Math 105, Topics in Mathematics

Introduction	9.1 The Philosophy of Testing Hypotheses		9.2 Developing a Test
9.3 Testing on Single Population		9.4 About Population Proportion	Homework

Introduction

For this lesson on testing hypotheses you need the help of an advanced calculator. We discuss the concepts and theory and then use a calculator to solve the problems.

In this lesson we will test a hypothesis H₀, to be called Null hypothesis, against another hypothesis H_A, to be called the alternative hypothesis. Only one of these two hypotheses is true. Based on the collected sample and testing criterion that we set up, we will accept only one of them and reject the other one.

9.1 The Philosophy of Testing Hypotheses

Example 9.1.1. We may like to test the hypothesis that the disparity between the wages (annual income) of working men and women does not exist any more. Let μ₁ be the mean annual income of the men and μ₂ be the mean annual income of the working women. Our Null hypothesis H₀ and the alternative hypothesis H_A may be written as

H0  :	µ1- µ2	> 0
HA  :	µ1- µ2	= 0

Example 9.1.2. A TV commentator mentioned that in the early 1990s the average life expectancy of a human being was 75, and now it has increased substantially. We would like to test the claim of this commentator. We let μ be the average life expectancy of a human being. We set up our Null and alternative hypotheses as follows:

follows:

H0   :	µ	=  75
HA   :	µ	>   75

1. Definition. A statistical hypothesis is a statement, claim, or a proposition regarding a population. Most often, they are about the values of the population parameters. In the above two examples, all these H₀, H_A are statistical hypotheses.
2. One of our important considerations is whether the hypothesis is null or alternative, in a given context. Essentially, one negates the other.
3. The null hypothesis H₀ represents the status quo; it is something that you believed for a long time, or it is some assumption or method that has been working, reliably enough, for a long time. You want to hold on to the null hypothesis, unless there is very strong evidence, in the collected data, that the alternative hypothesis is better.
4. 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 that his/her product is better. We would be skeptical about the alternative hypothesis and would accept it only if strong evidence, in the collected data, favors it.
5. 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. Our test is based on the value of a test statistic. The rule is also called the decision rule or a test of significance.
6. Two types of errors: In this process of testing, we may commit two types of errors.
   1. If we reject H₀ when it is in fact true, then it is called a type one error.
      
   2. If we accept H₀ when it is in fact false, then it is called a type two error.
      
   3. Type one error and type two error are analogous to false positive and false negative outcome in any kind of test. For example, in a pregnancy test, result may indicate that one is not pregnant while one actually is and conversely, one may not be pregnant and the results may indicate that one is pregnant. Similar analogy is possible regarding any other kind of medical examination.
      
   4. The probability of committing a type one error is called the level of significance and is, normally, denoted by α. Usually, α will be 0.1, 0.05, 0.01 or a small number.

9.2 Developing a Test

Let X be a random variable with mean μ and standard deviation σ. Some of the hypotheses testing that we will be doing would look like the following:

H0   :	µ	=  75
HA   :	µ	≠   75

or

H0   :	µ	=  75
HA   :	µ	>   75

or

H0   :	µ	=  75
HA   :	µ	<   75

More generally, we would test hypotheses like

H0   :	µ	=   µ0
HA   :	µ	≠   µ0

or

H0   :	µ	=   µ0
HA   :	µ	>   µ0

or

H0   :	µ	=   µ0
HA   :	µ	<   µ0

To Develop a Test:

Suppose we have a random variable X with mean µ and standard deviation σ. We want to develop a test procedure for the following null and alternative hypotheses:

H0   :	µ	=   µ0
HA   :	µ	≠   µ0

We will take a sample X₁,X₂, ... , X_m of size m from the X population and let X be the sample mean.

1. We assume that sample size m is large enough, so that we have by the central limit theorem, that X has

   N(µ, σ_X)
   
   distribution, where
   
   σ_X   =   σ/m^1/2.

