent among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would REJECT the null hypothesis?

Solution:
A Chi SquareTest will be performed.

The null and alternative hypotheses are:

H₀ : σ² = 64
H_A : σ² > 64.

We summarize the given data:

The sample size n = 14,
The Sample Variance S² = 99
Also, σ²₀ = 64
The Test Statistics
y = (n-1)S²/σ₀² = (14 - 1)99/64 = 20.1094
This is a Right Tail Test.
The degrees of freedom = df = n - 1 = 14 - 1 = 13.
p-value = P(Y < y) = 1 - Χ²cdf(0, y, n - 1) = 1 - Χ²cdf(0, 20.1094, 13) = = .0925
Three percent level of significance means α = .03. Since,
p-value = .0925 is not less than α = .03, we ACCEPT the null hypothesis at 3 percent level of significance.
In fact, .09 < p-value = .0925 < .10. Therefore, from these possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 10 percent would be the lowest level at which we would reject the null hypothesis.

9.5 A Significance Test to Compare Two Populations

Significance test for the equality of means would be a counter part of the confidence intervals of difference of means (Lesson 8). As in section 8.1, in this section it would be assumed that the population standard deviations of σ₁ and σ₂ of the two populations are known.

Let X and Y be two "similar" random variables. The mean and stadard deviation of X would, respectively, be denoted by μ₁ and σ₁. Similarly, the mean and stadard deviation of Y would, respectively, be denoted by μ₂ and σ₂. (For example, X and Y could be the height distrubutions of the US male and female populations.) A significance test for the equality (or inequality) of means μ₁, μ₂ would be considered. The Null hypothesis is given by

H₀ : μ₁ = μ₂

or equivalently

H₀ : μ₁- μ₂ = 0.

A sample X₁,X₂, …,X_m, of size m is drawn from the X-population and a sample Y₁,Y₂, …,Y_n, of size n, is drawn from the Y-population. Let X denote the sample mean of the X-sample and Y denote the sample mean of the Y-sample. Assume that the standard deviations σ₁, σ₂ are known. The test statistic would be

Z = (X-Y)/σ_d where σ_d = √( σ₁² /m + σ₂² /n )

If the Null hypothesis H₀ : μ₁- μ₂ = 0 is true, then Z has a N(0,1) distribution. As before, the decision rules are set as follows.

Two-tail test: The decision rule for testing the hypotheses

H₀ : μ₁ - μ₂= 0
H_A : μ₁ - μ₂≠ 0

is set as, at the level of significance α

Reject H₀ if z is not within [ -z_α/2, z_α/2] where z = (x - y) /σ_d

Accept H₀ otherwise.
Left-tail test: The decision rule for testing the hypotheses

H₀ : μ₁ - μ₂= 0
H_A : μ₁ - μ₂ < 0

is set as, at the level of significance α

Reject H₀if z < -z_α where z = (x - y) /σ_d

Accept H₀ otherwise.
Right-tail test: The decision rule for testing the hypotheses

H₀ : μ₁ - μ₂= 0
H_A : μ₁ - μ₂> 0

is set as, at the level of significance α

Reject H₀ if z > z_α where z = (x - y) /σ_d

Accept H₀ otherwise.

These are, informally, called the 2-Sample Z-Tests.

Remark. If sample sizes m, n are large, the sample standard deviations S₁, S₂ are used as estimates of σ₁, σ₂ in the above expression for Z.

p-Value based Decision Rules

With thhe test statistics

Z = (X - Y) /σ_d = (X - Y) / √( σ₁² /m + σ₂²/n)

p-Values are defined as

For the two-tail test,
p=P(Z ∉[-|z|,|z|]) = 1 - P(-|z| < Z < |z|) = 1 - normalcdf(-|z|, |z|).
For the left-tail test,
p=P(Z < z) = normalcdf(-5, z).
For the right-tail test,
p=P(Z > z) = normalcdf(z, 5).

p-value based Decision Rules:

The p-values can be computed as above. Then, the above decision rules could, equivalently, be written as, at the level of significance α,

Reject H₀if p < α
Accept H₀ otherwise.

Problems on 9.5: Testing of Hypotheses to Compare Two Populations — σ₁, σ₂ Known

Exercise 9.5.1. The equality of means μ₁, μ₂ of two populations is to be compared. The standard deviations σ₁, σ₂, respectively, are known to be σ₁ = 8.1 and σ₂ = 11.3. A sample of size m = 64 was collected from the first population, and the sample mean was found to be x = 3.5. A sample of size n = 99 was collected from the second population, and the sample mean was found to be y = 7.9.

It is speculated that these two means are unequal: μ₁ ≠ μ₂?

Perform a significance test as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Compute the p-value.
At the 3 percent level of significance will you reject or accept the null hypothesis?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would REJECT the null hypothesis?

