Chapter Seven - MDC
An estimator is a sample statistic used to estimate a population parameter. For example, the sample mean[pic] is an estimator of the population mean μ and the sample proportion [pic] is an estimator of the population proportion p. The value(s) assigned to a population parameter based on the value of a sample statistic is called an estimate.
1. The value of a sample statistic used to estimate a population parameter is called a point estimate. Under this procedure, we assign a single value to the population parameter being estimated. In interval estimation, an interval is constructed around the point estimate. This interval is likely to contain the corresponding population parameter.
2. The sample mean [pic] is the point estimator of the population mean μ. The margin of error is [pic] or [pic].
3. The width of a confidence interval may be decreased by lowering the confidence level or by increasing the sample size. Since lowering the confidence level gives a less reliable estimate, it is preferable to increase the sample size.
4. The width of a confidence interval depends on [pic]or [pic]. Thus, when n increases, [pic] (or [pic]) decreases. As a result of this, the width of the confidence interval decreases.
Example: Suppose [pic]=50, σ =12, and n =36.
The 95% confidence interval for μ is:[pic] to 53.92
If n = 100, but all other values remain the same, the 95% confidence interval for μ is:
[pic] to 52.35
A comparison shows that the 95% confidence interval for μ is narrower when n is 100 than when n is 36.
5. The width of a confidence interval depends on [pic]or [pic]. A decrease in the confidence level decreases z, which reduces the width of the confidence interval.
Example: Suppose [pic] and [pic].
The 95% confidence interval for μ is:[pic] to 53.92
The 90% confidence interval for μ is :[pic] to 53.30
By comparison, the 90% confidence interval is narrower than the 95% confidence interval.
7. A confidence interval is an interval constructed around a point estimate. A confidence level indicates how confident we are that the confidence interval contains the population parameter.
8. The maximum error of estimate for μ is subtracted from and added to the sample statistic [pic] to obtain a confidence interval for μ. It is given by [pic]or [pic], where z is determined by the confidence level.
9. If we took all possible samples of a given size and constructed a 99% confidence interval for μ from each sample, we would expect about 99% of these confidence intervals would contain μ and 1% would not.
8.10 a. z = 1.65 b. z = 1.96 c. z = 2.05
d. z = 2.17 e. z = 2.33 f. z = 2.58
8.11 [pic], [pic] [pic], and [pic]
a. Point estimate of μ is: [pic]
b. Margin of error = [pic]
c. The 99% confidence interval for μ is:[pic] to 25.50
d. Maximum error of estimate for 99% confidence level is:[pic]
8.12 [pic], [pic] [pic], and [pic]
a. Point estimate of μ is: [pic]
b. Margin of error = [pic]
c. The 95% confidence interval for μ is: [pic]
= 47.20 to 49.30.
d. Maximum error of estimate for 95% confidence level is:[pic]
8.13 [pic], [pic][pic] and [pic]
a. The 90% confidence interval for μ is: [pic] to 79.01
b. The 95% confidence interval for μ is: [pic] to 79.80
c. The 99% confidence interval for μ is:[pic] to 81.38
d. Yes, the width of the confidence intervals increases as the confidence level increases. This occurs because as the confidence level increases, the value of z increases.
8.14 [pic], [pic][pic] and [pic]
a. The 99% confidence interval for μ is:[pic] to 147.54
b. The 95% confidence interval for μ is:[pic] to 146.62
c. The 90% confidence interval for μ is:[pic] to 146.16
d. Yes, the width of the confidence intervals decreases as the confidence level decreases. This occurs because as the confidence level decreases, the value of z decreases.
8.15 [pic] and [pic]
a. [pic], so [pic] The 99% confidence interval for μ is:
[pic] to 84.61
b. [pic], so [pic]
The 99% confidence interval for μ is: [pic] to 83.71
c. [pic], so [pic]
The 99% confidence interval for μ is: [pic] to 83.53
d. Yes, the width of the confidence intervals decreases as the sample size increases. This occurs because the standard deviation of the sample mean decreases as the sample size increases.
8.16 [pic] and [pic]
a. [pic], so [pic]
The 95% confidence interval for μ is: [pic] to 49.52
b. [pic], so [pic]
The 95% confidence interval for μ is:[pic] to 49.92
c. [pic], so [pic]
The 95% confidence interval for μ is:[pic] to 50.52
d. Yes, the width of the confidence intervals increases as the sample size decreases. This occurs because the standard deviation of the sample mean increases as the sample size decreases.
8.17 a. [pic] and [pic], so [pic]
The 90% confidence interval for μ is: [pic] to 56.71
b. [pic] and [pic], so [pic]
The 90% confidence interval for μ is: [pic] to 58.64
c. [pic] and [pic], so [pic]
The 90% confidence interval for μ is: [pic] to 57.55
d. The confidence intervals of parts a and c cover μ but the confidence interval of part b does not.
8.18 a. [pic] and [pic], so [pic]
The 97% confidence interval for [pic] is: [pic] to 93.77
b. [pic] and [pic], so [pic]
The 97% confidence interval for μ is: [pic] to 93.32
c. [pic] and [pic]; so [pic]
The 97% confidence interval for μ is: [pic] to 91.08
d. The confidence intervals of parts b and c cover μ but the confidence interval of part a does not.
8.19 [pic] and [pic]
[pic]
a. Point estimate of μ is: [pic]
b. Margin of error = [pic]
c. The 98% confidence interval for μ is:[pic]=37.30 to 39.38
d. Maximum error of estimate = [pic]
8.20 [pic] and [pic], so [pic]
a. Point estimate of μ is: [pic]
b. Margin of error = [pic]
c. The 99% confidence interval for μ is:[pic] to 81.73
d. Maximum error of estimate = [pic]
8.21 [pic] and [pic] so [pic]
The 99% confidence interval for [pic] is:
[pic] to $369,422.64.
8.22 [pic]hours, and [pic]hours; so [pic]
The 99% confidence interval for μ is:[pic]
=16.67 to 17.07 hours.
8.23 [pic] and [pic] so [pic]
The 99% confidence interval for μ is:[pic] to $4299.01
8.24 [pic] and [pic] so [pic]
The 95% confidence interval for μ is:
[pic] to $146,487.54
8.25 [pic] and [pic] so [pic]
a. Point estimate of [pic]
Margin of error [pic]
b. The 90% confidence interval for μ is:
[pic] to $999,342.54
8.26 [pic], and [pic] so [pic]
a. Point estimate of [pic]
Margin of error [pic]
b. The 98% confidence interval for μ is:
[pic] to $7400.82
8.27 [pic]visits, and [pic]visits, so [pic]
a. The 97% confidence interval for μ is:
[pic] to 2.33 visits.
b. The sample mean of 2.3 visits is an estimate of μ based on a random sample. Because of sampling error, this estimate might differ from the true mean, μ, so we make an interval estimate to allow for this uncertainty and sampling error.
8.28 [pic] hours, and [pic] hours, so [pic] hours
a. The 95% confidence interval for μ is:[pic]=65.57 to 74.43 hours
b. The sample mean of 70 hours is an estimate of μ based on a random sample. Because of sampling error, this estimate might differ from the true mean, μ, so we make an interval estimate to allow for this uncertainty and sampling error.
8.29 [pic] inches and [pic] inches, so [pic] inches
The 99% confidence interval for μ is:
[pic] to 36.06 inches
Since the upper limit, 36.06, is greater than 36.05, the machine needs an adjustment.
8.30 [pic]ounces, and, [pic] ounces, so [pic]ounces.
The 99% confidence interval for μ is:
[pic] to 32.01 ounces.
Since the upper limit, 32.01, is less than 32.15, and the lower limit, 31.87, is greater than 31.85, the machine does not need an adjustment.
8.31 a. [pic], and [pic], so [pic]
The 99% confidence interval for μ is:
[pic] to $453.92
b. The width of the confidence interval obtained in part a may be reduced by:
1. Lowering the confidence level
2. Increasing the sample size
The second alternative is better because lowering the confidence level lowers the probability that the confidence interval contains μ.
8.32 a. [pic], and [pic] so [pic]
The 97% confidence interval for μ is:
[pic] = $1501.23 to $1648.77
b. The width of the confidence interval obtained in part a may be reduced by:
1. Lowering the confidence level
2. Increasing the sample size
The second alternative is better because lowering the confidence level lowers the probability that the confidence interval contains μ.
