Question 1

Which of the following would be an appropriate alternative hypothesis?

Accepted Answer

A)  The population proportion is less than 0.65. 
B)  The sample proportion is less than 0.65. 
C)  The population proportion is equal to 0.65. 
D)  The sample proportion is equal to 0.65. 
A)  The population proportion is less than 0.65. 
B)  The sample proportion is less than 0.65. 
C)  The population proportion is equal to 0.65. 
D)  The sample proportion is equal to 0.65.

Question 2

The test statistic for p is approximately normal when ____________________ and ____________________ are both greater than 5.

Accepted Answer

np;n(1 - p

Question 3

The chi-squared test and estimator of the population variance ____________________(are/are not)valid if the population is slightly to moderately nonnormal.

Accepted Answer

The answer of The chi-squared test and estimator of the...

Question 4

In forming a 95% confidence interval for a population mean from a sample size of 20,the number of degrees of freedom from the t-distribution equals 20.

Accepted Answer

A) True 
 B)False

Question 5

A population is defined to be small if it is ____________________ than ____________________ times the size of the sample.

Accepted Answer

The answer of A population is defined to be small...

Question 6

Applicants' Grades
The grades of a sample of 10 applicants,selected at random from a large population,are 71,86,75,63,92,70,81,59,80,and 90.
-{Applicants' Grades Narrative} Construct a 90% confidence interval estimate for the population standard deviation.

Accepted Answer

LCL = @#LAT-DLM& = 8

Question 7

To produce a confidence interval estimator for the total,we multiply the lower and upper confidence limits of the interval estimator of p by ____________________.

Accepted Answer

N
the popu

Question 8

The t-distribution allows the calculation of confidence intervals for means for small samples when the population variance is not known,regardless of the shape of the distribution in the population.

Accepted Answer

A) True 
 B)False

Question 9

Single Mothers' Ages
A random sample of 10 single mothers was drawn from a Obstetrics Clinic.Their ages are 22,17,27,20,23,19,24,18,19,and 24 years.
-{Single Mothers' Ages Narrative} Estimate the population mean with 90% confidence.

Accepted Answer

@#LAT-DLM& .Thus,LCL

Question 10

To estimate the total number of successes in a large finite population of size N,using a sample of size n,the confidence interval estimator is:

Accepted Answer

A)  $$\tilde { p } \pm z _ { \alpha / 2 } \sqrt { \tilde
 { p } ( 1 - \tilde { p } ) / n }$$ 
B)  $$N \left[ \tilde  { p } \pm z _ { \alpha / 2 } \sqrt { \tilde  { p } ( 1 - \tilde { p } ) / n } \right]$$ 
C)  $$( N + n ) \left[ \tilde { p } \pm z _ { a / 2 } \sqrt { \tilde { p } ( 1 - \tilde { p } ) / n } \right]$$ 
D)  $$( N - n ) \left[ \tilde  { p } \pm z _ { \alpha / 2 } \sqrt { \tilde { p } ( 1 - \tilde { p } ) / n } \right]$$ 
A)  $$\tilde { p } \pm z _ { \alpha / 2 } \sqrt { \tilde
 { p } ( 1 - \tilde { p } ) / n }$$ 
B)  $$N \left[ \tilde  { p } \pm z _ { \alpha / 2 } \sqrt { \tilde  { p } ( 1 - \tilde { p } ) / n } \right]$$ 
C)  $$( N + n ) \left[ \tilde { p } \pm z _ { a / 2 } \sqrt { \tilde { p } ( 1 - \tilde { p } ) / n } \right]$$ 
D)  $$( N - n ) \left[ \tilde  { p } \pm z _ { \alpha / 2 } \sqrt { \tilde { p } ( 1 - \tilde { p } ) / n } \right]$$

Question 11

New Mousetrap
After a financial analysis,the general manager of a large company decided that if more than 8% of potential buyers of a new mousetrap purchase that product,the company would show a profit.In a preliminary survey of 500 potential buyers,56 people say that they will buy the mousetrap.
-{New Mousetrap Narrative} Is there sufficient evidence at the 5% significance level that the mousetrap will produce a profit?