Solution:
Since σ₁ and σ₂ are known, a 2-Sample Z-Test will be performed. The null and alternative hypotheses are:

H₀ : μ₁ - μ₂ = 0
H_A : μ₁ - μ₂ ≠ 0

The given data is summarized as follows:

	Population I (X)	Population II (Y)
Population St. deviation	σ₁ = 8.1	σ₂ = 11.3
Sample Mean	X= 3.5	Y= 7.9
Sample size	m = 64	n = 99

The Test Statistics

Z = (X - Y) / √( σ₁² /m + σ²/n) = (3.7 - 4.1) / √( (8.1)² /64 + (11.3)₂²/99) = -2.8919
This is a Two Tail Test. So,
p-value = P(Z is not within [-|z|, |z|]) = 1 - P( -|z| < Z < |z|) = 1 - P( -2.8919 < Z < 2.8919) 1 - normalcdf(-2.8919, 2.8919) = 1 - .9962 = .0038
Three percent level of significance means α = .03. Since,
p-value = .0038 < α = .03, we REJECT the null hypothesis at 3 percent level of significance.
In fact, .003 < p-value = .0038 < .005.
Therefore, from this possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; .5 percent would be the lowest level at which we would reject the null hypothesis.

a p-Value Demo The problem may have changed.

Exercise 9.5.2. The birth weight of babies in two hospitals would have to be compared. The birth-weight distributions X and Y of these two hospitals are normal with means μ₁, μ₂ and standard deviations σ₁, σ₂, respectively. It is known the standard deviations σ₁ = 2.3 pounds and σ₂ = 2.9 pounds. A sample of size m = 35 babies from the first hospitals was collected, and the sample mean birth weight was found to be X = 8.9 pounds. A sample of size n = 48 babies from the second hospital was collected, and the sample mean birth weight was found to be y = 7.6 pounds.

Due to economic disparities between these two neighborhoods, it is speculated that the mean μ₁ is higher than the mean μ₂. Perform a significance test as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Compute the p-value.
At the 3 percent level of significance would you accept this speculation that μ₁ is higher than μ₂.?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would Reject the null hypothesis?

Solution:
Since σ₁ and σ₂ are known, a 2-Sample Z-Test will be performed.
The alternative hypothesis is that the X-mean μ₁ would be higher than the Y-mean μ₂. So. the null and alternative hypotheses are:

H₀ : μ₁ - μ₂ = 0
H_A : μ₁ - μ^{2 > 0}

The given data is summarized as follows:

	Population I (X)	Population II (Y)
Population St. deviation	σ₁ = 2.3	σ₂ = 2.9
Sample Mean	X= 8.9	Y= 7.6
Sample size	m = 35	n = 48

The Test Statistics

Z = (X - Y) / √( σ₁² /m + σ₂²/n) = (8.9 - 7.6) / √( (2.3)²/35 + (2.9)_2/48) = 2.2756
This is a Right Tail Test. So,
p-value = P(z < Z) = P(2.2756 < Z) = normalcdf(2.2756, 5)) = .0114
Three percent level of significance means α = .03. Since,
p-value = .0114 < α = .03, we REJECT the null hypothesis at 3 percent level of significance.
In fact, .01 < p-value = .0114 < .02.
Therefore, from this possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 2 percent would be the lowest level at which we would reject the null hypothesis.

a p-Value Demo. The problem may have changed.

Exercise 9.5.3. Elephants in different parts of the world are different in height, weight, and length of ear and tusk. The mean heights μ₁ and μ₂ of elephants in two different regions would be compared. It is assumed that the height distributions X and Y of the elephants in these two regions are normally distributed. The standard deviation of X and Y are μ₁, σ₁= 1.5 feet and σ₂= 1.3 feet, respectively. A sample of size 25 was collected from region-I, and the sample mean height was found to be x = 9.9 feet. A sample of size 28 was collected from the region-II was collected, and the sample mean height was found to be y = 9.1 feet. It is beleived that mean height μ₁ is higher than μ₂. Perform a significance test as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Compute the p-value.
At the 3 percent level of significance would you accept that μ₁ is higher than μ₂.?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would Reject the null hypothesis?

H₀ : μ₁ - μ₂ = 0
H_A : μ₁ - μ₂ > 0

The given data is summarized as follows:

	Population I (X)	Population II (Y)
Population St. deviation	σ₁ = 1.5	σ₂ = 1.3
Sample Mean	X= 9.9	Y= 9.1
Sample size	m = 25	n = 28

The Test Statistics

Z = (X - Y) / √( σ₁² /m + σ₂²/n) = (9.9 - 9.1) / √( (1.5)²/25 + (1.3)²/28 ) = 2.0631
This is a Right Tail Test. So,
p-value = P(z < Z) = P(2.0631 < Z) = normalcdf(2.0631, 5)) = .0196
Three percent level of significance means α = .03. Since,
p-value = .0196 < α = .03, we REJECT the null hypothesis at 3 percent level of significance.
In fact, .01 < p-value = .0196 < .02.
Therefore, from this possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 2 percent would be the lowest level at which we would reject the null hypothesis.

a p-Value demo . The problem may have changed.

Exercise 9.5.4. It is speculated that the mean weight μ₁ of King salmon in Kenai is lower than the mean weight μ₂ of King salmon in Anchor River. The standard deviation weight of the Kings in Kenai is σ₁ = 7.7 pounds. The standard deviation weight of the Kings in Anchor is σ₂ = 9.1 pounds. A sample of 51 King from Kenai had a mean X = 30.5 pounds. A sample of 63 King from Anchor had a mean Y = 33 pounds. Perform a significance test as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Compute the p-value.
At the 3 percent level of significance would you accept that μ₁ is lower than μ₂.?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would Reject the null hypothesis?

Solution:
Since σ₁ and σ₂ are known, a 2-Sample Z-Test will be performed.
The alternative hypothesis is that the μ₁ would be lower than μ₂. So. the null and alternative hypotheses are:

H₀ : μ₁ - μ₂ = 0
H_A : μ₁ - μ₂ < 0

The given data is summarized as follows:

	Population I (X) Kenai	Population II (Y) Anchor
Population St. deviation	σ₁ = 7.7	σ₂ = 9.1
Sample Mean	X= 30.5	Y= 33
Sample size	m = 51	n = 63

The Test Statistics

Z = (X - Y) / √( σ₁² /m + σ₂²/n) = (30.5 - 33) / √( (7.7)²/51 + (9.1)²/63 ) = -1.5885
This is a Left Tail Test. So,
p-value = P(Z < z) = P(Z < -1.5885) = normalcdf(-5, -1.5885)) = .0561
Three percent level of significance means α = .03. Since,
p-value = .0561 is not less than α = .03, we ACCEPT the null hypothesis at 3 percent level of significance. That means we do not accept that the mean weight of Kings in Kenai is less than that of Anchor, at 3 percent level of significance.
In fact, .05 < p-value = .0561 < .06.
Therefore, from this possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 6 percent would be the lowest level at which we would reject the null hypothesis.

Exercise 9.5.5. There is a speculation circulating that the mean percent scores μ₁ in fall semester grades is higher than the mean percent scores μ₂ in spring semester grades. The standard deviation of fall percent scores is σ₁ = 27 percent and the standard deviation of spring percent scores is σ₂ = 23 percent. A sample of 87 students in fall had a sample mean score X = 76 percent. A sample of 77 students in spring had a sample mean score Y = 69 percent. Perform a significance test as follows.

Formulate the null and alternative hypotheses. ( This would be a Right Tail.)
Compute the value of the test statistic.
Compute the p-value.
At the 3 percent level of significance would you accept that μ₁ is higher than μ₂.?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would Reject the null hypothesis?

Exercise 9.5.6. it is speculated that the mean annual salary μ₁ of the professors in a State University (I) is higher than the mean annual salary μ₂ of the professors in the State University (II). The standard deviation of the annual salary in the University -I σ₁ = $16,000 and the standard deviation of the annual salary in the University-II σ₂ = $11,500. A sample of 47 professors in University-I had a mean salary X = $77,000. A sample of 58 professors in University-II had a mean salary Y = $71,500 Perform a significance test as follows.

Formulate the null and alternative hypotheses. ( This would be a Right Tail.)
Compute the value of the test statistic.
Compute the p-value.
At the 3 percent level of significance would you accept that μ₁ is higher than μ₂.?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would Reject the null hypothesis?

9.6 Compare Means of Two Populations: σ₁, σ₂ Unknown

Analogous to 2-Sample T-intervals (section 8.2), 2-Sample T-Test for the difference of means will be considered, in this section. Let X be a random variable with mean μ₁ and standard deviation σ₁. Let Y be a random variable with mean μ₂ and standard deviation σ₂. As in section 9.5, a significance test for the null hypothesis H₀ : μ₁ - μ₂ = 0 will be considered. Unlike section 9.5, in this case the standard deviations σ₁, σ₂ would be unknown. As a price, it is assumed that X and Y have normal distribution. It is further assumed the standard deviations are equal and a common notation is used:

σ₁ = σ₂ = σ.

A sample X₁,X₂, …,X_m, of size m is drawn from the X-population and a sample Y₁,Y₂, …,Y_n, of size n, is drawn from the Y-population. Let X denote the sample mean and S_X denote the sample standard deviation of the X-sample and Y denote the sample mean and S_Y denote the sample standard deviation of the Y-sample. As in section 8.2, the pooled estimator S_p of σ is defined as

S_p = √ ([(m-1)S_X²+(n-1)S_Y² ]/ [m+n-2]).

The test statistic would be

T = (X-Y) /[S_p √( 1/m+1/n) ]

If the Null hypothesis H₀ : μ₁- μ₂ = 0 is true, then T has a t-distribution with degrees of freedom m+n-2. The decision rules are as follows.

Two-tail test: The decision rule for testing the hypotheses

H₀ : μ₁ - μ₂= 0
H_A : μ₁ - μ₂≠ 0

is set as, at the level of significance α,

Reject H₀ if t is not within [-t_m+n-2,α/2, t_m+n-2,α/2] where t = (x-y) / [s_p √( 1/m + 1/n )]

Accept H₀ otherwise.
Left-tail test: The decision rule for testing the hypotheses

H₀ : μ₁ - μ₂= 0
H_A : μ₁ - μ₂< 0

is set as, at the level of significance α,

Reject H₀ if t < -t_m+n-2,α where t = (x-y) / [s_p √ ( 1/m + 1/n )]

Accept H₀ otherwise.
Right-tail test: The decision rule for testing the hypotheses

H₀ : μ₁ - μ₂= 0
H_A : μ₁ - μ₂> 0

is set as, at the level of significance α,

Reject H₀if t > t_m+n-2,α where t = (x-y) / [s_p √ ( 1/m + 1/n )]

Accept H₀ otherwise.

These tests are informally called as 2-sample T-Tests.

p-Value based Decision Rules

With thhe test statistics

T = (X-Y) /[S_p √( 1/m+1/n) ]

Let t be the observed value of T corresponding to the collected samples. p-Values are defined as

For the two-tail test,
p=P(T not within [-|t|,|t|]) = 1 - P(-|t| < Z < |t|) = 1 - tcdf(-|t|, |t|, m+n-2).
For the left-tail test,
p=P(T < t) = tcdf(-5, t, m+n-2) + (1- tcdf(-5, 5, m+n-2))/2 ≈ tcdf(-5, t, m+n-2) when m+n-2 is large.
For the right-tail test,
p=P(t < T) = tcdf(t, 5, m+n-2) + (1- tcdf(-5, 5, m+n-2))/2 ≈ tcdf(t, 5, m+n-2) when m+n-2 is large.

p-value based Decision Rules:

The p-values can be computed as above. Then, the above decision rules could, equivalently, be written as, at the level of significance α,

Reject H₀if p < α
Accept H₀ otherwise.

Problems on 9.6: Comparing Means of Two Populations — σ₁, σ₁ Unknown:

Exercise 9.6.1. Suppose that two "similar" normal populations have means μ₁, μ₂ respectively and same standard deviations σ. It is believed that μ₁ and μ₂ are not equal. A sample of size m = 11 from the first population the sample mean was found to be x = 13.5 and the sample standard deviation s₁ = 2.33. A sample of size n = 13 was collected from the second population that had a sample mean y = 11.5 and sample standard deviation s₂ = 2.73. Perform a significance test as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Compute the p-value.
At the 3 percent level of significance would you accept that μ₁ is not equal to μ₂.?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would Reject the null hypothesis?

Solution:
Since σ₁ and σ₂ are unknown, a 2-Sample T-Test will be performed.
The null and alternative hypotheses are:

H₀ : μ₁ - μ₂ = 0
H_A : μ₁ - μ₂ ≠ 0

The given data is summarized as follows:

	Population I (X)	Population II (Y)
Sample size	m = 11	n = 13
Sample Mean	X= 13.5	Y= 11.5
Sample St. deviation	s₁ = 2.33	s₂ = 2.73

The pooled estimate of σ is given by

S_p = √ ([(m-1)S_X²+(n-1)S_Y² ]/ [m+n-2])
= √ ([(11-1)(2.33)²+(13-1)(2.73)²]/ [11+13-2]) = 2.5560

The degrees for freedom
df= m+n -2 =11+13-2 =22

The Test Statistics

T = t = (X-Y) /[S_p √( 1/m+1/n) ] = (13.5 - 11.5) /[2.5560√ (1/11+1/13) ] = 1.9100
This is a Two Tail Test. So,
p-value = P(T not within [-|t|, |t|]) = 1- P(-|t| < T, |t|) = 1- tcdf(-|t|, |t|, m+n-2)
= 1- tcdf(-1.9100, 1.9100, 22) = 1 - .9307 = .0693
Three percent level of significance means α = .03. Since,
p-value = .0693 is not less than α = .03, we ACCEPT the null hypothesis at 3 percent level of significance.
In fact, .06 < p-value = .0693 < .07.
Therefore, from this possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 7 percent would be the lowest level at which we would reject the null hypothesis.

a p-Value demo. The problem may have changed.

Exercise 9.6.2. The means μ₁, μ₂ of two normal random variables X and Y would have to be compared. They have equal standard deviation σ. it is believed that μ₁ is lower than μ₂. A sample of size m = 64 was collected from the X-population and the sample mean and standard deviation were found to be x = 1.8, s₁ = 9.2 . A sample of size n = 99 was collected from the Y-population and the sample mean and standard deviation were y = 4.4, s₂ = 8.7. Perform a significance test as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Compute the p-value.
At the 3 percent level of significance would you accept that μ₁ is lower than μ₂?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would Reject the null hypothesis?

Solution:
Since σ₁ and σ₂ are unknown, a 2-Sample T-Test will be performed.
The null and alternative hypotheses are:

H₀ : μ₁ - μ₂ = 0
H_A : μ₁ - μ₂ < 0

The given data is summarized as follows:

	Population I (X)	Population II (Y)
Sample size	m = 64	n = 99
Sample Mean	X= 1.8	Y= 4.1
Sample St. deviation	s₁ = 9.2	s₂ = 8.7

The pooled estimate of σ is given by

S_p = √ ([(m-1)S_X²+(n-1)S_Y² ]/ [m+n-2])
= √ ([(64-1)(9.2)²+(99-1)(8.7)²]/ [64+99-2]) = 8.8990

The degrees for freedom
df= m+n -2 = 64+99-2 = 161

The Test Statistics

T = t = (X-Y) /[S_p √( 1/m+1/n) ] = (1.8 - 4.1) /[8.8990√ (1/64+1/99) ] = -1.6114
This is a Left Tail Test. So,
p-value = P(T < t) = tcdf(-5, t, m+n-2)
= tcdf(-5, -1.6114, 161) = .0545
Three percent level of significance means α = .03. Since,
p-value = .0545 is not less than α = .03, we ACCEPT the null hypothesis at 3 percent level of significance.
In fact, .05 < p-value = .0545 < .06.
Therefore, from this possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 6 percent would be the lowest level at which we would reject the null hypothesis.

a p-Value demo. The problem may have changed.

Exercise 9.6.3. The difference in mean monthly water consumption in two adjacent towns has to be compared. It is speculated that the mean monthly consumption μ₁ in Town-I is lower than the mean monthly consumption μ₂ in Town-II. A sample 37 household in the Town-I had a sample mean 6500 gallons and standard deviation 450 gallons. A sample 49 household in the Town-II had a sample mean 6800 gallons and standard deviation 650 gallons. Assume that the standard deviations are equal. Perform a significance test as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Compute the p-value.
At the 3 percent level of significance would you accept that μ₁ is lower than μ₂?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would Reject the null hypothesis?

Solution:
Since σ₁ and σ₂ are unknown, a 2-Sample T-Test will be performed.
The alternative hypothesis is that mean consumption in Town-I is lower than that of Town-II. The null and alternative hypotheses are:

H₀ : μ₁ - μ₂ = 0
H_A : μ₁ - μ₂ < 0

The given data is summarized as follows:

	Population I (X)	Population II (Y)
Sample size	m = 37	n = 49
Sample Mean	X= 6500	Y= 6800
Sample St. deviation	s₁ = 450	s₂ = 650

The pooled estimate of σ is given by

S_p = √ ([(m-1)S_X²+(n-1)S_Y² ]/ [m+n-2])
= √ ([(37-1)(450)²+(49-1)(650)²]/ [37+49-2]) = 572.8990

The degrees for freedom
df= m+n -2 = 37+49-2 = 84

The Test Statistics

T = t = (X-Y) /[S_p √( 1/m+1/n) ] = (6500 - 6800) /[572.8990√ (1/37+1/49) ] = -2.4043
This is a Left Tail Test. So,
p-value = P(T < t) = tcdf(-5, t, m+n-2)
= tcdf(-5, -2.4043, 84) = .0092
Three percent level of significance means α = .03. Since,
p-value = .0092 < α = .03, we REJECT the null hypothesis at 3 percent level of significance.
That means we ACCEPT that the mean monthly consumption in Town-I is lower than that of town-II.
In fact, .005 < p-value = .0092 < .01.
Therefore, from this possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 1 percent would be the lowest level at which we would reject the null hypothesis.

Exercise 9.6.4. The birth weight of babies in two hospitals are normally distributed with mean μ₁, μ₂ and equal standard deviation σ. Due to the economic disparities of the neighborhoods, it is speculated that μ₁ is higher than μ₂. The following data about birth weight in from these two hospitals were collected.

8.8	8.1	6.3	9.7	6.3
7.1	5.3	7.7	9.1	8.1
8.2	7.9	8.3	8.9	9.0
10.1	9.9	8.8	7.8	5.2
7.2

6.3	5.2	8.3	5.9	5.5
7.1	8.1	7.9	6.3	6.9
9.1	8.1	7.0	4.9	5.3
6.3	7.1	6.3	6.1	5.8
5.7	6.8	8.3	7.7

Perform a significance test as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Compute the p-value.
At the 3 percent level of significance would you accept that μ₁ is higher than μ₂?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would Reject the null hypothesis?

Solution:
Since σ₁ and σ₂ are unknown, a 2-Sample T-Test will be performed.
The alternative hypothesis is that mean birth weight in Hospital-I is higher than that in Hospital-II. The null and alternative hypotheses are:

H₀ : μ₁ - μ₂ = 0
H_A : μ₁ - μ₂ > 0

Use TI-84, as in Lesson 2, and summarize the give data as follows:

	Hospital I (X)	Hospital II (Y)
Sample size	m = 21	n = 24
Sample Mean	X= 7.9905	Y= 6.75
Sample St. deviation	s₁ = 1.3758	s₂ = 1.1417

The pooled estimate of σ is given by

S_p = √ ([(m-1)S_X²+(n-1)S_Y² ]/ [m+n-2])
= √ ([(21-1)(1.3758)²+(24 - 1)(1.1417)²]/ [21+24-2]) = 1.2560

The degrees for freedom
df= m+n -2 = 21+24-2 = 43

The Test Statistics

T = t = (X-Y) /[S_p √( 1/m+1/n) ] = (7.9905 - 6.75) /[1.2560√ (1/21+1/24) ] = 3.3053
This is a Right Tail Test. So,
p-value = P(t < T) = tcdf(t, 5, m+n-2)
= tcdf(3.3053, 5, 43) = 9.6030*10^-4
Three percent level of significance means α = .03. Since,
p-value = 9.6030*10^-4 < α = .03, we REJECT the null hypothesis at 3 percent level of significance.
That means we ACCEPT that the mean birth weight in Hospital-I is higher than that in Hospital-II.
In fact, 0 < p-value = 9.6030*10^-4 < .001.
Therefore, from this possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; .1 percent would be the lowest level at which we would reject the null hypothesis.

a p-Value demp . The problem may have changed.

Exercise 9.6.5. Elephants in different parts of the world are different in height, weight, and length of ear and tusk. It is speculated that mean height μ ₁ of elephants in a region is higher than the mean height μ ₂ in another region. It would be reasonable to assume that the height distributions X and Y of elephants in these two regions are normally distributed and they have equal standard deviations σ. The following data were collected on the height of the elephants from these two regions:

10.9	11.7	9.3	9.9	11.5
8.8	12.9	11.7	9.1	11.1
9.1	8.7	10.5	11.3	12.3
13.1	12.9	9.5	10.7	11.3

8.1	9.3	9.2	10.1	11.3
10.3	10.7	9.9	9.8	10.1
8.9	10.9	10.2	9.8	9.1
9.7	9.8	10.3	11. 1	10.9
10.9

Perform a significance test as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Compute the p-value.
At the 3 percent level of significance would you accept that μ₁ is higher than μ₂?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would Reject the null hypothesis?

Solution:
Since σ₁ and σ₂ are unknown, a 2-Sample T-Test will be performed.
The alternative hypothesis is that the mean height of elephants in Region-I is higher than the mean height the Region-II. The null and alternative hypotheses are:

H₀ : μ₁ - μ₂ = 0
H_A : μ₁ - μ₂ > 0

Use TI-84, as in Lesson 2, and summarize the give data as follows:

	Region I (X)	Region II (Y)
Sample size	m = 20	n = 24
Sample Mean	X= 10.815	Y= 10.0190
Sample St. deviation	s₁ = 1.4162	s₂ = .8072

The pooled estimate of σ is given by

S_p = √ ([(m-1)S_X²+(n-1)S_Y² ]/ [m+n-2])
= √ ([(20-1)(1.4162)²+(21 - 1)(.8072)²]/ [20+21-2]) = 1.1451

The degrees for freedom
df= m+n -2 = 20+21-2 = 39

The Test Statistics

T = t = (X-Y) /[S_p √( 1/m+1/n) ] = (10.815 - 10.0190) /[1.1451√ (1/20+1/21) ] = 2.2249
This is a Right Tail Test. So,
p-value = P(t < T) = tcdf(t, 5, m+n-2)
= tcdf(2.2249, 5, 39) = .0160
Three percent level of significance means α = .03. Since,
p-value = .0160 < α = .03, we REJECT the null hypothesis at 3 percent level of significance.
That means we ACCEPT that the mean mean height of elephants in Region-I is higher than that in Region-II.
In fact, .01 < p-value = 9.0160 < .02.
Therefore, from this possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9,10 percent; 2 percent would be the lowest level at which we would reject the null hypothesis.

a p-Value demo. The problem may have changed.

9.7 Comparing Proportions p₁, p₂ of Two Populations

The counter part of 1-Proportion Z-interval (section 8.3) would be a significance test for equality of two proportions p₁ and p₂ of an attribute A present, respectively, in two different populations (Population-I and Population-II). The Null hypothesis would be

H₀ : p₁-p₂ = 0.

A sample of size m is drawn from Population-I. Let X denote the number of the sample members that have this attribute A, and X = X/m be the proportion (of success). Similarly, a sample of size n is drawn from Population-II. Let Y denote the number of the sample members that have this attribute A and Y = Y/n be the sample proportion. (So, X, Y are proportion of "success" of these two samples.)

Write

P=(X+Y)/(m+n).

If the null hypothesis, H₀ : p₁ - p₂ = is true, then p would be a natural estimate for p₁ = p₂.

The test statistic to be used would be

Z = (X-Y) /s_D where s_D = √ [P(1-P)(1/m + 1/n) ]

If H₀ : p₁-p₂ = 0 is true, then Z would have, approximately, N(0,1) distribution. The decision rules are set as follows.

Two-tail test: The decision rule for testing the hypotheses

H₀ : p₁ - p₂= 0
H_A : p₁ - p₂≠ 0

is set as, at the level of significance α

Reject H₀ if z is not within [-z_α/2, z_α/2] where z = (X-Y)/s_D

Accept H₀ otherwise.
Left-tail test: The decision rule for testing the hypotheses

H_O : p₁ - p₂ = 0
H_A : p₁ - p₂ < 0

is set as, at the level of significance α

Reject H₀ if z < -z_α where z = (X-Y)/s_D

Accept H₀ otherwise.
Right-tail test: The decision rule for testing the hypotheses

H₀ : p₁ - p₂= 0
H_A : p₁ - p₂ > 0

is set as, at the level of significance α

Reject H₀ if z > z_α where z = (X-Y)/s_D

Accept H₀ otherwise.

These are, informally, called the 2-Proportion Z-Tests.

p-Value based Decision Rules

The test statistics would be

Z = (X-Y) /s_D = (X-Y) /√ [P(1-P)(1/m + 1/n)].

Let z be the observed value of Z, corresponding to the sample.

p-Values would be defined as

For the two-tail test,
p=P(Z ∉[-|z|,|z|]) = 1 - P(-|z| < Z < |z|) = 1 - normalcdf(-|z|, |z|).
For the left-tail test,
p=P(Z < z) = normalcdf(-5, z).
For the right-tail test,
p=P(Z > z) = normalcdf(z, 5).

p-value based Decision Rules:

The p-values can be computed as above. Then, the above decision rules could, equivalently, be written as, at the level of significance α,

Reject H₀if p < α
Accept H₀ otherwise.

Problems on 9.7: 2-Proportion Z-Tests

Exercise 9.7.1. The proportions p₁,p₂ , respectively, of an attribute A present in two populations would have to be compared. It is believed that p₁ is higher than p₂. A sample of size m = 117 was drawn from the first population and x = 70 had the attribute A. Similarly, a sample of size n = 79 was drawn from the second second population and y = 37 had the attribute A.

Perform a significance test as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Decide if it is a Two Tail, Left Tail or Right Tail Test and compute the p-value.
At the 3 percent level of significance would you accept that p₁ is higher than p₂?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would Reject the null hypothesis?

Solution:

A 2-Proportion Z-Test will be performed.

The null and alternative hypotheses are:

H₀ : p₁ - p₂ = 0
H_A : p₁ - p₂ > 0

We summarize the given data:

	Population I (X)	Population II (Y)
Numbr of Success	X = 70	Y = 37
Sample size	m = 117	n = 79
Sample Proportion	X = X/m = 70/117 = .5983	Y= 39/79 = .4684
Grand Sample Proportion	P = (X + Y)/(m + n) = (70+ 37)/(117 + 79) = .5459

The Test Statistics
Z = (X-Y) /√ [P(1-P)(1/m + 1/n)] = (.5983 - .4684) /√ [.5459(1-.5459)(1/117 + 1/79)] = 1.7917
This is a Right Tail Test. So,
p-value = P(z < Z) = normalcdf(z, 5) = normalcdf(1.7917, 5) = .0366
Three percent level of significance means α = .03. Since,
p-value = .0366 is not less than α = .03, we ACCEPT the null hypothesis at 3 percent level of significance.
That means, at three percent level of significance, we DO NOT accept the alternative.
In fact, .03 < p-value = .0366 < .04.
Therefore, from this possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9, 10 percent; 4 percent would be the lowest level at which we would reject the null hypothesis.

a p-Value demo. The problem may have changed.

Exercise 9.7.2. To compare the proportions p₁, p₂ of defective lamps produced by new production center and old the production center, respectively, samples were collected. In a sample of 157 lamps from the new center, 26 were found to be defective; and in a sample of 141 lamps from the old center, 32 were defective. Perform a significance test that the new center is performing better, based on proportion of defective lamps, as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Decide if it is a Two Tail, Left Tail or Right Tail Test and compute the p-value.
At the 3 percent level of significance would you accept that p₁ is lower than p₂?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would Reject the null hypothesis?

Solution:

A 2-Proportion Z-Test will be performed.

The alternative hypothesis is that the new production center is performing better, which means p₁ < p₂. The null and alternative hypotheses are:

H₀ : p₁ - p₂ = 0
H_A : p₁ - p₂ < 0

We summarize the given data:

	Population I (X) New Center	Population II (Y) Old Center
Numbr of Success	X = 26	Y = 32
Sample size	m = 157	n = 141
Sample Proportion	X = X/m = 26/157 = .1656	Y= 32/141 = .2270
Grand Sample Proportion	P = (X + Y)/(m + n) = (26 + 32)/(157 + 141) = .1946

The Test Statistics
Z = (X-Y) /√ [P(1-P)(1/m + 1/n)] = (.1656 - .2270) /√ [.1946(1-.1946)(1/157 + 1/141)] = -1.3367
This is a Left Tail Test. So,
p-value = P(Z < z) = normalcdf(-5, z)) = normalcdf(-5, -1.3367) = .0907
Three percent level of significance means α = .03. Since,
p-value = .0907 is not less than α = .03, we ACCEPT the null hypothesis at 3 percent level of significance.
That means, at three percent level of significance, we DO NOT accept that the new center is performing better.
In fact, .09 < p-value = .0907 < .10.
Therefore, from this possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9, 10 percent; 9 percent would be the lowest level at which we would reject the null hypothesis.

a p-Value demo. The problem may have changed.

Exercise 9.7.3. Data was collected to compare the proportions p₁,p₂ of men and women, respectively, who watch football. In a sample of 199 men, 83 said that they watch football; and in a sample of 161 women, 51 said they watch football. (These are not real data).

Perform a significance test that the proportion of men who watch football is higher, as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Decide if it is a Two Tail, Left Tail or Right Tail Test and compute the p-value.
At the 3 percent level of significance would you accept that p₁ is higher than p₂?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would Reject the null hypothesis?

Solution:

A 2-Proportion Z-Test will be performed.

The alternative hypothesis is that the proportion of men who watch football is higher, which means p₁ > p₂. The null and alternative hypotheses are:

H₀ : p₁ - p₂ = 0
H_A : p₁ - p₂ > 0

We summarize the given data:

	Population I (X) Men	Population II (Y) Women
Numbr of Success	X = 83	Y = 51
Sample size	m = 199	n = 161
Sample Proportion	X = X/m = 83/199 = .4171	Y= 51/161 = .3168
Grand Sample Proportion	P = (X + Y)/(m + n) = (83 + 51)/(199 + 161) = .3722

The Test Statistics
Z = (X-Y) /√ [P(1-P)(1/m + 1/n)] = (.4171 - .3168) /√ [.3722(1-.3722)(1/199 + 1/161)] = 1.9574
This is a Right Tail Test. So,
p-value = P(z < Z) = normalcdf(z, 5)) = normalcdf(1.9574, 5) = .0251
Three percent level of significance means α = .03. Since,
p-value = .0251 < α = .03, we REJECT the null hypothesis at 3 percent level of significance.
That means, at three percent level of significance, we ACCEPT that the proportion of men who watch football is higher than that of women.
In fact, .02 < p-value = .0251 < .03.
Therefore, from this possibilities of .1, .5, 1 ,2, 3, 4, 5,4, 7,8, 9, 10 percent; 3 percent would be the lowest level at which we would reject the null hypothesis.

a p-Value demo. The problem may have changed.

Exercise 9.7.4. Two varieties of grapes are compared. To compare the proportions p₁, p₂ of acceptable grapes of these two varieties, respectively, samples were drawn. In a sample of 131 grapes from the variety I, 112 were acceptable. In a sample of 143 grapes from the variety II, 113 were acceptable. Perform a significance test that the proportion p₁ acceptable grapes of variety-I is higher than that of the variety-II, as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Decide if it is a Two Tail, Left Tail or Right Tail Test and compute the p-value.
At the 3 percent level of significance would you REJECT the null hypothesis (that means that the proportion p₁ of acceptable grapes of variety-I is higher)?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would Reject the null hypothesis?

