The following MINITAB output presents a $95 \%$ confidence interval. $\begin{array}{ccccc} \hline {\text { The assumed sigma }=10.5215} & \\ \text { Variable } & \mathrm{N} & \text { Mean } & \text { SE Mean } & 95 \% \mathrm{CI} \\ \mathbf{x} & 41 & 31.580 & 1.6432 & (28.359,34.801) \\ \hline \end{array}$ Find the sample size needed so that the $95 \%$ confidence interval will have a margin of err $1.2$. rror of

A) 511 B) 18 C) 416 D) 296 A) 511 B) 18 C) 416 D) 296

The following MINITAB output presents a $95 \%$ confidence interval. $\begin{array}{crrcc} \hline{\text { The assumed sigma }=9.9860} & \\ \text { Variable } & \mathrm{N} & \text { Mean } & \text { SE Mean } & 95 \% \mathrm{CI} \\ \mathrm{x} & 37 & 31.799 & 1.6417 & (28.581,35.017) \\ \hline \end{array}$ Use the appropriate critical value along with the information in the computer output to construct a $99 \%$ confidence interval.

A) (30.461, 33.137) B) (28.581, 35.017) C) (27.980, 35.618) D) (27.570, 36.028) A) (30.461, 33.137) B) (28.581, 35.017) C) (27.980, 35.618) D) (27.570, 36.028)

Six measurements were made of the magnesium ion concentration (in parts per million, o in a city's municipal water supply, with the following results. It is reasonable to assume th population is approximately normal. $\begin{array}{llllll} \hline 170 & 201 & 199 & 202 & 173 & 153 \\ \hline \end{array}$ Based on a $95 \%$ confidence interval for the mean magnesium ion concentration, is it reasonable to believe that the mean magnesium ion concentration may be greater than 208 ? (Hint: you should first calculate the $95 \%$ confidence interval for the mean magnesium ion concentration.) )

A) The likelihood cannot be determined. B) No C) Yes A) The likelihood cannot be determined. B) No C) Yes

Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in ounces, of a sample of 12 boxes. It is reasonable to assume that the population is approximately normal. $\begin{array}{llllll} \hline 21.72 & 21.75 & 21.62 & 21.92 & 22.10 & 22.13 \\ \hline 22.25 & 22.26 & 22.04 & 21.88 & 22.02 & 22.15 \\ \hline \end{array}$ Construct a $95 \%$ confidence interval for the mean weight.

A) $21.782 < \mu < 22.192$ B) $21.853 < \mu < 22.120$ C) $21.790 < \mu < 22.183$ D) $21.847 < \mu < 22.126$ A) $21.782 < \mu < 22.192$ B) $21.853 < \mu < 22.120$ C) $21.790 < \mu < 22.183$ D) $21.847 < \mu < 22.126$

The following MINITAB output presents a confidence interval for a proportion. $\begin{array}{crrcc} \hline \text { Variable } & \mathrm{X} & \mathrm{N} & \text { Sample p } & 95 \% \mathrm{CI} \\ \mathrm{x} & 117 & 159 & 0.735849 & (0.667319,0.804379) \\ \hline \end{array}$ Use the information in the display to construct a $98 \%$ confidence interval for $p$.

A) (0.655, 0.817) B) (0.678, 0.793) C) (0.667, 0.804) D) (0.646, 0.826 A) (0.655, 0.817) B) (0.678, 0.793) C) (0.667, 0.804) D) (0.646, 0.826

Exam 7: Confidence Intervals and Sample Size

The average score for 100 teenage boys playing a certain computer game was 80,000 points per player. If the standard deviation of the population is 20,000 points, find the $95 \%$ confidence interval of the mean score for all teenage boys.

(Short Answer)

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Question 101

A survey of 800 women shoppers found that 17% of them shop on impulse. What is the 98% confidence interval for the true proportion of women shoppers who shop on Impulse?

(Multiple Choice)

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Question 102

When computing a confidence interval for a population mean using raw data, round off to two more decimal places than the number of decimal places in the original data.

(True/False)

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Question 103

A study of 95 apple trees showed that the average number of apples per tree was 525 The standard deviation of the population is 200. Which of the following is the 80% Confidence interval for the mean number of apples per tree for all trees?

(Multiple Choice)

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Question 104

$\text { Identify the degree of confidence displayed in the confidence interval shown below. }$ $\text { Identify the degree of confidence displayed in the confidence interval shown below. }$

(Multiple Choice)

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Question 105

$\text { What is the value of } \alpha \text { used in describing the confidence interval shown below. }$ $\text { What is the value of } \alpha \text { used in describing the confidence interval shown below. }$

(Multiple Choice)

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Question 106

The following MINITAB output presents a $95 \%$ confidence interval. The assumed sigma =10.5215 Variable Mean SE Mean 95\% 41 31.580 1.6432 (28.359,34.801) Find the sample size needed so that the $95 \%$ confidence interval will have a margin of err $1.2$ . rror of

(Multiple Choice)

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Question 107

The following display from a TI-84 Plus calculator presents a $95 \%$ confidence interval. $The following display from a TI-84 Plus calculator presents a 95 \% confidence interval. Fill in the blanks: We are ____ confident that the population mean is between _ and _.$ Fill in the blanks: We are confident that the population mean is between _ and _____.

(Multiple Choice)

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Question 108

The following MINITAB output presents a $95 \%$ confidence interval. The assumed sigma =9.9860 Variable Mean SE Mean 95\% 37 31.799 1.6417 (28.581,35.017) Use the appropriate critical value along with the information in the computer output to construct a $99 \%$ confidence interval.

(Multiple Choice)

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Question 109

$\text { The } t \text {-distribution has a variance that is greater than one. }$

(True/False)

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Question 110

John Davis, a manager of a supermarket, wants to estimate the proportion of customers who use food stamps at his store. He has no initial estimate of what the sample Proportion will be. How large a sample is required to estimate the true proportion to Within 3 percentage points with 98% confidence?

(Multiple Choice)

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Question 111

Six measurements were made of the magnesium ion concentration (in parts per million, o in a city's municipal water supply, with the following results. It is reasonable to assume th population is approximately normal. 170 201 199 202 173 153 Based on a $95 \%$ confidence interval for the mean magnesium ion concentration, is it reasonable to believe that the mean magnesium ion concentration may be greater than 208 ? (Hint: you should first calculate the $95 \%$ confidence interval for the mean magnesium ion concentration.) )

(Multiple Choice)

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Question 112

For a random sample of 23 European countries, the variance on life expectancy was years. What is the 95% confidence interval for the variance of life expectancy in all of Europe?

(Multiple Choice)

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Question 113

Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in ounces, of a sample of 12 boxes. It is reasonable to assume that the population is approximately normal. 21.72 21.75 21.62 21.92 22.10 22.13 22.25 22.26 22.04 21.88 22.02 22.15 Construct a $95 \%$ confidence interval for the mean weight.

(Multiple Choice)

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Question 114

A sample of 125 tobacco smokers who recently completed a new smoking-cessation program were asked to rate the effectiveness of the program on a scale of 1 to 10, with 10 corresponding to "completely effective" and 1 corresponding to "completely Ineffective". The average rating was 4.1 and the standard deviation was 4.0. Construct a 95% confidence interval for the mean score.

(Multiple Choice)

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Question 115

The following MINITAB output presents a confidence interval for a proportion. Variable Sample p 95\% 117 159 0.735849 (0.667319,0.804379) Use the information in the display to construct a $98 \%$ confidence interval for $p$ .

(Multiple Choice)

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Question 116

The following display from a TI- 84 Plus calculator presents a $99 \%$ confidence interval for proportion. $The following display from a TI- 84 Plus calculator presents a 99 \% confidence interval for proportion. a Use the information in the display to construct a 95% confidence interval for p.$ a Use the information in the display to construct a 95% confidence interval for p.

(Multiple Choice)

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Question 117

The percentage rates of home ownership for 8 randomly selected states are listed below. Estimate the population variance and standard deviation for the percentage rate of home ownership with 99% confidence. Assume the variable is normally distributed. 64.7 76.2 68.9 75.0 61.8 69.9 71.5 67.3

(Multiple Choice)

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Question 118

What is the 95% confidence interval for the standard deviation of birth weights at County General Hospital, if the standard deviation of the last 25 babies born there was 1)1 pounds?

(Multiple Choice)

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Question 119

What is the 90% confidence interval for the variance of exam scores for 28 algebra students, if the standard deviation of their last exam was 12.7?

(Multiple Choice)

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Question 120

Showing 101 - 120 of 122

The average score for 100 teenage boys playing a certain computer game was 80,000 points per player. If the standard deviation of the population is 20,000 points, find the $95 \%$ confidence interval of the mean score for all teenage boys.

A survey of 800 women shoppers found that 17% of them shop on impulse. What is the 98% confidence interval for the true proportion of women shoppers who shop on Impulse?

When computing a confidence interval for a population mean using raw data, round off to two more decimal places than the number of decimal places in the original data.

A study of 95 apple trees showed that the average number of apples per tree was 525 The standard deviation of the population is 200. Which of the following is the 80% Confidence interval for the mean number of apples per tree for all trees?

$\text { Identify the degree of confidence displayed in the confidence interval shown below. }$ $\text { Identify the degree of confidence displayed in the confidence interval shown below. }$

$\text { What is the value of } \alpha \text { used in describing the confidence interval shown below. }$ $\text { What is the value of } \alpha \text { used in describing the confidence interval shown below. }$

The following MINITAB output presents a $95 \%$ confidence interval. The assumed sigma =10.5215 Variable Mean SE Mean 95\% 41 31.580 1.6432 (28.359,34.801) Find the sample size needed so that the $95 \%$ confidence interval will have a margin of err $1.2$ . rror of

The following MINITAB output presents a $95 \%$ confidence interval. The assumed sigma =9.9860 Variable Mean SE Mean 95\% 37 31.799 1.6417 (28.581,35.017) Use the appropriate critical value along with the information in the computer output to construct a $99 \%$ confidence interval.

$\text { The } t \text {-distribution has a variance that is greater than one. }$

John Davis, a manager of a supermarket, wants to estimate the proportion of customers who use food stamps at his store. He has no initial estimate of what the sample Proportion will be. How large a sample is required to estimate the true proportion to Within 3 percentage points with 98% confidence?

For a random sample of 23 European countries, the variance on life expectancy was years. What is the 95% confidence interval for the variance of life expectancy in all of Europe?

The following MINITAB output presents a confidence interval for a proportion. Variable Sample p 95\% 117 159 0.735849 (0.667319,0.804379) Use the information in the display to construct a $98 \%$ confidence interval for $p$ .

The percentage rates of home ownership for 8 randomly selected states are listed below. Estimate the population variance and standard deviation for the percentage rate of home ownership with 99% confidence. Assume the variable is normally distributed. 64.7 76.2 68.9 75.0 61.8 69.9 71.5 67.3

What is the 95% confidence interval for the standard deviation of birth weights at County General Hospital, if the standard deviation of the last 25 babies born there was 1)1 pounds?

What is the 90% confidence interval for the variance of exam scores for 28 algebra students, if the standard deviation of their last exam was 12.7?

The Nature of Probability and Statistics

Frequency Distributions and Graphs

Data Description

Probability and Counting Rules

Discrete Probability Distributions

The Normal Distribution

Hypothesis Testing

Testing the Difference Between Two Means, Two Variances, and Two Proportions

Correlation and Regression

Other Chi-Square Tests

Analysis of Variance

Nonparametric Statistics

Sampling and Simulation

Filters

Exam 7: Confidence Intervals and Sample Size

The average score for 100 teenage boys playing a certain computer game was 80,000 points per player. If the standard deviation of the population is 20,000 points, find the 95%95 \%95% confidence interval of the mean score for all teenage boys.

A survey of 800 women shoppers found that 17% of them shop on impulse. What is the 98% confidence interval for the true proportion of women shoppers who shop on Impulse?

When computing a confidence interval for a population mean using raw data, round off to two more decimal places than the number of decimal places in the original data.

A study of 95 apple trees showed that the average number of apples per tree was 525 The standard deviation of the population is 200. Which of the following is the 80% Confidence interval for the mean number of apples per tree for all trees?

Identify the degree of confidence displayed in the confidence interval shown below. \text { Identify the degree of confidence displayed in the confidence interval shown below. } Identify the degree of confidence displayed in the confidence interval shown below.

What is the value of α used in describing the confidence interval shown below. \text { What is the value of } \alpha \text { used in describing the confidence interval shown below. } What is the value of α used in describing the confidence interval shown below.

The following MINITAB output presents a 95%95 \%95% confidence interval. The assumed sigma =10.5215 Variable Mean SE Mean 95\% 41 31.580 1.6432 (28.359,34.801) Find the sample size needed so that the 95%95 \%95% confidence interval will have a margin of err 1.21.21.2 . rror of

The following display from a TI-84 Plus calculator presents a 95%95 \%95% confidence interval. Fill in the blanks: We are ________ confident that the population mean is between _______ and _______.

The following MINITAB output presents a 95%95 \%95% confidence interval. The assumed sigma =9.9860 Variable Mean SE Mean 95\% 37 31.799 1.6417 (28.581,35.017) Use the appropriate critical value along with the information in the computer output to construct a 99%99 \%99% confidence interval.

The t-distribution has a variance that is greater than one. \text { The } t \text {-distribution has a variance that is greater than one. } The t-distribution has a variance that is greater than one.

John Davis, a manager of a supermarket, wants to estimate the proportion of customers who use food stamps at his store. He has no initial estimate of what the sample Proportion will be. How large a sample is required to estimate the true proportion to Within 3 percentage points with 98% confidence?

For a random sample of 23 European countries, the variance on life expectancy was years. What is the 95% confidence interval for the variance of life expectancy in all of Europe?

The following MINITAB output presents a confidence interval for a proportion. Variable Sample p 95\% 117 159 0.735849 (0.667319,0.804379) Use the information in the display to construct a 98%98 \%98% confidence interval for ppp .

The following display from a TI- 84 Plus calculator presents a 99%99 \%99% confidence interval for proportion. a Use the information in the display to construct a 95% confidence interval for p.

The percentage rates of home ownership for 8 randomly selected states are listed below. Estimate the population variance and standard deviation for the percentage rate of home ownership with 99% confidence. Assume the variable is normally distributed. 64.7 76.2 68.9 75.0 61.8 69.9 71.5 67.3

What is the 95% confidence interval for the standard deviation of birth weights at County General Hospital, if the standard deviation of the last 25 babies born there was 1)1 pounds?

What is the 90% confidence interval for the variance of exam scores for 28 algebra students, if the standard deviation of their last exam was 12.7?

The Nature of Probability and Statistics

Frequency Distributions and Graphs

Data Description

Probability and Counting Rules

Discrete Probability Distributions

The Normal Distribution

Hypothesis Testing

Testing the Difference Between Two Means, Two Variances, and Two Proportions

Correlation and Regression

Other Chi-Square Tests

Analysis of Variance

Nonparametric Statistics

Sampling and Simulation

Filters

The average score for 100 teenage boys playing a certain computer game was 80,000 points per player. If the standard deviation of the population is 20,000 points, find the $95 \%$ confidence interval of the mean score for all teenage boys.

$\text { Identify the degree of confidence displayed in the confidence interval shown below. }$ $\text { Identify the degree of confidence displayed in the confidence interval shown below. }$

$\text { What is the value of } \alpha \text { used in describing the confidence interval shown below. }$ $\text { What is the value of } \alpha \text { used in describing the confidence interval shown below. }$

The following MINITAB output presents a $95 \%$ confidence interval. The assumed sigma =10.5215 Variable Mean SE Mean 95\% 41 31.580 1.6432 (28.359,34.801) Find the sample size needed so that the $95 \%$ confidence interval will have a margin of err $1.2$ . rror of

The following MINITAB output presents a $95 \%$ confidence interval. The assumed sigma =9.9860 Variable Mean SE Mean 95\% 37 31.799 1.6417 (28.581,35.017) Use the appropriate critical value along with the information in the computer output to construct a $99 \%$ confidence interval.

$\text { The } t \text {-distribution has a variance that is greater than one. }$

The following MINITAB output presents a confidence interval for a proportion. Variable Sample p 95\% 117 159 0.735849 (0.667319,0.804379) Use the information in the display to construct a $98 \%$ confidence interval for $p$ .