NARRBEGIN: Socialist Voters Socialist Party Voters A pollster in Italy wants to challenge a claim that 5% of the registered voters in his country are Socialists; he thinks the percentage is lower than that. In a test of hypothesis, HS1U1B10S1U1B0: p = 0.05 vs. HS1U1B11S1U1B0: p < 0.05, his random sample of size 1,000 registered voters revealed that the number of Socialists was 40.NARREND -{Socialist Voters Narrative} Construct a 95% confidence interval estimate of the population proportion and explain how to use it to test the hypotheses.

@#LAT-DLM& . Thus, LCL = 0.028, and UCL

Exam 12: Inference About a Population

NARRBEGIN: Socialist Voters Socialist Party Voters A pollster in Italy wants to challenge a claim that 5% of the registered voters in his country are Socialists; he thinks the percentage is lower than that. In a test of hypothesis, H₀: p = 0.05 vs. H₁: p < 0.05, his random sample of size 1,000 registered voters revealed that the number of Socialists was 40.NARREND -{Socialist Voters Narrative} Construct a 95% confidence interval estimate of the population proportion and explain how to use it to test the hypotheses.

(Essay)

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Question 61

When a population is small, it is necessary to include the ____________________ factor in our hypothesis tests and confidence interval estimators for p.

(Short Answer)

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Question 62

A race car driver tested his car for time from 0 to 60 mph, and in 20 tests obtained an average of 48.5 seconds with a standard deviation of 1.47 seconds. A 95% confidence interval for the 0 to 60 time is 45.2 seconds to 51.8 seconds.

(True/False)

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Question 63

Target marketing focuses on satisfying the demands of all segments of the entire market.

(True/False)

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Question 64

The ____________________ confidence limit of the confidence interval estimator of the population variance is $\frac { ( n - 1 ) s ^ { 2 } } { \chi _ { \alpha / 2 } ^ { 2 } }$ .

(Short Answer)

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Question 65

A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90, a random sample of 100 doctors' results in 83 who indicate that they recommend aspirin. The value of the test statistic in this problem is approximately equal to:

(Multiple Choice)

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Question 66

NARRBEGIN: Energy Drink Consumption 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.NARREND -{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.

(Essay)

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Question 67

NARRBEGIN: Mystic Pizza Mystic Pizza Mystic Pizza in Mystic, Connecticut, advertises that they deliver your pizza within 15 minutes of placing an order or it is free. A sample of 25 customers is selected at random. The average delivery time in the sample was 13 minutes with a sample standard deviation of 4 minutes.NARREND -{Mystic Pizza Narrative} We want to know whether Mystic can make this claim or not. Test to determine if we can infer at the 5% significance level that the population mean delivery time is less than 15 minutes.

(Essay)

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Question 68

The use of the standard normal distribution for constructing confidence interval estimate for the population proportion p requires:

(Multiple Choice)

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Question 69

When the population standard deviation is ____________________ and the population is normal, the test statistic for testing hypotheses about $\mu$ is the t-distribution with n -1 degrees of freedom.

(Short Answer)

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Question 70

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 .

(Short Answer)

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Question 71

NARRBEGIN: Socialist Voters Socialist Party Voters A pollster in Italy wants to challenge a claim that 5% of the registered voters in his country are Socialists; he thinks the percentage is lower than that. In a test of hypothesis, H₀: p = 0.05 vs. H₁: p < 0.05, his random sample of size 1,000 registered voters revealed that the number of Socialists was 40.NARREND -{Socialist Voters Narrative} Test the hypotheses at the 5% significance level.

(Essay)

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Question 72

The t-distribution is used in a confidence interval for a mean when the actual standard error is not known.

(True/False)

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Question 73

A professor of statistics refutes the claim that the proportion of Republican voters in Michigan is at most 44%. To test the claim, the hypotheses: H₀: p = 0.44, H₁: p $\neq$ 0.44, should be used.

(True/False)

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Question 74

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

(Short Answer)

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Question 75

The test statistic used to test hypotheses about the population variance is given by $\chi$ ² = (n -1)s² / $\sigma$ ², which is chi-squared distributed with n -1 degrees of freedom when the population is normally distributed with variance equal to $\sigma$ ².

(True/False)

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Question 76

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:

(Multiple Choice)

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Question 77

NARRBEGIN: Sinus Drug Sinus Drug A company claims that 10% of the users of a certain sinus drug experience drowsiness. In clinical studies of this sinus drug, 81 of the 900 subjects experienced drowsiness NARREND -{Sinus Drug Narrative} Compute the p-value of the test.

(Short Answer)

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Question 78

If the lower and upper confidence limits of the population proportion p, using a sample of size 1500, are 0.184 and 0.238, respectively, then the lower and upper confidence limits of the total number of successes in the population, given that the population size is 750,000, are respectively:

(Multiple Choice)

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Question 79

The degrees of freedom for the test statistic for $\mu$ when $\sigma$ is unknown is:

(Multiple Choice)

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Question 80

Showing 61 - 80 of 149

When a population is small, it is necessary to include the ____________________ factor in our hypothesis tests and confidence interval estimators for p.

A race car driver tested his car for time from 0 to 60 mph, and in 20 tests obtained an average of 48.5 seconds with a standard deviation of 1.47 seconds. A 95% confidence interval for the 0 to 60 time is 45.2 seconds to 51.8 seconds.