Exercise 9.7.5. To compare the proportions p₁, p₂ of students, respectively, in two state universities who pay more than $15 K tuition per year, samples were collected. In a sample of 217 students in the university I, 129 paid more than $15 K. In a sample of 313 students in the university II, 158 paid more than $15 K. It is speculated that the university I is more expensive than university-II and p₁is higher than p₂. Perform a significance test that the proportion p₁ of those in university-I that pay more than $15 K in tuition is higher that the proportion p₂ of those in university-II, as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Decide if it is a Two Tail, Left Tail or Right Tail Test and compute the p-value.
At the 3 percent level of significance would you REJECT the null hypothesis (that means that the proportion p₁ is higher)?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would Reject the null hypothesis?

Exercise 9.7.6. To compare the proportions p₁, p₂ of college graduates who earn more than $50 K, in two states, data was collected. In a sample of 444 college graduates in the state I, 354 earn more than $50 K. In a sample of 546 college graduates in the state II, 414 earn more than $50 K. It is speculated that higher proportion of graduates in state-I earn more than $50 K. Perform a significance test that the proportion p₁ higher than p₂, as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Compute the p-value.
At the 3 percent level of significance would you REJECT the null hypothesis (that means that the proportion p₁ is higher)?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would Reject the null hypothesis?

Exercise 9.7.7. It is believed that women are safer drivers than men. Let p₁, p₂ denote the proportions of women and men drivers, respectively, who were involved in an auto accident in a year period. In a sample of a size 770 women drivers 39 were involved in auto accident during this period. During the same period, in a sample of size 1215 men 79 were involved in auto accident in a year. It is speculated that proportion p₁ of women drivers who were involved in auto accidents last year is lower than that p₂ of men. Perform a significance test that the proportion p₁ lower than p₂I, as follows.

Formulate the null and alternative hypotheses.
Compute the value of the test statistic.
Decide if it is a Two Tail, Left Tail or Right Tail Test and compute the p-value.
At the 3 percent level of significance would you REJECT the null hypothesis (that means that the proportion p₁ is lower)?
What would be the lowest level of significance, among .1, .5, 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10 percent, at which you would Reject the null hypothesis?

9.8 Paired t-test

This section is to read only. While situation is different from the T-Test (section 9.2) or 2-Sample T-Test (section 9.5), the problem reduces to a T-Test.

Once again, equality of two means μ₁, μ₂ would be tested and the Null Hypothesis would be

H₀ : μ₁- μ₂ = 0.

Continue to denote the first population random variable by X and the second population random variable by Y. It is also assumed that X and Y have normal distributions, and that they are independent.

In certain situations, it is natural to collect samples in "pairs" (X,Y) from the two populations and consider the difference D = X-Y. So, D has mean

μ_D = μ₁- μ₂

and our Null hypothesis can be written as

H₀ : μ_D = 0.

Also D has

N(μ_D, σ_D)-distribution where σ_D = √( σ₁² + σ₂² ).

Samples (X₁,Y₁), …,(X_n,Y_n) are collected in pairs and produces D-sample:

D₁ = X₁-Y₁, …, D_n = X_n-Y_n.

Then, D would denote the mean of the D-sample and S_D would denote the sample standard deviation of the D-sample. The test statistic would be

T = (D√n) /S_D.

If the Null hypothesis H₀ : μ_D = μ₁- μ₂ = 0 is true, then T has a t-distribution with degrees of freedom n-1. Now, the same decision rules of the T-Test (section 9.2) can be used for the D-sample. These would be called paired T-Tests.

Example. Suppose two models of cars would have to be comapred to see how fast they accelerate. In this case, to avoid any variation due to individual drivers, a sample of n drivers drive one of each model. So, (x_i,y_i) are the accelerations of the first and second model driven by i^th-driver. Thus, there would be a sample n pairs of observations d₁ = x₁ - y₁, d₂ = x₂ - y₂, d₃ = x₃ - y₃, …, d_n = x_n - y_n. Now, it can be treaated as a T-Test.

Remark. The same approach of paired T-test could also be used to compute confidence intervals. A (1-α)100 percent confidence interval for μ_D = μ₁- μ₂ is given by

d-t_n-1,α/2s_d ≤ μ₁- μ₂ ≤ d-t_n-1,α/2s_d .

Math 365, Elementary Statistics

Lesson 9 :Testing Hypotheses

The Philosophy of Testing Hypotheses

Definitions and Terminologies.

9.1 A Significance Test for mean μ, when σ is known

Develop a Significance Test

The Hypothesis Test for the mean μ, when σ is known

p-Value based Decision Rules

9.2 Significance Test for μ, Case of σ Unknown

p-Value based Decision Rules

9.3 Population Proportion

p-Value based Decision Rules

9.4 Testing Hypotheses on Variance σ2

p-Value based Decision Rules

9.5 A Significance Test to Compare Two Populations

p-Value based Decision Rules

9.6 Compare Means of Two Populations: σ1, σ2 Unknown

p-Value based Decision Rules

9.7 Comparing Proportions p1, p2 of Two Populations

p-Value based Decision Rules

9.8 Paired t-test

9.4 Testing Hypotheses on Variance σ²

9.6 Compare Means of Two Populations: σ₁, σ₂ Unknown

9.7 Comparing Proportions p₁, p₂ of Two Populations