Accepted Answer

Rejection region: z > z_0.05 = 1.64

Question 12

Concert Tickets
A simple random sample of 100 concert tickets was drawn from a normal population.The mean and standard deviation of the sample were $120 and $25,respectively.
-{Concert Tickets Narrative} Explain how to use the confidence interval to test the hypotheses at $\alpha$ = 0.10.

Accepted Answer

Since the hypothesized value @#LAT-DLM&<

Question 13

In selecting the sample size to estimate the population proportion p,if we have no knowledge of even the approximate values of the sample proportion $\hat { P }$ ,we:

Accepted Answer

A)  take another sample and estimate. 
B)  take two more samples and find the average of their. 
C)  let= 0.50. 
D)  let= 0.95. 
A)  take another sample and estimate. 
B)  take two more samples and find the average of their. 
C)  let= 0.50. 
D)  let= 0.95.

Question 14

Marketing segmentation separates consumers of a product into different groups in such a way that members of each group are ____________________ and there are differences between ____________________.

Accepted Answer

The answer of Marketing segmentation separates consumers of a product...

Question 15

A major electronics store chain is interested in estimating the average amount its credit card customers spent on their first visit to the chain's new store in the mall.Fifteen credit card accounts were randomly sampled and analyzed with the following results: $\bar { x }$ = $50.50 and s2 = 400.A 95% confidence interval for the average amount the credit card customers spent on their first visit to the chain's new store in the mall is:

Accepted Answer

A)  $50.50 -$9.09. 
B)  $50.50 - $10.12. 
C)  $50.50 - $11.08. 
D)  None of these choices. 
A)  $50.50 -$9.09. 
B)  $50.50 - $10.12. 
C)  $50.50 - $11.08. 
D)  None of these choices.

Question 16

A random sample of size 15 taken from a normally distributed population revealed a sample mean of 75 and a sample variance of 25.The upper limit of a 95% confidence interval for the population mean would equal:

Accepted Answer

A)  77.77 
B)  72.23 
C)  88.85 
D)  77.27 
A)  77.77 
B)  72.23 
C)  88.85 
D)  77.27

Question 17

Hourly Fees
A random sample of 15 hourly fees for car washers (including tips)was drawn from a normal population.The sample mean and sample standard deviation were $\bar { x }$ = $14.9 and s = $6.75.
-{Hourly Fees Narrative} Can we infer at the 5% significance level that the population mean is greater than 12,assuming that you know the population standard deviation is equal to 6.75?

Accepted Answer

Rejection region: z > z_0.05 = 1.64

Question 18

A pharmaceutical company has just developed a new capsule to alleviate the symptoms of allergies and colds.However,they are concerned about the variability in the amount of time until the drug becomes effective.In a random sample of 10 individuals who suffer from allergies,the amount of time (in hours)for the capsule to take effect was recorded and listed as follows: 5,7,6,10,9,12,8,17,4,and 16.Estimate with 90% confidence the variance of the time for the drug to become effective.

Accepted Answer

\[\begin{array} { l } 
\mathrm { LCL } =

Question 19

Energy Drink Consumption
A researcher at the University of Washington medical school believes that energy drink consumption may increase heart rate.Suppose it is known that heart rate (in beats per minute)is normally distributed with an average of 70 bpm for adults.A random sample of 25 adults was selected and it was found that their average heartbeat was 73 bpm after energy drink consumption,with a standard deviation of 7 bpm.
-{Energy Drink Consumption Narrative} Test the hypotheses in the previous question at the 10% significance level to determine if we can infer that energy drink consumption increases heart rate.

Accepted Answer

Rejection region: t > t_0.10,24 = 1.31

Question 20

Based on sample data,the 90% confidence interval limits for the population mean are LCL = 170.86 and UCL = 195.42.If the 10% level of significance were used in testing the hypotheses H0: $\mu$ = 201 vs.H1: $\mu$$\neq$ 201,the null hypothesis:

Accepted Answer

A)  would be rejected. 
B)  would be accepted. 
C)  would fail to be rejected. 
D)  would become H0: $\mu$$\neq$201 
A)  would be rejected. 
B)  would be accepted. 
C)  would fail to be rejected. 
D)  would become H0: $\mu$$\neq$201

Exam 12: Inference About a Population

Which of the following would be an appropriate alternative hypothesis?

