Question 1

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: $H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T$ 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

The level of significance given in the scenario is 0.10, not 1%. Therefore, the null hypothesis would be rejected if the probability of committing a Type I error is less than or equal to 0.10, not 0.01.

Question 2

The larger the p-value, the more likely you are to reject the null hypothesis.

Accepted Answer

This statement is incorrect. A larger p-value indicates weaker evidence against the null hypothesis, which means that we are less likely to reject the null hypothesis. We typically reject the null hypothesis when the p-value is small (e.g. less than 0.05), indicating strong evidence against the null hypothesis.

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: $H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T$ 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 p-value (0.0086) is less than the level of significance (0.10), indicating strong evidence against the null hypothesis. Therefore, the null hypothesis would be rejected and the manager can conclude that the mean number of defective bulbs per case is indeed greater than 20 during the morning shift.

Question 4

The smaller the p-value, the stronger is the evidence against the null hypothesis.

Accepted Answer

A p-value represents the probability of obtaining a test statistic at least as extreme as the one observed, assuming that the null hypothesis is true. A small p-value indicates strong evidence against the null hypothesis, as it suggests that the observed result is highly unlikely to occur by chance alone. Therefore, the smaller the p-value, the stronger is the evidence against the null hypothesis.

Question 5

The statement of the null hypothesis always contains an equality.

Accepted Answer

The null hypothesis is a statement that there is no significant difference between certain variables, and it always includes an equality, such as "no difference" or "equal to". The alternative hypothesis, on the other hand, may not contain an equality and can state a direction of difference (e.g. greater than or less than).

Question 6

A sample is used to obtain a 95% confidence interval for the mean of a population.The confidence interval goes from 15 to 19.If the same sample had been used to test the null hypothesis that the mean of the population is equal to 20 versus thealternative hypothesis that the mean of the population differs from 20, the null hypothesis could be rejected at a level of significance of 0.05.

Accepted Answer

Since the confidence interval does not include 20, it is reasonable to reject the null hypothesis that the population mean is equal to 20 at a level of significance of 0.05. Therefore, the statement is true.

Question 7

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/or the direction of the alternative hypothesis. A one-tail test is used when the alternative hypothesis specifies a direction of effect (e.g., the mean is greater than a certain value), while a two-tail test is used when the alternative hypothesis does not specify a particular direction (e.g., the mean is different from a certain value). Choosing the wrong type of test can lead to incorrect conclusions.

Question 8

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 is0.01.

Accepted Answer

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

Question 9

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: $H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T$ 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

The information provided in the scenario is not sufficient to determine with certainty whether the mean number of defective bulbs per case is greater than 20. While the output from Microsoft Excel may provide some indications, more information, such as the sample mean and standard deviation, as well as the calculated test statistic and p-value, are needed to make a conclusion.

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: $H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T$ 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 a...

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: $H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T$ 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

The Microsoft Excel output shows a t-value of -2.15 and a p-value of 0.018. Since the p-value is less than the level of significance of 0.10, the null hypothesis can be rejected. However, the statement about the probability of incorrectly rejecting the true null hypothesis is incorrect - it should be "no more than a 10% probability" since the level of significance is 0.10.

Question 12

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 would be approximately ±2.58. Since 2.04 does not exceed 2.58, the null hypothesis should not be rejected based on this level of significance.

Question 13

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: $H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T$ 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

The p-value in the Excel output is 0.027, which is less than the level of significance of 0.10. Therefore, the null hypothesis can be rejected and 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.

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: $H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T$ 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

In Scenario 9-1, if a 5% probability of committing a Type I error is allowed, it means that the significance level (alpha) is set at 0.05. If the p-value obtained from the statistical test is less than or equal to 0.05, the null hypothesis would be rejected, indicating that the results are statistically significant at the 5% level.

Question 15

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

Accepted Answer

This is true. The test statistic is calculated by comparing the sample statistic (such as the sample mean or proportion) to the hypothesized population parameter (typically stated in the null hypothesis) and measuring how far apart the two are. It tells us how likely it is that the difference between the sample statistic and the population parameter is due to chance or random sampling error.

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: $H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T$ 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 output shows that the sample mean of defective bulbs per case is 21.02 and the p-value for a one-tailed test with null hypothesis of mean=20 is 0.042, which is less than the level of significance of 0.10. Therefore, the null hypothesis can be rejected and it can be concluded that there is sufficient evidence to show that the mean number of defective bulbs per case is greater than 20 during the morning shift.

Question 17

A sample is used to obtain a 95% confidence interval for the mean of a population.The confidence interval goes from 15 to 19.If the same sample had been used to test the null hypothesis that the mean of the population is equal to 18 versus thealternative hypothesis that the mean of the population differs from 18, the null hypothesis could be rejected at a level of significance of 0.05.

Accepted Answer

Since the 95% confidence interval for the mean includes 18, it means that at a 0.05 level of significance, there is not enough evidence to reject the null hypothesis that the mean of the population is 18.

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: $H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T$ 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-tailed test, the p-value represents the probability of observing a test statistic as extreme as, or more extreme than, the observed value under the null hypothesis. A p-value of 0.042 indicates that there is a 4.2% chance of observing such extreme results if the null hypothesis were true, which is a direct outcome of performing the test with the given data.

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: $H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T$ 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