1. The chi-square and F-distributions are used primarily to make inferences about population ___________.
a. means
b. variances
c. medians
d. modes
e. proportions
2. When we replace with the sample standard devia
...
1. The chi-square and F-distributions are used primarily to make inferences about population ___________.
a. means
b. variances
c. medians
d. modes
e. proportions
2. When we replace with the sample standard deviation (s), we introduce a new source of variability and the sampling
distribution becomes the:
a. t -distribution b. F- distribution
c. chi-square distribution d. normal distribution
3. The t-distribution for developing a confidence interval for a mean has degrees of freedom
a. n + 2
b. n +1
c. n
d. n – 1
e. n - 2
4. As the sample size increases, the t-distribution becomes more similar to the ________ distribution.
a. normal
b. exponential
c. multinominal
d. chi-square
e. binomial
5. A parameter such as is sometimes referred to as a ________ parameter, because many times we need its value even
though it is not the parameter of primary interest.
a. special
b. random
c. nuisance
d. independent
e. dependent
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6. Two independent samples of sizes 50 and 50 are randomly selected from two populations to test the difference between
the population means, . The sampling distribution of the sample mean difference is:
a. normally distributed
b. approximately normal
c. t - distributed with 98 degrees of freedom
d. chi-squared distributed with 99 degrees of freedom
7. The general form of a confidence interval is:
a. Point Estimate = Multiple × Standard Error
b. Point Estimate = Multiple +Standard Error
c. Point Estimate ± Multiple × Standard Error
d. Point Estimate = Multiple ± Standard Error
8. The approximate standard error of the point estimate of the population total is:
a. b.
c. d.
9. A confidence interval is an interval estimate for which there is a specified degree of certainty that the actual true value of
the population parameter will fall within the interval.
a. True
b. False
10. When samples of size n are drawn from a population, then the sampling distribution of the sample mean is
approximately normal, provided that n is reasonably large.
a. True
b. False
11. The t-distribution and the standard normal distribution are practically indistinguishable as the degrees of freedom
increase.
a. True
b. False
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THE FOLLOWING ITEMS REQUIRE THE USE OF EXCEL:
12. (A) Compute has a t-distribution with 15 degrees of freedom.
(B) Compute has a t-distribution with 150 degrees of freedom.
(C) How do you explain the difference between the results obtained in (A) and (B)?
(D) Compute where Z is a standard normal random variable.
(E) Compare the results of (D) to the results obtained in (A) and (B). How do you explain the difference in these
probabilities?
13. What is the probability of a t-value smaller than 1.00?
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14. What is the probability of a t-value larger than 1.20?
15. What is the probability of a t-value between –1.40 and +1.40?
16. What would be the t-value where 0.05 of the values are in the upper tail?
17. What would be the t-values where 0.10 of the values are in both tails (sum of both tails)?
18. SWER: - 2.1What would be the t-values where 0.95 of the values would fall within this interval?
19. If you increase the confidence level, the confidence interval:
a. decreases
b. increases
c. stays the same
d. may increase or decrease, depending on the sample data
20. If you are constructing a confidence interval for a single mean, the confidence interval will with an increase in the
sample size.
a. decrease
b. increase
c. stay the same
d. increase or decrease, depending on the sample data
21. The interval estimate 18.5 2.5 was developed for a population mean when the sample standard deviation s was 7.5.
Had s equaled 15, the interval estimate would be 37 5.0.
a. True
b. False
22. The 95% confidence interval for the population mean , given that the sample size n = 49 and the population standard
deviation
a. True
b. False
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23. In order to construct a confidence interval estimate of the population mean , the value of must be given.
a. True
b. False
24. In general, increasing the confidence level will narrow the confidence interval, and decreasing the confidence level
widens the interval.
a. True
b. False
25. If a sample has 20 observations and a 95% confidence estimate for is needed, the appropriate value of t-multiple is
2.093
a. True
b. False
A sample of 40 country CD recordings of Willie Nelson has been examined. The average playing time of these
recordings is 51.3 minutes, and the standard deviation is 5.8 minutes.
26. (A) Construct a 95% confidence interval for the mean playing time of all Willie Nelson recordings.
