Satyagopal Mandal
Department of Mathematics
University of Kansas
Office: 624 Snow Hall  Phone: 785-864-5180
  • e-mail: mandal@math.ukans.edu
  • © Copy right Laws Apply. My Students have the permission to copy.

    Estimation 2

    Estimation of m when s is unknown

    When s is unknown we can not use the formula given last time (or Z-interval). We will give you two methods.

    Method 1: If the sample size n is large, then the sample standard deviation s can be used as an estimate for s and use the same formula as before to get an approximate (1- a)100 percent confidence interval for m. So, if the sample size is large, an approximate, (1 - a)100 percent confidence interval for m is given by

    x - za/2(s/n1/2) < m < x + za/2(s/n1/2).

    Use Your Calculator: If you have the values of x and s, use the Z-Interval as before and for s give the value of s.

     

    Problems:

    Ex. 1: Suppose that a sample of size n = 40 pumpkins collected from a farm had a mean weight x = 18.6 pounds and standard deviation s = 7.3 pounds. Give a 99 percent, approximate, confidence interval for the mean weigh m of the pumpkins in the farm.

    Solution: Here 1-a = 0.99 and za/2 = z 0.005 = 2.58. So, a 99 percent confidence interval is given by

    x - za/2(s/n1/2) < m < x + za/2(s/n1/2)

    which is

    18.6 - 2.58(7.3/40 1/2) < m < 18.6 + 2.58(7.3/40 1/2)

    which is

    15.62<m <21.92

    Alternately, you could use Z-Interval in the calculator, substituting s for s.

     

    Ex.2: A factory pays the workers depending on the number of units they produce. A sample of 72 workers produced a mean x = 13.4 units and standard deviation s = 2.1 units. Compute a 95 percent confidence interval for mean number m of units produced by the workers.

     

    Ex.3: The mean m daily number of classified ads published in a newspaper needs to be estimated. A sample over 84 days produced a mean x = 9910 ads and standard deviation s = 1105 ads. Give a 90 percent confidence interval for m.

     

     

    The T-Interval

    Method 2: If the parent population X is normal then we can give a exact (1 - a)100 percent confidence interval for m. This method depends on the distribution of

    T= (X- m)/(S/n1/2),

    where X is normal with mean m and standard deviation s.

     

    The distribution of T is called a t-distribution with degree of freedom n-1, which we have not discussed. The distribution of t-random variable is very similar to that of a Z-random variable (i.e. standard normal random variable). One of the main differences is that for each positive integer n, there is a t-random variable. This n is called the degree of freedom. As we did for the Z-random variable, we next give some of the properties of t-distribution:

     

    1. Given a positive integer n, there is a random variable T= tn that is said to have a t-distribution with degree of freedom n.

    2. A t-random variable is continuous.

    3. The mean of a t- random variable is ZERO.

    4. The distribution of a t-random variable is symmetric around the vertical axis and has a perfect bell shape (like Z).

    5. Suppose the degree of freedom n of a t-random variable T is large. Then it can be approximated by a Z-random variable.

    6. For a number 0 < a < 1 and any positive integer n, we define a number tn,a by the equation

      7. P(T> tn,a) = a

      where T has t-distribution with degree of freedom n. (This number tn,a is similar to za.

    7. For each degrees of freedom n, there is a table that gives probabilities.

    8. As I have said before, the distribution of a t-random variable is very similar to that of a Z-random variable. Besides, we do not plan to go deeper into t-random variable. As far as t-random variables are concerned, we rely entirely on the calculator.

     

    Theorem: Suppose X is a normal random variable with mean m and standard deviation s. Suppose

    X1, X2, … , Xn

    is a sample of size n from the X-population. Then

    T= (X- m)/(S/n1/2)

    has a t-distribution with degree of freedom n-1.

     

    So,

    P (- tn-1,a/2 < (X- m)/(S/n1/2) < tn-1,a/2) = 1-a.

     

    If we simplify, we get

    P(X - tn-1,a/2 (S/n1/2) < m < X + tn-1,a/2 (S/n1/2))= 1-a.

     

     

    Therefore, a (1-a)100 percent Confidence Interval for m is given by

    x-tn-1,a/2 (s/n1/2) < m < x + tn-1,a/2 (s/n1/2).

     

    Use your Calculator: In the calculator, this confidence interval is denoted by T-Interval.

    1. You will select TESTS and enter.

    2. Depending on whether you fed data in a list or you have statistics x and s, you select Data or Stats (and enter). Feed in the list or x and s, c-level.

    3. Now select calculate and enter.

     

    Problems on T-Interval: You have to use T-Interval if you do not know s.

    Ex.1: Assume that we have normal population X with mean m and standard deviation s. We have a sample of size n=18 that has sample mean x = 170.5 and sample standard deviation s = 13.3. Compute a 99 percent confidence interval for m.

     

    Use the T-Interval of the calculator.

     

    Ex.2: Suppose that the time X taken to complete a problem in the Math 106 test has a normal distribution with mean m and standard deviation s. A sample of size 23 was taken and sample mean and standard deviation was found to be x =4.7 and s = 0.47. Estimate the mean time m taken to complete a problem using a 95 percent confidence interval.

     

    Ex.3: It is assumed that the lifetime X (in hours) of lamps produced in factory has a normal distribution with mean m and standard deviation s. To estimate m following data was collected on the lifetime of 20 lamps:

    5110

    4671

    6441

    3331

    5055

    5270

    5335

    4973

    1837

    5487

    7783

    4560

    6074

    4777

    4707

    5263

    4978

    5418

    5123

    5017

    Compute a 95 percent confidence interval for m.

     

    Ex.4: 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

    		  

    Compute a 99 percent confidence interval for the sample mean m.

    Ex.5: Suppose we collect a sample from a normal population of size 40 with sample mean x =18.6 and standard deviation s=9.486. Construct a 95 percent confidence interval for mean m.

    Ex.6: Fifteen individuals were sampled to determine their daily intake X of wheat selenium (mg/day), which is considered essential as a constituent of erythrocyte glutathione peroxide. We assume that X has a normal distribution.

    The data are presented as follows.

    81.7

    93.2

    85.1

    78.0

    81.0

    87.7

    79.9

    84.8

    88.8

    77.9

    87.1

    83.9

    90.7

    90.1

    83.2

    		  

    Compute a 99 percent confidence interval for the mean m.

     

    Ex.7: The total soluble sugar X (g/100 ml) in a brand of canned soft drink was determined through high-performance liquid chromatography. The data from ten samples tested are presented as follows:

    8.4

    8.7

    10.1

    11.1

    9.9

    11.2

    10.7

    9.1

    9.9

    11.0

     

     

     

     

    Confidence Interval of the Population Proportion p

    Suppose you have Bernoulli trial with probability of success p in each trial. Now you want to estimate p. You have seen some of the examples:

    1. You may be interested in estimating the proportion p of Bank customers who would visit driveway counters.
    2. You may be interested in estimating the proportion p of households who own a pressure cooker.
    3. You may be interested in estimating the proportion p of voters in favor of a candidate.
    4. You may be interested in estimating the proportion p of eggs that are fertilized successfully in a clinic.
    5. You may be interested in estimating the proportion p of defective computer chips produced in a factory.
    6. You may be interested in estimating the proportion p of students who have access to Internet.
    7. You may be interested in estimating the proportion p of the population who favors medical use of marijuana.
    8. You may be interested in estimating the proportion p of middle income families that invest in the stock market.
    9. You may be interested in estimating the proportion p of the insurance-policy holders in a company who will survive more than 10 years.

     

    To estimate p we take a sample of n trials from the Bernoulli(p) population and let X be the number of success in these n trials and let X = X/n be the proportion of success in the sample. We have seen that X = X/n is approximately normally distributed with mean m = p and standard deviation s = (p(1-p)/n)1/2.

    So,

     

    P (- za/2 < (X-p)/(p(1-p)/n)1/2 < za/2 ) = 1-a.

     

    If we simplify, we get

    P(X - za/2 (p(1-p)/n)1/2 < p < X + za/2 (p(1-p)/n)1/2)= 1-a.

     

    Since p is not known, we use X as an estimate for p and conclude from above that an approximate (1-a)100 percent confidence interval for p is given by

    x - za/2 (x (1-x)/n)1/2 < p < x + za/2 (x (1-x)/n)1/2

     

     

    Remarks:

    1) As in the estimation of the margin of error is defined as

    E = za/2 (x (1-x)/n)1/2

    2) One can prove that E < za/2 (1/4n)1/2. That is why we define the conservative margin of error as CE = za/2 (1/4n)1/2.

    1. Some evening you might have hard a TV news reader read:

    President Clinton has 64 percent job approval rating. The poll has a margin of error plus or minus 3.1 percentage points. The poll surveyed 972 people.

     

    The TV reader meant that the sample proportion x of people who "approved" President's job performance is x = 0.64. Normally they would not tell us the level of confidence 1-a. Usually, they take 1-a = 0.95. That means that they give 95 percent confidence interval. Although they say "margin of error", they in fact give as the conservative margin of error. In this case, we can compute

    CE = za/2 (1/4n)1/2 = z0.025 (1/4n)1/2=1.96(1/4x972)1/2 = 0.031.

     

     

     

    Problems: Estimation of p

    Ex.1: In a sample of 197 apples from a lot, 19 were found to be sour. Set a 99 percent confidence interval for the proportion p of sour apples in the lot. Solution: Here n = 197 and the number of success x = 19. So, the sample proportion x = 19/197 = 0.096. Also since (1-a) = 0.99 we have za/2 = z 0.005 = 2.58. So, a 99 percent confidence interval for p is given by

    x - za/2 (x (1-x)/n)1/2 < p < x + za/2 (x (1-x)/n)1/2

    which is

    0.096 - 2.58(0.096(1-0.096)/197) 1/2 < p

    < 0.096 + 2.58(0.096(1-0.096)/197) 1/2

    which is

    0.096 - 0.004 < p < 0.096 + 0.004

    0.092 < p <0.100

    Note we are working with fractional numbers. So, take at least up to 3 decimal points. Alternately, you could use the "1-propZInterval".

    Ex.2: A new vaccine was tried on 147 randomly selected individuals and it was determined that 97 of them developed immunity. Find a 95 percent confidence interval for the proportion p of individuals in the population for whom the vaccine would help.

     

    Ex.3: For the coming congressional election in a seat, a poll was conducted. Out of 887 randomly selected voters interviewed, 389 said that they would vote for Candidate ME, 359 said that they would vote for Candidate YOU.

    a) Construct a 99 percent confidence interval for the proportion p of voters who would vote for ME.

    b) Construct a 95 percent confidence interval for the proportion q of voters who would vote for YOU.

     

    Ex.4: In a poll (read on October 28,1998), it was revealed that 60 percent of the 1,013 Americans interviewed want President Clinton rebuked but not impeached. Construct a 95 percent confidence interval for the proportion p of Americans who thought that way.