2. Both type one and type two errors can be controlled by increasing the sample size m. But once the sample size is fixed, it is not possible to control both simultaneously. If you want to reduce the probability of type one error, then the probability of type two error goes up and conversely. Because we are more concerned about type one error, we will try to minimize the probability of type one error, which is also called the level of significance. We would like to develop a test at the level of significance α.
   
   
3. Because X is a good estimator for μ and because the alternative hypothesis is
   
   

   HA   :

   µ

   ≠   µ0

   
   we reject our null hypothesis H₀ only if X and μ₀ are far apart, that is if
   

   | X - μ₀| is large.

4. Also, if H₀ is true, then µ = µ₀ and
   

   Z=(X-µ₀) /σ _X

   has N(0,1) distribution, where
   

   σ_X   =   σ/m^1/2.

   This expression Z above will be called a test statistic, and we will be accepting H₀ if the observed (absolute) value |z| of |Z| is small and reject H₀ if the observed value |z| of |Z| is large.

5. If H₀ is true, then
   

   P(Z is not in ( -z_α/2 , z_α/2 ))  =  α.

6. At the level of significance α, our decision rule is as follows:

   Reject H₀ if z is not in ( -z_α/2, z_α/2) where z = (x-µ₀) /σ _X
   
   
   Accept H₀ otherwise.

7. The above decision rule works only if we know the value of σ.
   
   

Some Hypotheses and Decision Rules. We assume that the value of σ is known.

1. Two-tail test: Suppose we are testing

   H0   :	µ	=   µ0
   HA   :	µ	≠   µ0

   At the level of significance α, our decision rule is as follows:

   Reject H₀ if z is not in ( -z_α/2 , z_α/2 )where z = (x-µ₀) /σ _X
   
   
   Accept H₀ otherwise.

2. Left-tail test: Suppose we are testing
   
   

   H0   :	µ	=   µ0
   HA   :	µ	<   µ0

   At the level of significance α, our decision rule is as follows:

   Reject H₀ if z < -z_αwhere z = (x-µ₀) /σ _X
   
   
   Accept H₀ otherwise.

3. Right-tail test: Suppose we are testing

   H0   :	µ	=   µ0
   HA   :	µ	>   µ0

   At the level of significance α, our decision rule is as follows:

   Reject H₀ if z >   z_α where z = (x-µ₀) /σ _X
   
   
   Accept H₀ otherwise.

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.

Definition. Suppose we have a test statistic T to test H₀ against H_A. Let the observed value of T = t. The P-value 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 that we have given, the test statistic that we are talking about is

Z = (X-µ₀) /σ _X.

If Z = z is the observed value of Z, then we have:

1. For the two-tail test the P-value is given by
   
   

   p=P(Z is not in (-|z|,|z|)).

2. For the left-tail test the P-value is given by
   
   

   p=P(Z < z).

3. For the right-tail test the P-value is given by
   
   

   p=P(Z > z).

Use of Calculators and P-values:

1. When we use calculators (say TI-83) for testing of hypotheses, the calculator will give us z-values and p-values.
2. We can use the z-values with the above decision rules to do the testing of hypotheses.
3. Alternately, at the level of significance α, if the P-value=p, then

   Reject H₀ if p   <   α
   
   Accept H₀ otherwise.

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.

Definition. Suppose we have a test statistic T to test H₀ against H_A. Let the observed value of T = t. The P-value 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 that we have given, the test statistic that we are talking about is

Z = (X-µ₀) /σ _X.

If Z = z is the observed value of Z, then we have:

1. For the two-tail test the P-value is given by
   
   

   p=P(Z is not in (-|z|,|z|)).

2. For the left-tail test the P-value is given by
   
   

   p=P(Z < z).

3. For the right-tail test the P-value is given by
   
   

   p=P(Z > z).