(B) Interpret the confidence interval you constructed in (A).
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11 $77,600 31 $67,300 51 $44,800
12 $47,600 32 $63,800 52 $57,400
13 $62,700 33 $70,600 53 $48,100
14 $46,200 34 $49,800 54 $52,700
15 $64,300 35 $51,300 55 $57,400
16 $56,000 36 $56,600 56 $65,500
17 $53,400 37 $49,600 57 $59,600
18 $56,800 38 $67,400 58 $62,000
19 $51,200 39 $53,700 59 $49,700
20 $59,000 40 $48,700 60 $54,400
27. (A) Use Excel to obtain a simple random sample of size 10 from this frame.
(B) Using the sample generated in (A), construct a 95% confidence interval for the mean average annual household
income level of citizens in
the selected U.S. cities. Assume that the population consists of all average annual household income levels in the
given frame.
(C) Interpret the 95% confidence interval constructed in (B).
(D) Does the 95% confidence interval contain the actual population mean? If not, explain why not. What proportion of
many similarly constructed
confidence intervals should include the true population mean value?
Chapter 08
(B)
(C) We are 95% confident that the average annual household income level of all citizens is approximately
between $52,737 and $66,243.
(D) This confidence interval easily captures the true population mean of $57,043. Approximately 95% of
the confidence intervals constructed in this way should contain the true population mean.
The personnel department of a large corporation wants to estimate the family dental expenses of its employees to
determine the feasibility of providing a dental insurance plan. A random sample of 12 employees in 2004 reveals the
following family dental expenses (in dollars): 115, 370, 250, 93, 540, 225, 177, 425, 318, 182, 275, and 228. Use
StatTools for your calculations.
28. (A) Construct a 90% confidence interval estimate of the mean family dental expenses for all employees of this
corporation.
(B) What assumption about the population distribution must be made to answer (A)?
(C) Interpret the 90% confidence interval constructed in (A).
(D) Suppose you used a 95% confidence interval in (A). What would be your answer?
(E) Suppose the fourth value were 593 instead of 93. What would be your answer to (A)? What effect does this change
have on the confidence interval?
(F) Construct a 90% confidence interval estimate for the standard deviation of family dental expenses for all employees
of this corporation.
(G) Interpret the 90% confidence interval constructed in (E).
be approximately normally distributed.
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(C) We are 90% confident that the mean family dental expenses for all employees of this corporation is
between $199.26 and $333.74.
(D)
(E)
The additional $500 in dental expenses, divided across the sample of 12, raises the mean by $41.67 and
increases the standard deviation by nearly $18.20. The interval width increases over $23 in the process.
(F)
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The widths of 100 elevator rails have been measured. The sample mean and standard deviation of the elevator rails are
2.05 inches and 0.01 inch.
29. (A) Construct a 95% confidence interval for the average width of an elevator rail. Do we need to assume that the width of
elevator rails follows a normal distribution?
(B) How large a sample of elevator rails would we have to measure to ensure that we could estimate, with 95%
confidence, the average diameter of an elevator rail within 0.01 inch?
30. (A) Construct a 95% confidence interval for the mean of the average annual credit account balances.
(B) Interpret the 95% confidence interval constructed in (A).
(C) Use the confidence interval constructed for (A) to help the store evaluate its criteria for whether or not the credit card
program is worthwhile.
Chapter 08
31. If you decrease the confidence level, the confidence interval .
a. decreases
b. increases
c. stays the same
d. may increase or decrease, depending on the sample data
32. If you are constructing a confidence interval for a single mean, the confidence interval will with a decrease in the sample
size.
a. decrease
b. increase
c. stay the same
d. increase or decrease, depending on the sample data
33. Suppose there are 500 accounts in a population. You sample 50 of them and find a sample mean of $500. What would
be your estimate for the population total?
a. $5,000
b. $50,000
c. $250,000
d. $500,000
e. None of these choices
34. We can form a confidence interval for the population total T by finding a confidence interval for the population mean
in the usual way, and then multiplying the lower and upper limits the confidence interval by the population size N.
a. True
b. False
35. A 90% confidence interval estimate for a population mean is determined to be 72.8 to 79.6. If the confidence level is
reduced to 80%, the confidence interval for becomes narrower.
a. True
b. False
36. The approximate standard error of the point estimate of the population total is .
a. True
b. False
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Auditors of Independent Bank are interested in comparing the reported value of all 1775 customer savings account
balances with their own findings regarding the actual value of such assets. Rather than reviewing the records of each
savings account at the bank, the auditors randomly selected a sample of 100 savings account balances from the frame.
The sample mean and sample standard deviations were $505.75 and 360.95, respectively.
37. (A) Construct a 90% confidence interval for the total value of all savings account balances within this bank. Assume
that the population consists of all savings account balances in the frame.
(B) Interpret the 90% confidence interval constructed in (A).
38. The standard error of the sampling distribution of the sample proportion , when the sample size n = 50 and the
population proportion p = 0.25, is 0.00375.
a. True
b. False
39. As a general rule, the normal distribution is used to approximate the sampling distribution of the sample proportion
only if the sample size n is greater than 30.
a. True
b. False
40. If the standard error of the sampling distribution of the sample proportion is 0.0324 for samples of size 200, then the
population proportion must be 0.30.
a. True
b. False
41. The lower limit of the 95% confidence interval for the population proportion p, given that n = 300; and = 0.10 is
0.1339.
a. True
b. False
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42. If a random sample of size 250 is taken from a population, where it is known that the population proportion p = 0.4, then
the mean of the sampling distribution of the sample proportion is 0.60.
a. True
b. False
43. The upper limit of the 90% confidence interval for the population proportion p, given that n = 100; and = 0.20 is
0.2658.
a. True
b. False
44. The mean of the sampling distribution of the sample proportion , when the sample size n = 100 and the population
proportion p = 0.15, is 15.0.
a. True
b. False
45. You are told that a random sample of 150 people from Iowa has been given cholesterol tests, and 60 of these people
had levels over the “safe” count of 200. Construct a 95% confidence interval for the population proportion of people in
Iowa with cholesterol levels over 200.
Chapter 08
A marketing research consultant hired by Coca-Cola is interested in determining the proportion of customers who favor
Coke over other soft drinks. A random sample of 400 consumers was selected from the market under investigation and
showed that 53% favored Coca-Cola over other brands.
46. (A) Compute a 95% confidence interval for the true proportion of people who favor Coke. Do the results of this poll
convince you that a majority of people favors Coke?
(B) Suppose 2,000 (not 400) people were polled and 53% favored Coke. Would you now be convinced that a majority of
people favor Coke? Why might your answer be different than in (A)?
(C) How many people would have to be surveyed to be 95% confident that you can estimate the fraction of people who
favor Coca-Cola within 1%?
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An automobile dealer wants to estimate the proportion of customers who still own the cars they purchased six years
ago. A random sample of 200 customers selected from the automobile dealer’s records indicates that 88 still own cars
that were purchased six years earlier.
47. (A) Construct a 95% confidence interval estimate of the population proportion of all customers who still own the cars
they purchased six years ago
(B) How can the result in (A) be used by the automobile dealer to study satisfaction with cars purchased at the
dealership?
48. The shape of a chi-square distribution
a. is symmetric b. is skewed to the left
c. is skewed to the right d. depends on the sample data
49. In developing a confidence interval for the population standard deviation , we make use of the fact that the sampling
distribution of the sample standard deviation s is not the normal distribution or the t-distribution, but rather a rightskewed distribution called the chi-square distribution, which (for this procedure) has n – 1 degrees of freedom.
a. True
b. False
50. The degrees of freedom for the t and chi-square distributions is a numerical parameter of the distribution that defines the
precise shape of the distribution.
a. True
b. False
51. The confidence interval for the population standard deviation s is centered at the point estimate, the sample standard
deviation s.
a. True
b. False
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The employee benefits manager of a medium size business would like to estimate the proportion of full-time employees
who prefer adopting plan A of three available health care plans in the coming annual enrollment period. A reliable frame
of the company’s employees and their tentative health care preferences are available. Using Excel, the manager chose
a random sample of size 50 from the frame. There were 17 employees in the sample who preferred plan A.