Target marketing focuses on satisfying the demands of all segments of the entire market.

The ____________________ confidence limit of the confidence interval estimator of the population variance is $\frac { ( n - 1 ) s ^ { 2 } } { \chi _ { \alpha / 2 } ^ { 2 } }$ .

The use of the standard normal distribution for constructing confidence interval estimate for the population proportion p requires:

When the population standard deviation is ____________________ and the population is normal, the test statistic for testing hypotheses about $\mu$ is the t-distribution with n -1 degrees of freedom.

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 .

The t-distribution is used in a confidence interval for a mean when the actual standard error is not known.

A professor of statistics refutes the claim that the proportion of Republican voters in Michigan is at most 44%. To test the claim, the hypotheses: H₀: p = 0.44, H₁: p $\neq$ 0.44, should be used.

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

The test statistic used to test hypotheses about the population variance is given by $\chi$ ² = (n -1)s² / $\sigma$ ², which is chi-squared distributed with n -1 degrees of freedom when the population is normally distributed with variance equal to $\sigma$ ².

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:

NARRBEGIN: Sinus Drug Sinus Drug A company claims that 10% of the users of a certain sinus drug experience drowsiness. In clinical studies of this sinus drug, 81 of the 900 subjects experienced drowsiness NARREND -{Sinus Drug Narrative} Compute the p-value of the test.

If the lower and upper confidence limits of the population proportion p, using a sample of size 1500, are 0.184 and 0.238, respectively, then the lower and upper confidence limits of the total number of successes in the population, given that the population size is 750,000, are respectively:

The degrees of freedom for the test statistic for $\mu$ when $\sigma$ is unknown is:

What Is Statistics

Graphical Descriptive Techniques I

Graphical Descriptive Techniques II

Numerical Descriptive Techniques

Data Collection and Sampling

Probability

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 Optional

Simple Linear Regression and Correlation

Multiple Regression

Filters

Exam 12: Inference About a Population

When a population is small, it is necessary to include the ____________________ factor in our hypothesis tests and confidence interval estimators for p.

A race car driver tested his car for time from 0 to 60 mph, and in 20 tests obtained an average of 48.5 seconds with a standard deviation of 1.47 seconds. A 95% confidence interval for the 0 to 60 time is 45.2 seconds to 51.8 seconds.

Target marketing focuses on satisfying the demands of all segments of the entire market.

The ____________________ confidence limit of the confidence interval estimator of the population variance is (n−1)s2χα/22\frac { ( n - 1 ) s ^ { 2 } } { \chi _ { \alpha / 2 } ^ { 2 } }χα/22​(n−1)s2​ .

The use of the standard normal distribution for constructing confidence interval estimate for the population proportion p requires:

When the population standard deviation is ____________________ and the population is normal, the test statistic for testing hypotheses about μ\muμ is the t-distribution with n -1 degrees of freedom.

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 ____________________.

The t-distribution is used in a confidence interval for a mean when the actual standard error is not known.

A professor of statistics refutes the claim that the proportion of Republican voters in Michigan is at most 44%. To test the claim, the hypotheses: H0: p = 0.44, H1: p ≠\neq= 0.44, should be used.

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

The test statistic used to test hypotheses about the population variance is given by χ\chiχ 2 = (n -1)s2 / σ\sigmaσ 2, which is chi-squared distributed with n -1 degrees of freedom when the population is normally distributed with variance equal to σ\sigmaσ 2.

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:

NARRBEGIN: Sinus Drug Sinus Drug A company claims that 10% of the users of a certain sinus drug experience drowsiness. In clinical studies of this sinus drug, 81 of the 900 subjects experienced drowsiness NARREND -{Sinus Drug Narrative} Compute the p-value of the test.

If the lower and upper confidence limits of the population proportion p, using a sample of size 1500, are 0.184 and 0.238, respectively, then the lower and upper confidence limits of the total number of successes in the population, given that the population size is 750,000, are respectively:

The degrees of freedom for the test statistic for μ\muμ when σ\sigmaσ is unknown is:

What Is Statistics

Graphical Descriptive Techniques I

Graphical Descriptive Techniques II

Numerical Descriptive Techniques

Data Collection and Sampling

Probability

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 Optional

Simple Linear Regression and Correlation

Multiple Regression

Filters

The ____________________ confidence limit of the confidence interval estimator of the population variance is $\frac { ( n - 1 ) s ^ { 2 } } { \chi _ { \alpha / 2 } ^ { 2 } }$ .

When the population standard deviation is ____________________ and the population is normal, the test statistic for testing hypotheses about $\mu$ is the t-distribution with n -1 degrees of freedom.

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 professor of statistics refutes the claim that the proportion of Republican voters in Michigan is at most 44%. To test the claim, the hypotheses: H₀: p = 0.44, H₁: p $\neq$ 0.44, should be used.

The test statistic used to test hypotheses about the population variance is given by $\chi$ ² = (n -1)s² / $\sigma$ ², which is chi-squared distributed with n -1 degrees of freedom when the population is normally distributed with variance equal to $\sigma$ ².

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:

The degrees of freedom for the test statistic for $\mu$ when $\sigma$ is unknown is: