Question 1

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?

Accepted Answer

A)  $1086 < \sigma ^ { 2 } < 269.6$ 
B)  $10.4 < \sigma ^ { 2 } < 164$ 
C)  $1237 < \sigma ^ { 2 } < 312.7$ 
D)  $122.8 < \sigma ^ { 2 } < 316.5$ 
A)  $1086 < \sigma ^ { 2 } < 269.6$ 
B)  $10.4 < \sigma ^ { 2 } < 164$ 
C)  $1237 < \sigma ^ { 2 } < 312.7$ 
D)  $122.8 < \sigma ^ { 2 } < 316.5$

Question 2

The formula for the confidence interval of the mean for a specific $\alpha$ is __________.

Accepted Answer

The answer of The formula for the confidence interval of...

Question 3

A 90% confidence interval around the variance was calculated. For n = 27, the confidence interval was 35.0-88.4. Give the variance of the sample.

Accepted Answer

The answer of A 90% confidence interval around the variance...

Question 4

Jennifer wants to find a 95% confidence interval for the time it takes her to get to work. She kept records
for 30 days and found her average time to commute to work was 20.5 minutes with a standard deviation
of 3.9 minutes. Jennifer's maximum error of estimate would be 1.4 minutes.

Accepted Answer

A) True 
 B)False

Question 5

The t-distribution has a variance that is greater than one.

Accepted Answer

A) True 
 B)False

Question 6

A chi-square variable cannot be negative, and the distributions are positively skewed.

Accepted Answer

A) True 
 B)False

Question 7

8 squirrels were found to have an average weight of 10.5 ounces with a sample standard deviation is 0.30. Find the 95% confidence interval of the

Accepted Answer

A)  (10.25, 10.75) 
B)  (9.53, 11.47) 
C)  (10.23, 10.77) 
D)  (10.39, 10.61) 
A)  (10.25, 10.75) 
B)  (9.53, 11.47) 
C)  (10.23, 10.77) 
D)  (10.39, 10.61)

Question 8

In order to find confidence intervals for variances and standard deviations, one must assume that the
variable is __________. 
________________________________________

Accepted Answer

normally d

Question 9

A confidence interval was constructed around a proportion. The interval was from 13.5% to 21.5%. The
proportion that was used to construct this interval was 17.5%.

Accepted Answer

A) True 
 B)False

Question 10

John Davis, a manager of a supermarket, wants to estimate the proportion of customers that will be using discount coupons at his store. How large a sample is required to estimate the

Accepted Answer

A)  756 
B)  1,849 
C)  1,067 
D)  1,509 
A)  756 
B)  1,849 
C)  1,067 
D)  1,509

Question 11

A retailer wants to estimate with 99% confidence the number of people who buy at his store. A previous
study showed that 24% of those interviewed had shopped at his store. He wishes to be accurate within 3%
of the

Accepted Answer

A) True 
 B)False

Question 12

In a study of size 8 where the variance is unknown, the distribution that should be used to calculate confidence intervals is

Accepted Answer

A)  a normal distribution 
B)  a t distribution with 7 degrees of freedom 
C)  a t distribution with 8 degrees of freedom 
D)  a t distribution with 9 degrees of freedom 
A)  a normal distribution 
B)  a t distribution with 7 degrees of freedom 
C)  a t distribution with 8 degrees of freedom 
D)  a t distribution with 9 degrees of freedom

Question 13

A chi-square distribution is negatively skewed.

Accepted Answer

A) True 
 B)False

Question 14

The Pizza Shop wanted to determine what proportion of its customers ordered only cheese pizza. Out of 80 customers surveyed, 15 ordered cheese pizza. What is the 99% confidence interval of the

Accepted Answer

A)  0.075 < p < 0.300 
B)  0.115 < p < 0.260 
C)  0.086 < p < 0.289 
D)  0.102 < p < 0.273 
A)  0.075 < p < 0.300 
B)  0.115 < p < 0.260 
C)  0.086 < p < 0.289 
D)  0.102 < p < 0.273

Question 15

The average number of miles that 90 truckers drove in a day was 540 with a standard deviation of 40
miles. What is the 99% confidence interval of the

Accepted Answer

The answer of The average number of miles that 90...

Question 16

The rounding rule for a confidence interval for a proportion is to round off to three decimal places.

Accepted Answer

A) True 
 B)False

Question 17

Using Table G, find the values for $\chi _ { \text {left } } ^ { 2 }$ and $\chi _ { \text {left } } ^ { 2 }$ for $\alpha = 05 \text { and } n = 27$

Accepted Answer

A)  15.379 and 38.885 
B)  16.151 and 40.113 
C)  16.5395 and 40.725 
D)  16.928 and 41.337 
A)  15.379 and 38.885 
B)  16.151 and 40.113 
C)  16.5395 and 40.725 
D)  16.928 and 41.337

Question 18

$$\text { The value of } \alpha / 2 \text { used in the figure below is } 0.025 \text {. }$$

Accepted Answer

A) True 
 B)False

Question 19

A recent study of 750 Internet users in Europe found that 35% of Internet users were women. What is the 95% confidence interval of the

Accepted Answer

A)  0.321 < p < 0.379 
B)  0.316 < p < 0.384 
C)  0.309 < p < 0.391 
D)  0.305 < p < 0.395 
A)  0.321 < p < 0.379 
B)  0.316 < p < 0.384 
C)  0.309 < p < 0.391 
D)  0.305 < p < 0.395

Question 20

The value for for a 95% confidence interval when is 26.119.

Accepted Answer

A) True 
 B)False

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 formula for the confidence interval of the mean for a specific $\alpha$ is __________.

A 90% confidence interval around the variance was calculated. For n = 27, the confidence interval was 35.0-88.4. Give the variance of the sample.

Jennifer wants to find a 95% confidence interval for the time it takes her to get to work. She kept records for 30 days and found her average time to commute to work was 20.5 minutes with a standard deviation of 3.9 minutes. Jennifer's maximum error of estimate would be 1.4 minutes.

The t-distribution has a variance that is greater than one.

A chi-square variable cannot be negative, and the distributions are positively skewed.

8 squirrels were found to have an average weight of 10.5 ounces with a sample standard deviation is 0.30. Find the 95% confidence interval of the

In order to find confidence intervals for variances and standard deviations, one must assume that the variable is . ______________________________

A confidence interval was constructed around a proportion. The interval was from 13.5% to 21.5%. The proportion that was used to construct this interval was 17.5%.

John Davis, a manager of a supermarket, wants to estimate the proportion of customers that will be using discount coupons at his store. How large a sample is required to estimate the

A retailer wants to estimate with 99% confidence the number of people who buy at his store. A previous study showed that 24% of those interviewed had shopped at his store. He wishes to be accurate within 3% of the

In a study of size 8 where the variance is unknown, the distribution that should be used to calculate confidence intervals is

A chi-square distribution is negatively skewed.

The Pizza Shop wanted to determine what proportion of its customers ordered only cheese pizza. Out of 80 customers surveyed, 15 ordered cheese pizza. What is the 99% confidence interval of the

The average number of miles that 90 truckers drove in a day was 540 with a standard deviation of 40 miles. What is the 99% confidence interval of the

The rounding rule for a confidence interval for a proportion is to round off to three decimal places.

Using Table G, find the values for $\chi _ { \text {left } } ^ { 2 }$ and $\chi _ { \text {left } } ^ { 2 }$ for $\alpha = 05 \text { and } n = 27$

$\text { The value of } \alpha / 2 \text { used in the figure below is } 0.025 \text {. }$ $\text { The value of } \alpha / 2 \text { used in the figure below is } 0.025 \text {. }$

A recent study of 750 Internet users in Europe found that 35% of Internet users were women. What is the 95% confidence interval of the

The value for for a 95% confidence interval when is 26.119.

The Nature of Probability and Statistics

Frequency Distributions and Graphs

Data Description

Discrete Probability Distributions

The Normal Distribution

Hypothesis Testing

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

Correlation and Regression

Chi-Square and Analysis of Variance Anova

Tables

Data Bank

Glossary

Filters

Exam 6: Confidence Intervals and Sample Size

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 formula for the confidence interval of the mean for a specific α\alphaα is __________.

A 90% confidence interval around the variance was calculated. For n = 27, the confidence interval was 35.0-88.4. Give the variance of the sample.

Jennifer wants to find a 95% confidence interval for the time it takes her to get to work. She kept records for 30 days and found her average time to commute to work was 20.5 minutes with a standard deviation of 3.9 minutes. Jennifer's maximum error of estimate would be 1.4 minutes.

The t-distribution has a variance that is greater than one.

A chi-square variable cannot be negative, and the distributions are positively skewed.

8 squirrels were found to have an average weight of 10.5 ounces with a sample standard deviation is 0.30. Find the 95% confidence interval of the

In order to find confidence intervals for variances and standard deviations, one must assume that the variable is __________. ________________________________________

A confidence interval was constructed around a proportion. The interval was from 13.5% to 21.5%. The proportion that was used to construct this interval was 17.5%.

John Davis, a manager of a supermarket, wants to estimate the proportion of customers that will be using discount coupons at his store. How large a sample is required to estimate the

A retailer wants to estimate with 99% confidence the number of people who buy at his store. A previous study showed that 24% of those interviewed had shopped at his store. He wishes to be accurate within 3% of the

In a study of size 8 where the variance is unknown, the distribution that should be used to calculate confidence intervals is

A chi-square distribution is negatively skewed.

The Pizza Shop wanted to determine what proportion of its customers ordered only cheese pizza. Out of 80 customers surveyed, 15 ordered cheese pizza. What is the 99% confidence interval of the

The average number of miles that 90 truckers drove in a day was 540 with a standard deviation of 40 miles. What is the 99% confidence interval of the

The rounding rule for a confidence interval for a proportion is to round off to three decimal places.

Using Table G, find the values for χleft 2\chi _ { \text {left } } ^ { 2 }χleft 2​ and χleft 2\chi _ { \text {left } } ^ { 2 }χleft 2​ for α=05 and n=27\alpha = 05 \text { and } n = 27α=05 and n=27

The value of α/2 used in the figure below is 0.025. \text { The value of } \alpha / 2 \text { used in the figure below is } 0.025 \text {. } The value of α/2 used in the figure below is 0.025.

A recent study of 750 Internet users in Europe found that 35% of Internet users were women. What is the 95% confidence interval of the

The value for for a 95% confidence interval when is 26.119.

The Nature of Probability and Statistics

Frequency Distributions and Graphs

Data Description

Discrete Probability Distributions

The Normal Distribution

Hypothesis Testing

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

Correlation and Regression

Chi-Square and Analysis of Variance Anova

Tables

Data Bank

Glossary

Filters

The formula for the confidence interval of the mean for a specific $\alpha$ is __________.

In order to find confidence intervals for variances and standard deviations, one must assume that the variable is . ______________________________

Using Table G, find the values for $\chi _ { \text {left } } ^ { 2 }$ and $\chi _ { \text {left } } ^ { 2 }$ for $\alpha = 05 \text { and } n = 27$

$\text { The value of } \alpha / 2 \text { used in the figure below is } 0.025 \text {. }$ $\text { The value of } \alpha / 2 \text { used in the figure below is } 0.025 \text {. }$