Question 1

The marketing manager for an automobile manufacturer is interested in determining the proportion of new compact-car owners who would have purchased a GPS navigation system if it had been available for an additional cost of $300.The manager believes from previous information that the proportion is 0.30.Suppose that a survey of 200 new compact-car owners is selected and 79 indicate that they would have purchased the GPS navigation system.If you were to conduct a test to determine whether there is evidence that the proportion is different from 0.30 and decided not to reject the null hypothesis, what conclusion could you reach?
A)There is sufficient evidence that the proportion is 0.30
B)There is not sufficient evidence that the proportion is 0.30.
C)There is sufficient evidence that the proportion is 0.30.
D)There is not sufficient evidence that the proportion is not 0.30.

Accepted Answer

Choosing not to reject the null hypothesis implies that there is not enough evidence to support a difference from the hypothesized proportion of 0.30, meaning there is not sufficient evidence to conclude that the actual proportion differs from 0.30.

Question 2

Suppose, in testing a hypothesis about a mean, the Z test statistic is computed to be 2.04.The null hypothesis should be rejected if the chosen level of significance is 0.01 and a two-tail test is used.

Accepted Answer

With a two-tail test at a significance level of 0.01, the critical Z values are approximately ±2.58. Since the computed Z test statistic of 2.04 is less than 2.58, the null hypothesis should not be rejected at this level of significance.

Question 3

SCENARIO 9-1 Microsoft Excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant.A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift.She will make her decision using a test with a level of significance of 0.10.The following information was extracted from the Microsoft Excel output for the sample of 46 cases: n = 46; Arithmetic Mean = 28.00; Standard Deviation = 25.92; Standard Error = 3.82; Null Hypothesis:  T Test Statistic = 2.09; One-Tail Test Upper Critical Value = 1.3006; p-value = 0.021; Decision = Reject.
-Referring to Scenario 9-1, the null hypothesis would be rejected if a 1% probability of committing a Type I error is allowed.

Accepted Answer

Without specific details from "Scenario 9-1," it's impossible to determine whether the null hypothesis would be rejected at a 1% significance level. Rejecting the null hypothesis depends on the p-value obtained from the statistical test in comparison to the significance level (alpha), not just the alpha level itself.

Question 4

SCENARIO 9-1 Microsoft Excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant.A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift.She will make her decision using a test with a level of significance of 0.10.The following information was extracted from the Microsoft Excel output for the sample of 46 cases: n = 46; Arithmetic Mean = 28.00; Standard Deviation = 25.92; Standard Error = 3.82; Null Hypothesis:  T Test Statistic = 2.09; One-Tail Test Upper Critical Value = 1.3006; p-value = 0.021; Decision = Reject.
-Referring to Scenario 9-1, the lowest level of significance at which the null hypothesis can be rejected is ______.

Accepted Answer

The answer of SCENARIO 9-1 Microsoft Excel was used on...

Question 5

A pizza chain is considering opening a new store in an area that currently does not have any such stores.The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area have a favorable view of its brain.It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have a favorable view.The pizza chain's conclusion from the hypothesis test using a 5% level of significance is:
A)to open a new store.
B)not to open a new store.
C)to delay opening a new store until additional evidence is collected.
D)we cannot tell what the decision should be from the information given.

Accepted Answer

The answer of A pizza chain is considering opening a...

Question 6

SCENARIO 9-1 Microsoft Excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant.A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift.She will make her decision using a test with a level of significance of 0.10.The following information was extracted from the Microsoft Excel output for the sample of 46 cases: n = 46; Arithmetic Mean = 28.00; Standard Deviation = 25.92; Standard Error = 3.82; Null Hypothesis:  T Test Statistic = 2.09; One-Tail Test Upper Critical Value = 1.3006; p-value = 0.021; Decision = Reject.
-Referring to Scenario 9-1, the evidence proves beyond a doubt that the mean number of defective bulbs per case is greater than 20 during the morning shift.

Accepted Answer

Without the specific details from "Scenario 9-1," it's impossible to determine the accuracy of the statement regarding the mean number of defective bulbs per case.

Question 7

SCENARIO 9-1 Microsoft Excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant.A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift.She will make her decision using a test with a level of significance of 0.10.The following information was extracted from the Microsoft Excel output for the sample of 46 cases: n = 46; Arithmetic Mean = 28.00; Standard Deviation = 25.92; Standard Error = 3.82; Null Hypothesis:  T Test Statistic = 2.09; One-Tail Test Upper Critical Value = 1.3006; p-value = 0.021; Decision = Reject.
-Referring to Scenario 9-1, if these data were used to perform a two-tail test, the p-value would be 0.042.

Accepted Answer

In a two-tail test, the p-value reported directly corresponds to the probability of observing a test statistic as extreme as the one calculated, in either direction, under the null hypothesis. If the scenario specifies a p-value of 0.042, it means that the probability of obtaining a result as extreme or more extreme than the observed, assuming the null hypothesis is true, is 4.2%, which is a valid outcome for a two-tail test.

Question 8

The marketing manager for an automobile manufacturer is interested in determining the proportion of new compact-car owners who would have purchased a GPS navigation system if it had been available for an additional cost of $300.The manager believes from previous information that the proportion is 0.30.Suppose that a survey of 200 new compact-car owners is selected and 79 indicate that they would have purchased the GPS navigation system.If you were to conduct a test to determine whether there is evidence that the proportion is different from 0.30, which test would you use?
A)Z-test of a population mean
B)Z-test of a population proportion
C)t-test of population mean
D)t-test of a population proportion

Accepted Answer

We would use a z-test of a population proportion because we are testing a proportion (proportion of new compact-car owners who would have purchased GPS navigation system) and we have a large sample size (n=200) which satisfies the conditions of a normal distribution.

Question 9

The test statistic measures how close the computed sample statistic has come to the hypothesized population parameter.

Accepted Answer

This statement is true. The test statistic measures the difference between the sample statistic and the hypothesized population parameter, taking into account the variability of the sample statistic. The closer the sample statistic is to the hypothesized population parameter, the smaller the test statistic will be, indicating stronger evidence in favor of the null hypothesis.

Question 10

SCENARIO 9-1 Microsoft Excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant.A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift.She will make her decision using a test with a level of significance of 0.10.The following information was extracted from the Microsoft Excel output for the sample of 46 cases: n = 46; Arithmetic Mean = 28.00; Standard Deviation = 25.92; Standard Error = 3.82; Null Hypothesis:  T Test Statistic = 2.09; One-Tail Test Upper Critical Value = 1.3006; p-value = 0.021; Decision = Reject.
-Referring to Scenario 9-1, the manager can conclude that there is sufficient evidence to show that the mean number of defective bulbs per case is greater than 20 during the morning shift with no more than a 1% probability of incorrectly rejecting the true null hypothesis.

Accepted Answer

Without the specific details of Scenario 9-1, such as statistical test results or p-values, it's impossible to accurately conclude the sufficiency of evidence regarding the mean number of defective bulbs per case or the probability of incorrectly rejecting the null hypothesis.

Question 11

SCENARIO 9-1 Microsoft Excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant.A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift.She will make her decision using a test with a level of significance of 0.10.The following information was extracted from the Microsoft Excel output for the sample of 46 cases: n = 46; Arithmetic Mean = 28.00; Standard Deviation = 25.92; Standard Error = 3.82; Null Hypothesis:  T Test Statistic = 2.09; One-Tail Test Upper Critical Value = 1.3006; p-value = 0.021; Decision = Reject.
-Referring to Scenario 9-1, the manager can conclude that there is sufficient evidence to show that the mean number of defective bulbs per case is greater than 20 during the morning shift using a level of significance of 0.10.

Accepted Answer

The decision to reject the null hypothesis with a p-value of 0.021 indicates that the sample mean of 28.00 is significantly greater than the hypothesized mean of 20. Therefore, there is sufficient evidence to conclude that the mean number of defective bulbs per case is greater than 20 during the morning shift, at a level of significance of 0.10.

Question 12

SCENARIO 9-1 Microsoft Excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant.A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift.She will make her decision using a test with a level of significance of 0.10.The following information was extracted from the Microsoft Excel output for the sample of 46 cases: n = 46; Arithmetic Mean = 28.00; Standard Deviation = 25.92; Standard Error = 3.82; Null Hypothesis:  T Test Statistic = 2.09; One-Tail Test Upper Critical Value = 1.3006; p-value = 0.021; Decision = Reject.
-Referring to Scenario 9-1, what critical value should the manager use to determine the rejection region?
A)1.6794
B)1.3011
C)1.3006
D)0.6800

Accepted Answer

The manager has chosen a one-tailed test with a level of significance of 0.10. Looking at a t-distribution table with 45 degrees of freedom and a level of significance of 0.10, the upper critical value is approximately 1.3006. Therefore, the critical value for the rejection region is 1.3006.

Question 13

Suppose, in testing a hypothesis about a mean, the p-value is computed to be 0.034.The null hypothesis should be rejected if the chosen level of significance is 0.01.

Accepted Answer

The null hypothesis should not be rejected at the 0.01 level of significance because the p-value of 0.034 is greater than 0.01, indicating insufficient evidence to reject the null hypothesis at this level.

Question 14

SCENARIO 9-1 Microsoft Excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant.A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift.She will make her decision using a test with a level of significance of 0.10.The following information was extracted from the Microsoft Excel output for the sample of 46 cases: n = 46; Arithmetic Mean = 28.00; Standard Deviation = 25.92; Standard Error = 3.82; Null Hypothesis:  T Test Statistic = 2.09; One-Tail Test Upper Critical Value = 1.3006; p-value = 0.021; Decision = Reject.
-Referring to Scenario 9-1, state the alternative hypothesis for this study.

Accepted Answer

The answer of SCENARIO 9-1 Microsoft Excel was used on...

Question 15

Suppose, in testing a hypothesis about a mean, the p-value is computed to be 0.043.The null hypothesis should be rejected if the chosen level of significance is 0.05.

Accepted Answer

Since the obtained p-value (0.043) is less than the chosen level of significance (0.05), the null hypothesis should be rejected. The p-value (0.043) is less than the chosen level of significance (0.05), indicating that the results are statistically significant, and therefore the null hypothesis should be rejected.

Question 16

SCENARIO 9-1 Microsoft Excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant.A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift.She will make her decision using a test with a level of significance of 0.10.The following information was extracted from the Microsoft Excel output for the sample of 46 cases: n = 46; Arithmetic Mean = 28.00; Standard Deviation = 25.92; Standard Error = 3.82; Null Hypothesis:  T Test Statistic = 2.09; One-Tail Test Upper Critical Value = 1.3006; p-value = 0.021; Decision = Reject.
-Referring to Scenario 9-1, the parameter the manager is interested in is:
A)the mean number of defective light bulbs per case produced at the plant.
B)the mean number of defective light bulbs per case among the 46 cases.
C)the mean number of defective light bulbs per case produced during the morning shift.
D)the proportion of cases with defective light bulbs produced at the plant.

Accepted Answer

The answer of SCENARIO 9-1 Microsoft Excel was used on...

Question 17

In a hypothesis test, it is irrelevant whether the test is a one-tail or two-tail test.

Accepted Answer

The choice between a one-tail or two-tail test depends on the research question and the directionality of the hypothesis being tested. A one-tail test is appropriate when the hypothesis specifies a direction for the effect (e.g. "the new drug will increase response times"). A two-tail test is appropriate when the hypothesis does not specify a direction (e.g. "the new drug will have an effect on response times"). Choosing the wrong type of test can lead to incorrect conclusions.

Question 18

SCENARIO 9-1 Microsoft Excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant.A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift.She will make her decision using a test with a level of significance of 0.10.The following information was extracted from the Microsoft Excel output for the sample of 46 cases: n = 46; Arithmetic Mean = 28.00; Standard Deviation = 25.92; Standard Error = 3.82; Null Hypothesis:  T Test Statistic = 2.09; One-Tail Test Upper Critical Value = 1.3006; p-value = 0.021; Decision = Reject.
-Referring to Scenario 9-1, the manager can conclude that there is sufficient evidence to show that the mean number of defective bulbs per case is greater than 20 during the morning shift with no more than a 5% probability of incorrectly rejecting the true null hypothesis.

Accepted Answer

Since the decision was to reject the null hypothesis, it means that the p-value was less than the level of significance of 0.10. Therefore, there is sufficient evidence to show that the mean number of defective bulbs per case is greater than 20 during the morning shift with no more than a 10% probability of incorrectly rejecting the true null hypothesis (not 5%, as stated in the question).

Question 19

SCENARIO 9-1 Microsoft Excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant.A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift.She will make her decision using a test with a level of significance of 0.10.The following information was extracted from the Microsoft Excel output for the sample of 46 cases: n = 46; Arithmetic Mean = 28.00; Standard Deviation = 25.92; Standard Error = 3.82; Null Hypothesis:  T Test Statistic = 2.09; One-Tail Test Upper Critical Value = 1.3006; p-value = 0.021; Decision = Reject.
-Referring to Scenario 9-1, the null hypothesis would be rejected.

Accepted Answer

The last sentence of the scenario states that the decision is to reject the null hypothesis.

Question 20

SCENARIO 9-1 Microsoft Excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant.A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift.She will make her decision using a test with a level of significance of 0.10.The following information was extracted from the Microsoft Excel output for the sample of 46 cases: n = 46; Arithmetic Mean = 28.00; Standard Deviation = 25.92; Standard Error = 3.82; Null Hypothesis:  T Test Statistic = 2.09; One-Tail Test Upper Critical Value = 1.3006; p-value = 0.021; Decision = Reject.
-Referring to Scenario 9-1, the null hypothesis would be rejected if a 5% probability of committing a Type I error is allowed.

Accepted Answer

The level of significance given in the scenario is 0.10, not 0.05. Therefore, the null hypothesis would be rejected at a 10% significance level, but we cannot determine if it would be rejected at a 5% level without additional information.

Quiz 9: Fundamentals of Hypothesis Testing: One-Sample Tests

Suppose, in testing a hypothesis about a mean, the Z test statistic is computed to be 2.04.The null hypothesis should be rejected if the chosen level of significance is 0.01 and a two-tail test is used.

The test statistic measures how close the computed sample statistic has come to the hypothesized population parameter.

Suppose, in testing a hypothesis about a mean, the p-value is computed to be 0.034.The null hypothesis should be rejected if the chosen level of significance is 0.01.

Suppose, in testing a hypothesis about a mean, the p-value is computed to be 0.043.The null hypothesis should be rejected if the chosen level of significance is 0.05.

In a hypothesis test, it is irrelevant whether the test is a one-tail or two-tail test.