52. (A) Construct a 99% confidence interval for the proportion of company employees who prefer plan A. Assume that the
population consists of the preferences of all employees in the frame.
(B) Interpret the 99% confidence interval constructed in (A)..
53. (A) Construct a 99% confidence interval for the standard deviation of the number of hours this firm’s employees spend
on work-related activities in a typical week.
(B) Interpret the 99% confidence interval constructed in (A).
(C) Given the target range of 40 to 60 hours of work per week, should senior management be concerned about the
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54. When the samples we want to compare are paired in some natural way, such as pretest/posttest for each person or
husband/wife pairs, a more appropriate form of analysis is to not compare two separate variables, but their .
a. difference
b. sum
c. ratio
d. total
e. product
55. Two independent samples of sizes 20 and 25 are randomly selected from two normal populations with equal variances.
In order to test the difference between the population means, the test statistic is:
a. a standard normal random variable
b. approximately standard normal random variable
c. t-distributed with 45 degrees of freedom
d. t-distributed with 43 degrees of freedom
56. The number of degrees of freedom needed to construct 90% confidence interval for the difference between means when
the data are gathered from paired samples, with 15 observations in each sample, is:
a. 30 b. 15
c. 28 d. 14
57. An example of a problem where the sample data would be paired is:
a. Difference between the means of appraised and sales house prices
b. Difference between the proportion of defective items from two suppliers
c. Difference in the mean life of two major brands of batteries
d. Difference in the mean salaries for graduates in two different academic fields at a university
e. None of these options
58. In general, the paired-sample procedure is appropriate when the samples are naturally paired in some way and there is
a reasonably large positive correlation between the pairs. In this case, the paired-sample procedure makes more
efficient use of the data and generally results in narrower confidence intervals.
a. True
b. False
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59. If two random samples of size 40 each are selected independently from two populations whose variances are 35 and
45, then the standard error of the sampling distribution of the sample mean difference, , equals 1.4142.
a. True
b. False
60. If two random samples of sizes 30 and 35 are selected independently from two populations whose means are 85 and
90, then the mean of the sampling distribution of the sample mean difference, , equals 5.
a. True
b. False
61. In developing confidence interval for the difference between two population means using two independent samples, we
use the pooled estimate in estimating the standard error of the sampling distribution of the sample mean difference
if the populations are normal with equal variances.
a. True
b. False
62. Samples of exam scores for employees before and after a training class would be examples of paired data
a. True
b. False
63. If two samples contain the same number of observations, then the data must be paired.
a. True
b. False
64. If we cannot make the strong assumption that the variances of two samples are equal, then we must use the pooled
standard deviation in calculating the standard error of a difference between the means.
a. True
b. False
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Q-Mart is interested in comparing its male and female customers. Q-Mart would like to know if its female charge
customers spend more money, on average, than its male charge customers. They have collected random samples of
25 female customers and 22 male customers. On average, women charge customers spend $102.23 and men charge
customers spend $86.46. Some information is shown below.
Summary statistics for two samples
Female Male
Sample sizes 25 22
Sample means 102.23 86.46
Sample standard deviations 93.393 59.695
Confidence interval for difference between means
Sample mean difference 15.77
Pooled standard deviation 79.466
Std error of difference 23.23
65. (A) Use a t - value of 2.014 to calculate a 95% confidence interval for the difference between the average female
purchase and the average male purchase. Would you conclude that there is a significant difference between females
and males in this case? Explain.
(B) What are the degrees of freedom for the t-multiple in this calculation? Explain how you would calculate the degrees
of freedom in this case.
(C) What is the assumption in this case that allows you to use the pooled standard deviation for this confidence
interval?
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A company employs two shifts of workers. Each shift produces a type of gasket where the thickness is the critical
dimension. The average thickness and the standard deviation of thickness for shift 1, based on a random sample of 40
gaskets, are 10.85 mm and 0.16 mm, respectively. The similar figures for shift 2, based on a random sample of 30
gaskets, are 10.90 mm and 0.19 mm. Let be the difference in thickness between shifts 1 and 2, and assume
that the population variances are equal.
66. (A) Construct a 95% confidence interval for .