The test statistic for p is approximately normal when ____________________ and ____________________ are both greater than 5.

The chi-squared test and estimator of the population variance ____________________(are/are not)valid if the population is slightly to moderately nonnormal.

In forming a 95% confidence interval for a population mean from a sample size of 20,the number of degrees of freedom from the t-distribution equals 20.

A population is defined to be small if it is ____________________ than ____________________ times the size of the sample.

Applicants' Grades The grades of a sample of 10 applicants,selected at random from a large population,are 71,86,75,63,92,70,81,59,80,and 90. -{Applicants' Grades Narrative} Construct a 90% confidence interval estimate for the population standard deviation.

To produce a confidence interval estimator for the total,we multiply the lower and upper confidence limits of the interval estimator of p by ____________________.

The t-distribution allows the calculation of confidence intervals for means for small samples when the population variance is not known,regardless of the shape of the distribution in the population.

Single Mothers' Ages A random sample of 10 single mothers was drawn from a Obstetrics Clinic.Their ages are 22,17,27,20,23,19,24,18,19,and 24 years. -{Single Mothers' Ages Narrative} Estimate the population mean with 90% confidence.

To estimate the total number of successes in a large finite population of size N,using a sample of size n,the confidence interval estimator is:

Concert Tickets A simple random sample of 100 concert tickets was drawn from a normal population.The mean and standard deviation of the sample were $120 and $25,respectively. -{Concert Tickets Narrative} Explain how to use the confidence interval to test the hypotheses at α\alphaα = 0.10.

In selecting the sample size to estimate the population proportion p,if we have no knowledge of even the approximate values of the sample proportion P^\hat { P }P^ ,we:

Marketing segmentation separates consumers of a product into different groups in such a way that members of each group are ____________________ and there are differences between ____________________.

A random sample of size 15 taken from a normally distributed population revealed a sample mean of 75 and a sample variance of 25.The upper limit of a 95% confidence interval for the population mean would equal:

Based on sample data,the 90% confidence interval limits for the population mean are LCL = 170.86 and UCL = 195.42.If the 10% level of significance were used in testing the hypotheses H0: μ\muμ = 201 vs.H1: μ\muμ≠\neq= 201,the null hypothesis:

What Is Statistics

Graphical Descriptive Techniques I

Graphical Descriptive Techniques II

A: Numerical Descriptive Techniques

B: Numerical Descriptive Techniques

C: Numerical Descriptive Techniques

Data Collection and Sampling

Probability

A: Random Variables and Discrete Probability Distributions

B: Random Variables and Discrete Probability Distributions

Continuous Probability Distributions

Sampling Distributions

Introduction to Estimation

Introduction to Hypothesis Testing

Inference About Comparing Two Populations

Analysis of Variance

Chi-Squared Tests

A: Simple Linear Regression and Correlation

B: Simple Linear Regression and Correlation

Multiple Regression

Model Building

Nonparametric Statistics

Time-Series Analysis and Forecasting

Statistical Process Control

Decision Analysis

Conclusion

Filters

The test statistic for p is approximately normal when and are both greater than 5.

A population is defined to be small if it is than times the size of the sample.

Concert Tickets A simple random sample of 100 concert tickets was drawn from a normal population.The mean and standard deviation of the sample were $120 and $25,respectively. -{Concert Tickets Narrative} Explain how to use the confidence interval to test the hypotheses at $\alpha$ = 0.10.

In selecting the sample size to estimate the population proportion p,if we have no knowledge of even the approximate values of the sample proportion $\hat { P }$ ,we:

Marketing segmentation separates consumers of a product into different groups in such a way that members of each group are and there are differences between .

Based on sample data,the 90% confidence interval limits for the population mean are LCL = 170.86 and UCL = 195.42.If the 10% level of significance were used in testing the hypotheses H₀: $\mu$ = 201 vs.H₁: $\mu$ $\neq$ 201,the null hypothesis: