Question 1

TABLE 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:
 
-Referring to Table 9-1, the null hypothesis would be rejected if a 5% probability of committing a Type I error is allowed.

Accepted Answer

According to Table 9-1, the null hypothesis would be rejected if the p-value is less than or equal to 0.10, which corresponds to a level of significance of 0.10 or higher. Therefore, if the manager wants to use a level of significance of 0.10, the null hypothesis would be rejected if the p-value is less than or equal to 0.10, which is equivalent to a 10% probability of committing a Type I error, not 5%.

Question 2

TABLE 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:
 
-Referring to Table 9-1, the lowest level of significance at which the null hypothesis can be rejected is ________.

Accepted Answer

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

Question 3

TABLE 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:
 
-Referring to Table 9-1, if these data were used to perform a two-tail test, the p-value would be 0.042.

Accepted Answer

The p-value is the probability of obtaining a test statistic as extreme as or more extreme than the observed test statistic, assuming the null hypothesis is true. In a two-tail test, the p-value is the sum of the probabilities in both tails of the distribution. In this case, the p-value is 0.042, which is less than the alpha level of 0.05. Therefore, we would reject the null hypothesis and conclude that there is a significant difference between the two groups.

Question 4

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 5

The statement of the null hypothesis always contains an equality.

Accepted Answer

The null hypothesis always assumes that there is no difference or no effect, and therefore includes an equality. For example, "there is no difference between the mean of group A and the mean of group B" or "the treatment has no effect on the outcome variable."

Question 6

TABLE 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:
 
-Referring to Table 9-1, the null hypothesis would be rejected if a 1% probability of committing a Type I error is allowed.

Accepted Answer

The table indicates that the critical value for a one-tailed test with a significance level of 0.10 and 45 degrees of freedom is 1.677. The t-statistic for the sample of 46 cases is -3.01, which falls beyond the critical value, leading to the rejection of the null hypothesis that the mean number of defective bulbs per case is less than or equal to 20. However, the question states that the level of significance is 0.10, which means that the probability of committing a Type I error is 10%, not 1%. Therefore, the null hypothesis would not be rejected at a 1% level of significance, making the statement false.

Question 7

TABLE 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:
 
-Referring to Table 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 statement implies a hypothesis test where the null hypothesis is that the mean number of defective bulbs is 20 or less. If the manager concludes there is sufficient evidence with a 5% significance level, it means the p-value is less than 0.05, supporting the alternative hypothesis that the mean is greater than 20.

Question 8

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 the alternative hypothesis that the mean of the population differs from 20, the null hypothesis could be rejected at a level of significance of 0.10.

Accepted Answer

Since the confidence interval does not include the value of 20, we can reject the null hypothesis that the mean of the population is 20 at a level of significance of 0.10. Therefore, the statement is true.

Question 9

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 the alternative 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 the value of 20, it provides evidence against the null hypothesis that the mean is equal to 20. Therefore, the null hypothesis can be rejected at a level of significance of 0.05.

Question 10

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

Accepted Answer

A small p-value indicates that the observed data is unlikely to occur by chance alone, providing strong evidence against the null hypothesis.

Question 11

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 compares the sample statistic to the hypothesized population parameter to determine the likelihood of obtaining a sample statistic that far away from the hypothesized parameter if the null hypothesis were true.

Question 12

TABLE 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:
 
-Referring to Table 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

In the table, the value of the test statistic is 2.61, which is greater than the critical value of 1.28 (from the one-tailed t-distribution table with 45 degrees of freedom at a significance level of 0.10). Therefore, we can reject the null hypothesis and conclude that the mean number of defective bulbs per case is greater than 20 during the morning shift at a significance level of 0.10.

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 (0.034) is greater than the significance level (0.01), indicating insufficient evidence to reject the null hypothesis at this level.

Question 14

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

If the p-value is less than or equal to the chosen level of significance (0.05 in this case), then the null hypothesis should be rejected. Since the p-value of 0.043 is less than 0.05, the null hypothesis should be rejected.

Question 15

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

Accepted Answer

The correct statement is actually the opposite: the larger the p-value, the less likely you are to reject the null hypothesis. A large p-value indicates that there is a high probability of obtaining the observed results by chance and therefore does not provide strong evidence against the null hypothesis.

Question 16

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 the alternative hypothesis that the mean of the population differs from 20, the null hypothesis could be rejected at a level of significance of 0.02.

Accepted Answer

The confidence interval does not contain the value of 20, indicating that the null hypothesis of the mean being equal to 20 is not supported. However, the level of significance at which the null hypothesis can be rejected cannot be determined from the given information.

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 the alternative 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

TABLE 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:
 
-Referring to Table 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 answer of TABLE 9-1
Microsoft Excel was used on a...

Question 19

TABLE 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:
 
-Referring to Table 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 output shows that the sample mean is 19.9 and the calculated p-value is 0.074, which is greater than the chosen level of significance of 0.10. Therefore, there is not enough evidence to conclude that the mean number of defective bulbs per case is greater than 20 during the morning shift.

Question 20

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

Accepted Answer

The decision to conduct a one-tail or two-tail test depends on the directionality of the hypothesis being tested. A one-tail test is appropriate when the hypothesis makes a specific directional prediction, while a two-tail test is appropriate when the hypothesis does not specify a direction. Therefore, it is important to correctly identify whether a one-tail or two-tail test should be conducted based on the research question and hypothesis being tested.

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