Use of Calculators and P-values:

1. When we use calculators (say TI-83) for testing of hypotheses, the calculator will give us z-values and p-values.
2. We can use the z-values with the above decision rules to do the testing of hypotheses.
3. Alternately, at the level of significance α, if the P-value=p, then

   Reject H₀ if p   <   α
   
   Accept H₀ otherwise.

Use of Calculators: Z-Test (TI-83)
Press stat and then select TESTS. Select Z-Test and enter. Input: you will have to select stats (not data) in this section. Feed in the values of m0, s, x, n. Select your Alternative Hypothesis. Select calculate and enter. The calculator will give you z and p. You accept or reject your Null Hypothesis H0 as follows: Reject H0 if  p <   α Accept H0 otherwise.

Problems on 9.2: σ Known

Use a calculator to solve the following problems. Flash-animated solutions may be helpful for conceptual reasons.

Exercise 9.2.1. Assume that you have a normal population with mean μ and standard deviation σ = 15. Suppose you have collected a sample of size 25 and the sample mean x was found to be 81. We want to test the null hypothesis

H0   :	µ	=   75
HA   :	µ	≠   75

At the 5 percent level of significance will you reject or accept the null hypothesis? Solution

Exercise 9.2.2. (Change the level of significance.) Assume that you have a normal population with mean μ and standard deviation σ = 15. Suppose you have collected a sample of size 25 and the sample mean x was found to be 81. We want to test the null hypothesis

H0   :	µ	=   75
HA   :	µ	≠   75

At the 1 percent level of significance will you reject or accept the null hypothesis?

Exercise 9.2.3. (Change the alternative hypothesis) Assume that you have a normal population with mean µ and standard deviation σ = 15. Suppose you have collected a sample of size 25 and the sample mean x was found to be 81. We want to test the null hypothesis

H0   :	µ	=   75
HA   :	µ	>   75

At the 5 percent level of significance will you reject or accept the null hypothesis? Solution

Exercise 9.2.4. The time taken by an athlete to run an event is normally distributed with mean µ and known standard deviation of σ = 3.5 second. The coach believes that his mean has improved from last year's mean of 34 seconds. To test the belief the athlete ran 16 times, and the sample mean was found to be x = 31 seconds.

1. Formulate the null and alternative hypotheses.
2. At 5 percent level of significance would the coach accept or reject the hypothesis that the athlete has improved?

Solution

Exercise 9.2.5. It is believed that the mean μ starting salary for the fresh KU graduates has increased from last year's mean of $51 K annually. It is known that the standard deviation of the starting salary is σ = 13 K. To test what you believe, you collect a sample of 15 fresh graduates, and the sample mean salary was found to be x= 54 K. At 1 percent level of significance, would you reject or accept that the starting salaries of the fresh KU graduates have increased? Solution

Exercise 9.2.6. The mean weight μ of babies at birth in the United States is believed to be higher than the worldwide mean birth weight of 112 ounces. The standard deviation of the birth weight in the United States is known to be 17 ounces. Data on 96 babies in the United States was collected, and the mean weight was found to be 115 ounces. At 5 percent level of significance, would you conclude that mean United States birth weight is higher than 112 ounces? Solution

Exercise 9.2.7. It is believed, due to pollution, the mean weight μ of salmon in a river is reduced from last year's mean of 23 pounds. It is known from past experience that standard deviation of the weight is σ = 6 pounds. To test such concerns about pollution, 46 fish were caught and the mean weight of the fish was x= 21.1 pounds. At 5 percent level of significance, would you accept or reject that the mean weight of salmon has been reduced?

Exercise 9.2.8. A supplier of light bulbs claims that the mean lifetime of his light bulbs is higher than that of the light bulbs available in the market. It is known that the mean lifetime of the light bulbs in the market is 3456 hours. To test the claim of the supplier you test a sample of 26 light bulbs, and the sample mean was found to be 3720 hours. It is known that the standard deviation of the lifetime of the light bulbs is σ = 1152 hours.

1. At 10 percent level of significance, would you accept the claim?
2. At 5 percent level of significance, would you accept the claim?
3. At 1 percent level of significance, would you accept the claim?
   
   

9.3 Testing on Single Population

In this section we assume that X is a N(μ,σ) random variable. In the last section, we assumed that σ was known. In this section we assume that σ is not known. We will be doing all the three tests as in the above section, assuming that the value of σ is not known.

Once again, we draw a sample X₁,X₂, ... ,X _m of size m from the X population. Let X and S² be the sample mean and variance, respectively. The test statistic we use is

T=((X-μ₀) /(S/m^1/2).

If H₀: μ = μ ₀ is true, then T has t-distribution with degrees of freedom m-1. Going through the same kind of arguments, we formulate the following decision rules.

1. Two-tail test: Suppose we are testing
   
   

   H0   :	µ	=   µ0
   HA   :	µ	≠   µ0

   At the level of significance α, our decision rule is as follows:

   Reject H₀ if t is not in ( -t_{m-1, α/2} , t_{m-1, α/2} ) where t =  (x-µ₀) /(s/m^1/2)
   
   
   Accept H₀ otherwise.

2. Left-tail test: Suppose we are testing
   
   

   H0   :	µ	=   µ0
   HA   :	µ	<   µ0

   At the level of significance α, our decision rule is as follows:

   Reject H₀ if t <   -t_{m-1, α} where t = (x-µ₀) /(s/m^1/2)
   
   
   Accept H₀ otherwise.

3. Right-tail test: Suppose we are testing

   H0   :	µ	=   µ0
   HA   :	µ	>   µ0

   At the level of significance α, our decision rule is as follows:

   Reject H₀ if t   >   t_{m-1, α} where t =  (x-µ₀) /(s/m^1/2)
   
   
   Accept H₀ otherwise.

Use of Calculators: T-Test (TI-83)
Press stat and then select TESTS. Select T-Test and enter. Input: you will have to select stats or data, as is the case. Feed in the values of µ0, s, x, n Or give the list number. Select your Alternative Hypothesis. Select calculate and enter. The calculator will give you t and p. You accept or reject your Null Hypothesis H0 as follows: Reject H0 if p   <   α Accept H0 otherwise.

Problems on 9.3 : σ Unknown

Use the calculator to solve the following problems. Flash-animated solutions may be helpful for conceptual reasons.

Exercise 9.3.1. It is assumed that the lifetime (in hours) of light bulbs produced in factory is normally distributed with mean μ and standard deviation σ. The mean lifetime for an average light bulb in the market is 6000 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 life expectancy of the light bulbs is more than average light bulbs in the market.

1. Formulate your null and alternative hypotheses.
2. Write down your decision rule.
3. At 1 percent level of significance what will you decide?

Solution

Exercise 9.3.2. To estimate the mean time taken by a group of athletes the following sample was collected:

24.7	23.8	28.2	25.3	21.8
35.3	33.1	31.3	22.5	22.3
21.8	31.5	34.5	24.2	21.3
22.6	29.5	23.1	33.3

Last year the mean of the same group was 25 seconds. You want to test if the mean time has changed significantly this year.

1. Formulate your null and alternative hypotheses.
2. Write down your decision rule.
3. At 1 percent level of significance what will you decide?

Solution

Exercise 9.3.3. It is believed that the mean length μ of babies at birth in the United States is higher than that of worldwide mean length of 18.7 inches. A sample of 26 U.S. babies was collected, and the sample mean and standard deviation were found to be x = 19 inches and s = 1 inch.

1. At 10 percent level of significance, would you accept the claim?
2. At 5 percent level of significance, would you accept the claim?
3. At 1 percent level of significance, would you accept the claim?

