Question 1

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 95% confidence interval from 15 to 19 does not include the value 20, which suggests that the null hypothesis (that the mean is equal to 20) would be rejected at the 0.05 level of significance. However, without more information, we cannot precisely determine if it would be rejected at the 0.02 level of significance, as the exact p-value depends on the sample size, variability, and the test used.

Question 2

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 3

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 95% confidence interval for the population mean does not include the value 20, we can reject the null hypothesis that the population mean is equal to 20 at the 0.05 significance level.

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

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:
 $$\begin{array} { l } 
n = 46 ; \text { Arithmetic Mean } = 28.00 ; \text { Standard Deviation } = 25.92 ; \text { Standard Error } = 3.82 \text {; } \
\text { Null Hypothesis: } H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T \text { Test Statistic } = 2.09 \text {; } \
\text { One-Tail Test Upper Critical Value } = 1.3006 ; p \text {-value } = 0.021 ; \text { Decision } = \text { Reject. }
\end{array}$$
-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 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:
 $$\begin{array} { l } 
n = 46 ; \text { Arithmetic Mean } = 28.00 ; \text { Standard Deviation } = 25.92 ; \text { Standard Error } = 3.82 \text {; } \
\text { Null Hypothesis: } H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T \text { Test Statistic } = 2.09 \text {; } \
\text { One-Tail Test Upper Critical Value } = 1.3006 ; p \text {-value } = 0.021 ; \text { Decision } = \text { Reject. }
\end{array}$$
-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

The given information shows that the null hypothesis is rejected and the p-value is less than the level of significance. Therefore, 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.

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:
 $$\begin{array} { l } 
n = 46 ; \text { Arithmetic Mean } = 28.00 ; \text { Standard Deviation } = 25.92 ; \text { Standard Error } = 3.82 \text {; } \
\text { Null Hypothesis: } H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T \text { Test Statistic } = 2.09 \text {; } \
\text { One-Tail Test Upper Critical Value } = 1.3006 ; p \text {-value } = 0.021 ; \text { Decision } = \text { Reject. }
\end{array}$$
-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 correctly interprets the results of the test and reflects the proper use of a level of significance of 0.10, a one-tailed test, and the decision to reject the null hypothesis based on the calculated test statistic and p-value.

Question 8

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

Accepted Answer

A p-value measures the strength of evidence against the null hypothesis. A smaller p-value indicates stronger evidence against the null hypothesis, in favor of the alternative hypothesis.

Question 9

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 p-value (0.043) is less than the chosen level of significance (0.05), the null hypothesis should be rejected.

Question 10

The statement of the null hypothesis always contains an equality.

Accepted Answer

This is true. The null hypothesis is a statement of no difference or no relationship between variables, and it is typically expressed as an equality (e.g. "there is no significant difference between group A and group B" or "there is no relationship between X and Y").

Question 11

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:
 $$\begin{array} { l } 
n = 46 ; \text { Arithmetic Mean } = 28.00 ; \text { Standard Deviation } = 25.92 ; \text { Standard Error } = 3.82 \text {; } \
\text { Null Hypothesis: } H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T \text { Test Statistic } = 2.09 \text {; } \
\text { One-Tail Test Upper Critical Value } = 1.3006 ; p \text {-value } = 0.021 ; \text { Decision } = \text { Reject. }
\end{array}$$
-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 12

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

Accepted Answer

It is relevant whether the test is a one-tail or two-tail test because each type of test has a different method for calculating the probability of getting the observed sample results under the null hypothesis. Knowing whether it is a one-tail or two-tail test affects the critical value or p-value used to make the decision about rejecting or failing to reject the null hypothesis.

Question 13

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 14

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 is calculated using the sample statistic and the hypothesized population parameter, and it represents how many standard errors the sample statistic is away from the population parameter. In other words, it measures the degree of agreement between the sample data and the null hypothesis.

Question 15

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:
 $$\begin{array} { l } 
n = 46 ; \text { Arithmetic Mean } = 28.00 ; \text { Standard Deviation } = 25.92 ; \text { Standard Error } = 3.82 \text {; } \
\text { Null Hypothesis: } H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T \text { Test Statistic } = 2.09 \text {; } \
\text { One-Tail Test Upper Critical Value } = 1.3006 ; p \text {-value } = 0.021 ; \text { Decision } = \text { Reject. }
\end{array}$$
-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 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 0.10.

Accepted Answer

Since the 95% confidence interval for the population mean does not include the value 20, it implies that a hypothesis test at a significance level of 0.10 (which is less stringent than the 0.05 level typically associated with a 95% confidence interval) would reject the null hypothesis that the population mean is 20.

Question 17

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 for a two-tail test with a test statistic of 2.05 and 28 degrees of freedom is 0.042.

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:
 $$\begin{array} { l } 
n = 46 ; \text { Arithmetic Mean } = 28.00 ; \text { Standard Deviation } = 25.92 ; \text { Standard Error } = 3.82 \text {; } \
\text { Null Hypothesis: } H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T \text { Test Statistic } = 2.09 \text {; } \
\text { One-Tail Test Upper Critical Value } = 1.3006 ; p \text {-value } = 0.021 ; \text { Decision } = \text { Reject. }
\end{array}$$
-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

The one-tail test upper critical value (with a level of significance of 0.10) is given as 1.3006, which corresponds to a probability of 0.05. Therefore, if a 5% probability of committing a Type I error is allowed, the critical value for the test would be even lower than 1.3006, making it easier to reject the null hypothesis. Since the calculated test statistic of 2.09 is greater than the critical value of 1.3006, the null hypothesis is rejected at a level of significance of 0.10 and would also be rejected at a level of significance of 0.05.

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:
 $$\begin{array} { l } 
n = 46 ; \text { Arithmetic Mean } = 28.00 ; \text { Standard Deviation } = 25.92 ; \text { Standard Error } = 3.82 \text {; } \
\text { Null Hypothesis: } H _ { 0 } : \mu \leq 20 ; \alpha = 0.10 ; \mathrm { df } = 45 ; T \text { Test Statistic } = 2.09 \text {; } \
\text { One-Tail Test Upper Critical Value } = 1.3006 ; p \text {-value } = 0.021 ; \text { Decision } = \text { Reject. }
\end{array}$$
-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 20

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

Accepted Answer

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

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