(B) Based on your answer to (A), are you convinced that the gaskets from shift 2 are, on average, wider than those from
shift 1? Why or why not?
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A market research consultant hired by Coke Classic Company is interested in estimating the difference between the
proportions of female and male customers who favor Coke Classic over Pepsi Cola in Chicago. A random sample of 200
consumers from the market under investigation showed the following frequency distribution.
Male Female
Coke 72 38 110
Pepsi 58 32 90
130 70 200
67. (A) Construct a 95% confidence interval for the difference between the proportions of male and female customers who
prefer Coke Classic over Pepsi Cola.
(B) Interpret the constructed confidence interval.
Chapter 08
A real estate agent has collected a random sample of 40 houses that were recently sold in Grand Rapids, Michigan.
She is interested in comparing the appraised value and recent selling price (in thousands of dollars) of the houses in
this particular market. The values of these two variables for each of the 40 randomly selected houses are shown below.
House Value Price House Value Price
1 140.93 140.24 21 136.57 135.35
2 132.42 129.89 22 130.44 121.54
3 118.30 121.14 23 118.13 132.98
4 122.14 111.23 24 130.98 147.53
5 149.82 145.14 25 131.33 128.49
6 128.91 139.01 26 141.10 141.93
7 134.61 129.34 27 117.87 123.55
8 121.99 113.61 28 160.58 162.03
9 150.50 141.05 29 151.10 157.39
10 142.87 152.90 30 120.15 114.55
11 155.55 157.79 31 133.17 139.54
12 128.50 135.57 32 140.16 149.92
13 143.36 151.99 33 124.56 122.08
14 119.65 120.53 34 127.97 136.51
15 122.57 118.64 35 101.93 109.41
16 145.27 149.51 36 131.47 127.29
17 149.73 146.86 37 121.27 120.45
18 147.70 143.88 38 143.55 151.96
19 117.53 118.52 39 136.89 132.54
20 140.13 146.07 40 106.11 114.33
68. (A) Use the sample data to generate a 95% confidence interval for the mean difference between the appraised values
and selling prices of the houses sold in Grand Rapids.
(B) Interpret the constructed confidence interval fin (A) for the real estate agent.
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Q-Mart is interested in comparing customer who used its own charge card with those who use other types of credit
cards. Q-Mart would like to know if customers who use the Q-Mart card spend more money per visit, on average, than
customers who use some other type of credit card. They have collected information on a random sample of 38 charge
customers as shown below. On average, the person using a Q-Mart card spends $192.81 per visit and customers using
another type of card spend $104.47 per visit.
Summary statistics for two samples
Q-Mart Other Charges
Sample sizes 13 25
Sample means 192.81 104.47
Sample standard deviations 115.243 71.139
Confidence interval for difference between means
Sample mean difference 88.34
Pooled standard deviation 88.323
Std error of difference 30.201
69. (A) Using a t - value of 2.0281, calculate a 95% confidence interval for the difference between the average Q-Mart
charge and the average charge on another type of credit card.
(B) What are the degrees of freedom for the t - multiple in this calculation? Explain how you would calculate the
degrees of freedom in this case.
(C) What is the assumption in this case that allows you to use the pooled standard deviation for this confidence
interval?
(D) Would you conclude that there is a significant difference between the two types of customers in this case? Explain.
70. In constructing confidence interval estimate for the difference between the means of two populations, where the
unknown population variances are assumed not to be equal, summary statistics computed from two independent
samples are as follows: Construct 90% confidence interval
for .
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A defensive driving training company is interested in evaluating the relative effectiveness of its two main modes of
training; online and traditional classroom. The company has collected a random sample of 300 customers in a
particular area, with the following results:
Online Classroom
Pass 143 128
Fail 17 12
Total 160 140
71. (A) Construct a 95% confidence interval for the difference between the proportions of online and classroom customers
who pass the final exam.
(B) Interpret the confidence interval obtained in (A).
72. When you calculate the sample size for a proportion, you use an estimate for the population proportion; namely . A
conservative value for n can be obtained by using =
a. 0.01
b. 0.05
c. 0.10
d. 0.50
e. 1.00
73. Confidence intervals are a function of which of the following three things?
a. The population, the sample, and the standard deviation
b. The sample, the variable of interest, and the degrees of freedom
c. The data in the sample, the confidence level, and the sample size
d. The sampling distribution, the confidence level, and the degrees of freedom
e. The mean, median, and mode
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74. After calculating the sample size needed to estimate a population proportion to within 0.05, you have been told that the
maximum allowable error (B) must be reduced to just 0.025. If the original calculation led to a sample size of 1000, the
sample size will now have to be:
a. 2000 b. 4000
c. 1000 d. 8000
75. For a given confidence level, the procedure for controlling interval length usually begins with the specification of
a. the point estimate b. the population standard deviation, s
c. the sample standard deviation, s d. the interval half-length, B
76. In determining the sample size n for estimating the population proportion p, a conservative value of n can be obtained by
using 0.50 as an estimate of p.
a. True
b. False
77. You are trying to estimate the average amount a family spends on food during a year. In the past, the standard deviation
of the amount a family has spent on food during a year has been approximately $1200. If you want to be 99% sure that
you have estimated average family food expenditures within $60, how many families do you need to survey?
78. You have been assigned to determine whether more people prefer Coke to Pepsi. Assume that roughly half the
population prefers Coke and half prefers Pepsi. How large a sample would you need to take to ensure that you could
estimate, with 95% confidence, the proportion of people preferring Coke within 3% of the actual value?.
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A sample of 9 production managers with over 15 years of experience has an average salary of $71,000 and a sample
standard deviation of $18,000.
79. (A) You can be 95% confident that the mean salary for all production managers with at least 15 years of experience is
between what two numbers (the t-multiple with 8 degrees of freedom is 2.306)? What assumption are you making about
the distribution of salaries?
(B) What sample size would be needed to ensure that we could estimate the true mean salary of all production
managers with more than 15 years of experience and have only 5 chances in 100 of being off by more than $4200?
80. (A) Find a 95% confidence interval for the mean account balance on this store’s credit card (the t-multiple with 39
degrees of freedom is 2.0227).
(B) What sample size would be needed to ensure that we could estimate the true mean account balance and have only
5 chances in 100 of being off by more than $100?,
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The percent defective for parts produced by a manufacturing process is targeted at 4%. The process is monitored daily
by taking samples of sizes n = 160 units. Suppose that today’s sample contains 14 defectives.
81. (A) Determine a 95% confidence interval for the proportion defective for the process today.
(B) Based on your answer to (A), is it still reasonable to think the overall proportion defective produced by today’s process is
actually the targeted 4%? Explain your reasoning.
(C) The confidence interval in (A) is based on the assumption of a large sample size. Is this sample size sufficiently large in this
example? Explain how you arrived at your answer.
(D) How many units would have to be sampled to be 95% confident that you can estimate the fraction of defective parts within
2% (using the information from today’s sample)?
82. You are trying to estimate the average amount a family spends on food during a year. In the past the standard deviation
of the amount a family has spent on food during a year has been approximately $800. If you want to be 95% sure that
you estimated average family food expenditures within $50, how many families do you need to survey?
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83. In past years, approximately 25% of all U.S. families purchased potato chips at least once a month. We are interested
in determining the fraction of all U.S. families that currently purchase potato chips at least once a month. How many
families must we survey if we want to be 99% sure that our estimate of the fraction of U.S. families currently purchasing
potato chips at least once a month is accurate within 2%?
ANSWER:
= 0.25, z – multiple = 2.575, B = 0.02
84. There are, generally speaking, two types of statistical inference. They are:
a. sample estimation and population estimation
b. confidence interval estimation and hypothesis testing
c. interval estimation for a mean and point estimation for a proportion
d. independent sample estimation and dependent sample estimation
e. None of these choices
85. From a sample of 500 items, 30 were found to be defective. The point estimate of the population proportion defective will
be:
a. 0.06 b. 30.0
c. 16.667 d. None of the above
86. The chi-square distribution for developing a confidence interval for a standard deviation has degrees of freedom.
a. n + 2
b. n +1
c. n
d. n – 1
e. n - 2
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Chapter 08
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