Solution

Exercise 9.3.4. A car manufacturer claims that the new model of the car will give more mileage per gallon than the old model. The old model gives a mean mileage of 33 miles per gallon. To test the claim, 9 cars from the new model were tested, and the sample mean was found to be x = 35 miles and standard deviation s = 2.2 miles.

1. At 10 percent level of significance, would you accept the claim?
2. At 5 percent level of significance, would you accept the claim?
3. At 1 percent level of significance, would you accept the claim?

Solution

Exercise 9.3.5. Satya claims that the mean time taken to complete an online homework assignment is less than that of traditional homework. You know from your experience that the mean time taken to complete a traditional assignment is 45 minutes. Now you collected a sample of 23 homework times and found that the sample mean time taken to complete this homework is x = 44 minutes and standard deviation s = 4 minutes. Pick the correct answer regarding acceptance or rejection of the claim of Satya.

1. At 10 percent level of significance, would you accept the claim?
2. At 5 percent level of significance, would you accept the claim?
3. At 1 percent level of significance, would you accept the claim?
   

9.4 About Population Proportion

Now let p be the population proportion that has a particular attribute A. We would like to test Null hypothesis

H₀ : p  =  p₀.

As usual, we draw (or interview) a sample of size m. Let X be the number of sample members that has this attribute and X = X/m be the sample proportion. (So, X is the sample proportion of "success".) The test statistic we use is

Z=(X-p₀) /σ_X

where

σ_X   =   [(p₀(1-p₀) /m)]^1/2.

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

1. Two-tail test: Suppose we are testing
   
   

   H0   :	p	=   p0
   HA   :	p	≠   p0

   At the level of significance α, our decision rule is as follows:

   Reject H₀ if z is not in ( -z_α/2 , z_α/2 ) where z = (x-p₀) /σ _X
   Accept H₀ otherwise.

2. Left-tail test: Suppose we are testing

   H0   :	p	=   p0
   HA   :	p	<   p0

   At the level of significance α, our decision rule is as follows:

   Reject H₀ if z   <   -z_α where z = (x-p₀) /σ _X
   Accept H₀ otherwise.

3. Right-tail test: Suppose we are testing

   H0   :	p	=   p0
   HA   :	p	>   p0

   At the level of significance α, our decision rule is as follows:

   Reject H₀ if z > z_α where z = (x-p₀) /σ _X
   Accept H₀ otherwise.

Use of Calculators: 1-PropZTest (TI-83)
Press stat and then select TESTS. Select 1-PropZTest and enter. Feed in the values of p0, x, n. Select your Alternative Hypothesis. Select calculate and enter. The calculator will give you z and p. You accept or reject your Null Hypothesis H0 as follows: Reject H0 if p <  α Accept H0 otherwise.

Problems on 9.4: Population Proportion p

Use the calculator to solve the following problems. Flash-animated solutions may be helpful for conceptual reasons.

Exercise 9.4.1. In a sample of 197 apples from a lot, 19 were found to be sour.

1. At 1 percent level of significance, would you conclude that more than 10 percent of the apples are sour?
2. At 5 percent level of significance, would you conclude that more than 10 percent of the apples are sour?
3. At 10 percent level of significance, would you conclude that more than 10 percent of the apples are sour?

Solution

Exercise 9.4.2. 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 50 percent of the population gets the virus.

1. At 1 percent level of significance, would you conclude that the vaccine is effective?
2. At 5 percent level of significance, would you conclude that the vaccine is effective?
3. At 10 percent level of significance, would you conclude that the vaccine is effective?

Solution

Exercise 9.4.3. For the coming congressional election, a poll was conducted. Out of 887 randomly selected voters interviewed, 389 said that they would vote for Candidate A, 359 said that they would vote for Candidate B.

1. At 1 percent level of significance would you conclude that more than 50 percent of the vote for A? Solution
2. At 5 percent level of significance would you conclude that more than 50 percent of the vote for A?
3. At 10 percent level of significance would you conclude that more than 50 percent of the vote for B? Solution