8.33 To estimate μ, the mean commuting time:
1. Take a random sample of 30 or more students from your school
2. Determine the commuting time for each student
3. Find the mean [pic] and the standard deviation s of this sample
4. Find [pic]
5. The 99% confidence interval for μ is obtained by using the formula [pic]
8.34 To estimate μ, the mean age of cars:
1. Take a random sample of 30 or more U.S. car owners
2. Determine age of each car
3. Find the mean [pic] and the standard deviation s of this sample
4. Find [pic]
5. The 95% confidence interval for μ is obtained by using the formula [pic]
8.35 The following are the similarities between the standard normal distribution and the t distribution:
1. Both distributions are symmetric (bell–shaped) about the mean.
2. Neither distribution meets the horizontal axis.
3. Total area under each of these curves is 1.0 or 100%.
4. The mean of both of these distributions is zero.
The following are the main differences between the standard normal distribution and the t distribution:
1. The t distribution has a lower height and a wider spread than the normal distribution.
2. The standard deviation of the standard normal distribution is 1 and that of the t distribution is [pic], which is always greater than 1.
3. The t distribution has only one parameter, the degrees of freedom, whereas the (standard) normal distribution has two parameters, μ and σ.
8.36 The normal distribution has two parameters: μ and σ. Given the values of these parameters for a normal distribution, we can find the area under the normal curve between any two points. The t distribution has only one parameter: the degrees of freedom. The shape of the t distribution curve is determined by the degrees of freedom.
8.37 The number of degrees of freedom is defined as the number of observations that can be chosen freely. As an example, suppose the mean score of five students in an examination is 81. Consequently, the sum of these five scores is 81(5) = 405. Now, how many scores, out of five, can we choose freely so that the sum of these five scores is 405? The answer is that we are free to choose 5 – 1 = 4 scores. Suppose we choose 87, 73, 69, and 94 as the four scores. Given these four scores and the information that the mean of the five scores is 81, the fifth score is: 405 – 87 – 73 – 69 – 94 = 82
Thus, once we have chosen four scores, the fifth score is automatically determined. Hence, the number of degrees of freedom for this example are: df = n – 1 = 5 – 1 = 4
We subtract 1 from n to obtain the degrees of freedom because we lose one degree of freedom to calculate the mean.
8.38 The t distribution is used to make a confidence interval for the population mean μ if the following three assumptions hold true.
1. The population from which the sample is taken is (approximately) normally distributed.
2. The sample size is small, that is, n < 30.
3. The population standard deviation σ is not known.
8.39 a. t = 1.782 b. df = n – 1 = 26 – 1 = 25 and t = –2.060
c. t =–3.579 d. df = n – 1 = 24 – 1 = 23 and t = 2.807
8.40 a. df = n – 1 = 21 – 1 = 20 and t =–1.325 b. df = n – 1 = 14 – 1 = 13 and t =2.160
c. t =3.686 d. t = –2.771
8.41 a. Area in the right tail = .01 b. Area in the left tail = .005
c. Area in the left tail = .025 d. Area in the right tail = .01
8.42 a. Area in the left tail = .05 b. Area in the right tail = .005
c. Area in the right tail = .10 d. Area in the left tail = .01
8.43 a. [pic] and [pic]
b. [pic] and t = 2.060
c. [pic] and [pic]
8.44 a. [pic] and t = 2.080
b. [pic] and [pic]
c. [pic] and t = 2.807
8.45 From the sample data: [pic]
[pic]
[pic]
[pic]
a. Point estimate of μ is: [pic]
b. [pic], [pic] and [pic]
The 99% confidence interval for μ is:
[pic] to 15.91.
c. Maximum error of estimate is: [pic]
8.46 From the sample data: [pic]
[pic]
[pic]
[pic]
a. Point estimate of μ is: [pic]
b. [pic], [pic] and [pic]
The 95% confidence interval for μ is:
[pic] to 49.58
c. Maximum error of estimate is: [pic]
8.47 a. [pic] and t = 2.131
[pic]
The 95% confidence interval for μ is: [pic] to 73.24.
b. [pic], [pic], and [pic]
The 90% confidence interval for μ is: [pic] to 72.40
The width of the 90% confidence interval for μ is smaller than that of the 95% confidence interval. This is so because the value of t for a 90% confidence level is smaller than that for a 95% confidence level (with df remaining the same).
c. [pic]
[pic] and t = 2.064
The 95% confidence interval for μ is: [pic] to 72.17
The width of the 95% confidence interval for μ is smaller with n = 25 than that of the 95% confidence interval with n = 16. This is so because the value of the standard deviation of the sample mean decreases as the sample size increases.
8.48 a. [pic] and t = 2.056
[pic]
The 95% confidence interval for μ is:[pic] to 27.44
b. [pic] and t = 2.779
The 99% confidence interval for μ is:[pic] to 28.12
The width of the 99% confidence interval for μ is larger than that of the 95% confidence interval. This is so because the value of t for a 99% confidence level is larger than that for a 95% confidence level.
c. [pic]
[pic] and t = 2.093
The 95% confidence interval for μ is:[pic] to 27.79
The width of the 95% confidence interval for μ is larger with n = 20 than that of the 95% confidence interval with n = 27. This is so because the value of the standard deviation of the sample mean is larger for a smaller sample size.
8.49 [pic] minutes, and [pic] minutes; then [pic] minutes
[pic] and t = 2.947
The 99% confidence interval for μ is:[pic] to 36.16 minutes
8.50 [pic] bushels per acre, and [pic] bushels per acre
[pic] bushels
[pic] and t = 1.729
The 90% confidence interval for μ is:
[pic] to 42.36 bushels per acre.
8.51 [pic], and [pic] so [pic]
[pic] and t = 2.064
The 95% confidence interval for μ is: [pic] to $6.13
8.52 [pic] ounces, and [pic] ounce; then [pic] ounce
[pic] and t = 2.131
The 95% confidence interval for μ is:[pic] to 32.12 ounces
8.53 [pic], and [pic] then [pic]
[pic] and t = 2.492
The 98% confidence interval for μ is:[pic] to $156.96
8.54 [pic]minutes, and [pic]minutes; so [pic]minutes.
[pic] and t = 2.797
The 99% confidence interval for μ is: [pic] to 25.36 minutes
8.55 a. [pic] MPG, and [pic] MPG; then [pic]
[pic] and t = 2.947
The 99% confidence interval for μ is:[pic] to 28.09 MPG
b. The width of the confidence interval obtained in part a can be reduced by:
1. Lowering the confidence level
2. Increasing the sample size
The second alternative is better because by lowering the confidence level we will simply lower the probability that our confidence interval includes μ.
8.56 a. [pic] hours, and [pic] hours, so [pic]
[pic] and t = 2.539
The 98% confidence interval for μ is:
[pic] to 25.13 hours
b. The width of the confidence interval constructed in part a may be reduced by:
1. Lowering the confidence level
2. Increasing the sample size
The second alternative is better because lowering the confidence level lowers the probability that our confidence interval contains μ.
8.57 [pic]
[pic] mph
[pic]
[pic] mph
[pic] and t = 1.833
The 90% confidence interval for μ is:
[pic] to 77.28 mph
8.58 [pic], and [pic] hours; then
[pic]hours and
[pic] hours.
[pic] and t = 2.306
The 95% confidence interval for μ is: [pic] to 11.12 hours
8.59 To estimate the mean time taken by a cashier to serve customers at a supermarket, we will follow the following procedure. (Assume that service times are normally distributed.)
1. Take a random sample of 10 customers served by this cashier
2. Collect the information on serving time for these customers
3. Calculate the sample mean, the standard deviation, and [pic]
4. Choose the confidence level and determine the t value
5. Make the confidence interval for μ using the t distribution
8.60 To estimate the mean waiting time μ (assuming that waiting times are normally distributed):
1. Take a random sample of 15 customers at the bank
2. Using a timing device, record the waiting time for each of the 15 customers
3. Find the mean and the standard deviation of these 15 waiting times
4. Calculate [pic]and t for 14 df and the chosen confidence level
5. The confidence interval for μ is [pic]
8.61 The normal distribution will be used to make a confidence interval for the population proportion if the sampling distribution of the sample proportion is (approximately) normal. We know from Chapter 7 (Section 7.7.3) that the sampling distribution of the sample proportion is (approximately) normal if:
[pic] and [pic]
Thus, the normal distribution can be used to make a confidence interval for the population proportion if:
[pic] and [pic]
8.62 The sample proportion [pic] is the point estimator of p. The margin of error for a point estimate of p
is [pic], where [pic].
8.63 a. [pic] and [pic]
Since [pic] and [pic], the sample size is large enough to use the normal distribution.
b. [pic] and [pic]
Since [pic], the sample size is not large enough to use the normal distribution.
c. [pic] and [pic]
Since[pic] and [pic], the sample size is large enough to use the normal distribution.
d. [pic] and [pic]
Since [pic], the sample size is not large enough to use the normal distribution.
8.64 a. [pic] and [pic]
Since [pic], the sample size is not large enough to use the normal distribution.
b. [pic] and [pic]
Since [pic], the sample size is not large enough to use the normal distribution.
c. [pic] and [pic]
Since[pic] and [pic], the sample size is large enough to use the normal distribution.
d. [pic] and [pic]
Since [pic] and [pic], the sample size is large enough to use the normal distribution.
8.65 a. [pic] and [pic]; then [pic]
The 95% confidence interval for p is: [pic] to .677
b. [pic] and [pic] then [pic]
The 95% confidence interval for p is: [pic] to .638
c. [pic] and [pic]; then [pic]
The 95% confidence interval for p is: [pic] to .716
d. The confidence intervals of parts a and c cover p, but the confidence interval of part b does not.
8.66 a. [pic] and [pic] then [pic]
The 90% confidence interval for p is: [pic] to .346
b. [pic] and [pic]; then [pic]
The 90% confidence interval for p is: [pic] to .386
c. [pic] and [pic]; then [pic]
The 90% confidence interval for p is: [pic] to .325
d. The confidence intervals of parts a and b cover p, but the confidence interval of part c does not.
8.67 [pic] and [pic]; [pic]
a. The 90% confidence interval for p is: [pic] to .714
b. The 95% confidence interval for p is: [pic] to .721
c. The 99% confidence interval for p is: [pic] to .734
d. Yes, the width of the confidence intervals increases as the confidence level increases. This occurs because as the confidence level increases, the value of z increases.
8.68 [pic] and [pic]; then [pic]
a. The 99% confidence interval for p is: [pic] to .351
b. The 97% confidence interval for p is: [pic] to .338
c. The 90% confidence interval for p is: [pic] to .322
d. Yes, the width of the confidence intervals decreases as the confidence level decreases. This occurs because as the confidence level decreases, the value of z decreases.
8.69 [pic], and [pic]
a. [pic] and [pic]
The 99% confidence interval for p is: [pic] to .845
b. [pic] and [pic]
The 99% confidence interval for p is: [pic] to .777
c. [pic] and [pic]
The 99% confidence interval for p is: [pic] to .760
d. Yes, the width of the confidence intervals decreases as the sample size increases. This occurs because increasing the sample size decreases the standard deviation of the sample proportion.
8.70 [pic] and [pic]
a. [pic] and [pic]
The 95% confidence interval for p is: [pic] to .336
b. [pic], [pic]
The 95% confidence interval for p is: [pic] to .351
c. [pic], [pic]
The 95% confidence interval for p is: [pic] to .411
d. Yes, the width of the confidence intervals increases as the sample size decreases. This occurs because decreasing the sample size increases the standard deviation of the sample proportion.
8.71 [pic] and [pic]; then [pic]
a. Point estimate of [pic]
Margin of error [pic]
b. The 99% confidence interval for p is: [pic] to .399
8.72 [pic] and [pic] then [pic]
a. Point estimate of [pic]
Margin of error [pic]
b. The 95% confidence interval for p is: [pic] to .284
8.73 [pic] and [pic] therefore [pic]
a. Point estimate of [pic]
Margin of error [pic]
b. The 95% confidence interval for p is: [pic] to .398
8.74 [pic] and [pic]; then [pic]
a. Point estimate of [pic] or 40%
Margin of error [pic] or 6.2%
b. The 97% confidence interval for p is:
[pic] to .469 or 33.1% to 46.9%.
8.75 [pic] and [pic]; then [pic]
a. The 98% confidence interval for p is:
[pic] to .851 or 54.9% to 85.1%.
b. The width of the confidence interval constructed in part a may be reduced by:
1. Lowering the confidence level
2. Increasing the sample size
The second alternative is better because lowering the confidence level lowers the probability that the confidence interval contains p.
8.76 a. [pic] and [pic] then [pic]
The 99% confidence interval for p is:
[pic] to .557 or 20.3% to 55.7%.
b. The width of the confidence interval constructed in part a may be reduced by:
1. Lowering the confidence level
2. Increasing the sample size
The second alternative is better because lowering the confidence level lowers the probability that the confidence interval contains p.
8.77 [pic] and [pic] therefore [pic]
a. The 99% confidence interval for p is:
[pic] to .198 or 14.2% to 19.8%.
b. The sample proportion of .17 is an estimate of p based on a random sample. Because of sampling error, this estimate might differ from the true proportion, p, so we make an interval estimate to allow for this uncertainty and sampling error.
8.78 [pic], and [pic] then [pic]
a. The 98% confidence interval for p is: [pic] to .696
b. The sample proportion of .64 is an estimate of p based on a random sample. Because of sampling error, this estimate might differ from the true proportion p, so we make an interval estimate to allow for this uncertainty and sampling error.
8.79 Nine of the 15 judges in the sample are in favor of the death penalty. Hence,
[pic], and [pic]. Therefore [pic]
a. Point estimate of [pic]
Margin of error [pic]
b. The 95% confidence interval for p is:
[pic] to .848 or 35.2% to 84.8%
8.80 Eight of the twenty–four policy holders in the sample have tried alternative treatments. Hence,
[pic]and [pic] Therefore [pic]
a. Point estimate of [pic]
Margin of error [pic]
b. The 99% confidence interval for p is:
[pic] to .581 or 8.5% to 58.1%
8.81 To estimate the proportion of students who hold off campus jobs:
1. Take a random sample of 40 students at your college.
2. Determine the number of students in this sample who hold off campus jobs
3. Calculate [pic] and [pic]
4. Choose the confidence level and find the required value of z from the normal distribution table
5. Obtain the confidence interval for p using the formula [pic]
8.82 To estimate the percentage of students who are satisfied with campus food services:
1. Take a random sample of 30 students who use campus food services.
2. Determine the number of students in this sample who are satisfied with the campus food services
3. Calculate [pic] and [pic]
4. Choose the confidence level and find the required value of z from the normal distribution table
5. Obtain the confidence interval for p using the formula [pic]
To obtain a confidence interval for the percentage, multiply the upper and lower limits of the confidence interval for p by 100.
8.83 a. [pic] and [pic] for 99% confidence level
[pic]
b. [pic] and [pic] for 96% confidence level
[pic]
8.84 a. [pic] and [pic] for 98% confidence level
[pic]
b. [pic] and [pic] for 95% confidence level
[pic]
8.85 a. [pic] and [pic] for 99% confidence level
[pic]
b. [pic] and [pic] for 95% confidence level
[pic]
c. [pic] and [pic] for 90% confidence level
[pic]
8.86 a. [pic] and [pic] for 99% confidence level
[pic]
b. [pic] and [pic] for 95% confidence level
[pic]
c. [pic] and [pic] for 90% confidence level
[pic]
8.87 [pic] [pic] and [pic] for 95% confidence level
[pic]
8.88 [pic] ounce, [pic] ounce, and [pic] for 99% confidence level
[pic]
8.89 [pic], [pic], and [pic] for 90% confidence level [pic]
8.90 [pic]hours, [pic]hours, and [pic] for 98% confidence level [pic]
8.91 a. [pic][pic]and [pic] for 99% confidence level
[pic]
b. [pic][pic] and [pic]for most conservative sample size
[pic]
8.92 a. [pic][pic]and [pic] for 98% confidence level
[pic]
b. [pic][pic] for most conservative sample size
[pic]
8.93 a. [pic]and [pic] for most conservative sample size
[pic]
b. [pic][pic] for most conservative sample size
[pic]
c. [pic][pic] for most conservative sample size
[pic]
8.94 a. [pic][pic]and [pic]
[pic]
b. [pic][pic]and [pic]
[pic]
c. [pic][pic]and [pic]
[pic]
8.95 [pic]
[pic]
8.96 [pic], and [pic]
[pic]
8.97 [pic], and [pic]
[pic]
8.98 [pic]
[pic]
8.99 [pic], and [pic] therefore [pic]
a. Point estimate of [pic]
Margin of error [pic]
b. The 95 % confidence interval for μ is: [pic] to $284.76
8.100 [pic] and [pic] then [pic]
a. Point estimate of [pic]
Margin of error [pic]
b. The 97% confidence interval for μ is: [pic] to $2765.43
8.101 [pic] inches, and [pic] inches; so [pic]
The 99% confidence interval for μ is:
[pic] to 24.041 inches.
Since the upper limit of the confidence interval is 24.041, which is greater than 24.025, the machine needs an adjustment.
8.102 [pic] inches, and [pic] inches; so [pic]
The 98% confidence interval for μ is:
[pic] to 4.003 inches.
Since the lower limit of the confidence interval is less than 3.98 inches, the machine needs an adjustment.
8.103 [pic]
[pic]minutes
[pic]minutes
[pic]minutes
a. Point estimate of [pic]minutes
Margin of error [pic]minutes
b. The 98% confidence interval for [pic] is:
[pic] to 71.04 minutes
8.104 [pic]
[pic] hours
[pic]
[pic] hour
a. Point estimate of [pic] hours
Margin of error [pic]hours
b. The 99% confidence interval for μ is: [pic] to 17.43 hours
8.105 [pic] and [pic] therefore [pic]
[pic] and t = 2.797
The 99% confidence interval for μ is:[pic] to $726.40.
8.106 [pic]minutes, and [pic] minutes; so [pic] minutes
[pic] and t = 2.110
The 95% confidence interval for [pic] is:
[pic] to 26.24 minutes
8.107 [pic]hours, and [pic] hours; so [pic] hour
[pic] and t = 1.729
The 90% confidence interval for [pic] is:
[pic] to 10.60 hours
8.108 [pic]hours, and [pic] hours; so [pic] hour
[pic] and t = 2.539
The 98% confidence interval for [pic] is:
[pic] to 4.93 hours.
8.109 [pic] and [pic]. This means [pic] hours,
[pic] hours, and
[pic] hours
[pic] and t = 2.201
The 95% confidence interval for [pic] is:
[pic] to 2.54 hours.
8.110 [pic] and [pic]; Then [pic] calories
[pic] calories, and
[pic]calories
[pic] and t = 3.250
The 99% confidence interval for [pic] is:
[pic] to 158.47 calories
8.111 [pic], and [pic] therefore [pic]
a. Point estimate for [pic] or 12%
Margin of error [pic] or [pic]
b. The 99% confidence interval for p is:
[pic] to .239 or .1% to 23.9%
8.112 [pic], and [pic]; therefore [pic]
a. Point estimate of [pic] or 44%
Margin of error [pic] or [pic]
b. The 90% confidence interval for p is:
[pic] to .477 or 40.3% to 47.7%
8.113 [pic] and [pic]; hence, [pic]
The 99% confidence interval for p is:
[pic] to .683 or 11.7% to 68.3%
8.114 [pic], and [pic] hence [pic]
The 97% confidence interval for p is:
[pic] to .565 or 6.1% to 56.5%
8.115 [pic] hours, [pic] hours, and [pic]; hence [pic]
8.116 [pic], [pic], and [pic]; therefore [pic]
8.117 [pic] and [pic] hence [pic]
8.118 [pic] and [pic]then [pic]
8.119 [pic] [pic] confidence interval: $8.46 to $9.86
95% confidence interval for a large sample: [pic]
a. [pic]
b. [pic]
Confidence level = 99%, i.e. z = 2.58 so the 99% confidence interval is [pic] to $10.08
8.120 a. Let x = time devoted to commercials per hour on Channel 66.
A sample of 30 twenty – minute time intervals yielded a mean of 4.68 minutes of commercials with a standard deviation of 1.30 minutes. This is equivalent to:
[pic] minutes and [pic] minutes for a 60–minute time interval.
[pic] minute.
The 90% confidence interval for [pic] is:
[pic] to 15.21 minutes per hour
b. [pic]
The point estimate of [pic] is (7/30)[pic]minutes per hour = 14 minutes per hour
c. Both estimates yield approximately 14 minutes per hour.
d. [pic] and [pic]
So the 90% confidence interval for[pic] is [pic] to .360. Hence the 90% confidence interval for [pic] × 60minutes per hour = 6.36 to 21.60 minutes per hour.
e. Maximum error of Waldo’s estimate:
[pic] minutes per hour = [pic] minutes per hour = 7.64 minutes per hour
Maximum error of class estimate: [pic] = 1.17 minutes per hour
f. We need a new [pic] such that [pic] minutes per hour; hence, [pic], so [pic]
It does not seem to be feasible to achieve the same accuracy as that of the class using Waldo’s method.
8.121 Let: p1= proportion of 12–18 year old females who expect a female president within 10 years
p2= proportion of 12–18 year old females who expect a female president within 15 years
p3= proportion of 12–18 year old females who expect a female president within 20 years
p4 = proportion of 12–18 year old females who do not expect a female president not within their
lifetime
[pic]
[pic] and [pic] exceed 5 for all these proportions, so the sample is considered large.
[pic]
The 95% confidence interval for p1 is:
[pic] to .427 or 37.3% to 42.7%
[pic]
The 95% confidence interval for p2 is:
[pic] to .274 or 22.6% to 27.4%
[pic]
The 95% confidence interval for p3 is:
[pic] to .233 or 18.7% to 23.3%
[pic]
The 95% confidence interval for p4 is:
[pic] to .159 or 12.1% to 15.9%
A confidence interval is a range of numbers (in this particular case proportions or percentages) which give an estimate for the true value (i.e. proportion of 12–18 year old females who feel this way). The 95% means that we are 95% confident that this interval actually contains the true value. A single percentage that we assign as an estimate would almost always differ from the true value, hence a range with the associated confidence level is more informative. We assume that the 1200 people are a random sample of 12–18 year old females.
8.122 [pic]and [pic]
a. [pic]
The 95% confidence interval for p is:
[pic] to .159 or 4.1% to 15.9%
b. No, 18% does not lie within the confidence interval in part a. This suggests that the vaccine is effective to some degree.
c. There are several things that may have distorted the outcome of this experiment: we only point out two. It is not mentioned, although it is reasonable to assume that all the dogs that were vaccinated did not have Lyme disease to begin with. Assuming this, the vaccinated dogs were exposed only 1 year to possible tick bites, whereas the other dogs in the area probably were exposed much longer. Also, the owners who agreed to participate in this experiment may be more concerned about their dogs’ health and restrict the area that the dogs can access, thereby decreasing the exposure to the ticks as well.
8.123 [pic]miles,[pic]and [pic].
[pic] may be estimated by s. Hence, [pic]
Thus, an additional 65–20=45 observations must be taken.
8.124 The major problem with this procedure is that the sample is not drawn from the whole target population. This introduces nonsampling error.
i. Households with no cars would be excluded from this sample. Furthermore, households with more than 2 cars could be counted more than once in the sample. Both of these problems would result in an upwardly biased estimate of p.
ii. Because of the characteristics of this particular gas station, the clientele may not be typical of all gas station clientele (with respect to multiple car ownership). Other problems may be present, such as the next 200 gasoline customers may not be typical of all customers, or there may be dependency between gasoline customers from the same household. If the attendant took a random sample of 200 households, and determined how many of these households owned more than 2 cars, the sampling error would still be present, but could be reduced by taking a larger sample.
125. a. Here, [pic], [pic], and [pic]and [pic]
Thus, the required sample size is [pic] or 20 days
Note that since n < 30, we must assume that the number of cars passing each day is approximately normally distributed. Or we may take a large [pic] sample.
b. Since [pic], then, [pic] which corresponds to a confidence level of approximately 90%.
c. Since, [pic], then, [pic]
Thus, they can be 99% confident that their point estimate is within 75 cars of the true average.
8.126 No, the student’s analysis does not make sense. The relevant parameter, p, is the proportion of all U.S. senators in favor of the bill. The value, .55, is not a sample proportion: instead it is the population proportion, p. Since, p = .55 is known there is no need to estimate it.
Self – Review Test for Chapter Eight
1. a. Estimation means assigning values to a population parameter based on the value of a sample statistic.
b. An estimator is the sample statistic used to estimate a population parameter.
c. The value of a sample statistic is called the point estimate of the corresponding population parameter.
2. b 3. a 4. a 5. d 6. b
7. [pic], [pic] and [pic] hence [pic]
a. Point estimate of [pic]
Margin of error [pic]
b. The 99% confidence interval for [pic] is:
[pic] to $170,610
8. [pic], [pic] and [pic]
[pic] and t = 2.064
[pic]
The 95% confidence interval for [pic] is:
[pic] to $441,310.70
9. [pic], [pic] and [pic] so [pic]
a. Point estimate of [pic]
Margin of error [pic]
b. The 95% confidence interval for p is: [pic] to .449
10. [pic]houses, [pic] houses, and [pic], then [pic] ≈ 45.
11. [pic], then [pic]
12. [pic][pic]and [pic] so [pic]
13. The width of the confidence interval can be reduced by:
1. Lowering the confidence level
2. Increasing the sample size
The second alternative is better because by lowering the confidence level we will simply lower the probability that our confidence interval includes [pic].
14. To estimate the mean number of hours that all students at your college work per week:
1. Take a random sample of 12 students from your college who hold jobs
2. Record the number of hours each of these students worked last week
3. Calculate [pic]and [pic] from these data
4. After choosing the confidence level, find the value for the t distribution with 11df and for an area of [pic] in the right tail.
5. Obtain the confidence interval for [pic] by using the formula [pic]
You are assuming that the hours worked by all students at your college have a normal distribution.
15. To estimate the proportion of people who are happy with their current jobs:
1. Take a random sample of 35 workers
2. Determine whether or not each worker is happy with his or her job
3. Calculate [pic]and [pic]
4. Choose the confidence level and find the required value of z from the normal distribution table
5. Obtain the confidence interval for p by using the formula [pic]
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