Question 1

$\text { TABLE 9-1 }$
Microsoft Excel was used on a set of data involving the number of parasites found on 46 Monarch butterflies captured in Pismo Beach State Park. A biologist wantsto know if the mean number of parasites per butterfly is over 20 . She will make her decision usinga test witha level of significance of $0.10$. The following informationwas extracted from the Microsoft Excel outputfor the sample of 46 Monarchbutterflies:

$\begin{array}{llcc} \hline n=46 ; \text { Arithmetic Mean }=28.00 ; \text { Standard Deviation }=25.92 ; \text { Standard Error }=3.82 ; \
\text { Null Hypothesis:} H_{0}: \mu \leq 20.000 ; \alpha=0.10 ; d f=45 ; T \text {  Test Statistic} = 2.09;\
\text { One-Tailed Test Upper Critical Value} =1.3006 ;\text {  p-value}  =0.021 ; \text {  Decision = Reject.}\
\hline
\end{array}$

-Referring to Table 9-1, the biologist can conclude that there is sufficient evidence to show that the average number of parasites per Monarch butterfly in Pismo Beach State Park is over 20 with no more than a 5% probability of incorrectly rejecting the true null hypothesis.

Accepted Answer

The p-value (0.021) is less than the level of significance (0.10), so the null hypothesis is rejected. This means that there is sufficient evidence to show that the average number of parasites per Monarch butterfly in Pismo Beach State Park is over 20.

Question 2

TABLE 9-6
The quality control engineer for a furniture manufacturer is interested in the mean amount of force necessary to produce cracks in stressed oak furniture. She performs a two-tailed test of the null hypothesis that the mean for the stressed oak furniture is 650. The calculated value of the Z test statistic is a positive number that leads to a p-value of 0.080 for the test.
-Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was greater than 650. Then if the test is performed with a level of significance of 0.10, the null hypothesis would be rejected.

Accepted Answer

In a one-tailed test where the alternative hypothesis is that the mean is greater than 650, a p-value of 0.080 is less than the significance level of 0.10, leading to the rejection of the null hypothesis.

Question 3

TABLE 9-5
A bank tests the null hypothesis that the mean age of the bank's mortgage holders is less than or equal to 45, versus an alternative that the mean age is greater than 45. They take a sample and calculate a p-value of 0.0202.
-Referring to Table 9-5, the null hypothesis would be rejected at a significance level of α = 0.05.

Accepted Answer

The critical value for a one-tailed test with α = 0.05 and degrees of freedom equal to the sample size minus 1 is 1.645. The calculated p-value of 0.0202 is less than the critical value of 1.645, indicating that we reject the null hypothesis at a significance level of 0.05. Therefore, the answer is A.

Question 4

TABLE 9-8
One of the biggest issues facing e-retailers is the ability to turn browsers into buyers. This is measured by the conversion rate, the percentage of browsers who buy something in their visit to a site. The conversion rate for a company's web site was 10.1% The web site at the company was redesigned in an attempt to increase its conversion rates. Samples of 200 browsers at the redesigned site were selected. Suppose that 24 browsers made a purchase. The company officials would like to know if there is evidence of an increase in conversion rate at the 5% level of significance.
-Referring to Table 9-8, the null hypothesis would be rejected.

Accepted Answer

The answer of TABLE 9-8
One of the biggest issues facing...

Question 5

$\text { TABLE 9-1 }$
Microsoft Excel was used on a set of data involving the number of parasites found on 46 Monarch butterflies captured in Pismo Beach State Park. A biologist wantsto know if the mean number of parasites per butterfly is over 20 . She will make her decision usinga test witha level of significance of $0.10$. The following informationwas extracted from the Microsoft Excel outputfor the sample of 46 Monarchbutterflies:

$\begin{array}{llcc} \hline n=46 ; \text { Arithmetic Mean }=28.00 ; \text { Standard Deviation }=25.92 ; \text { Standard Error }=3.82 ; \
\text { Null Hypothesis:} H_{0}: \mu \leq 20.000 ; \alpha=0.10 ; d f=45 ; T \text {  Test Statistic} = 2.09;\
\text { One-Tailed Test Upper Critical Value} =1.3006 ;\text {  p-value}  =0.021 ; \text {  Decision = Reject.}\
\hline
\end{array}$

-Referring to Table 9-1, the value of þ is 0.90.

Accepted Answer

The value of β (beta) is not given directly in the table. The value of β represents the probability of a Type II error, which is not the same as the significance level (α) of 0.10. The significance level α is the probability of a Type I error, not β.

Question 6

Just reporting the results of hypothesis tests that show statistical significance but omitting those for which there is insufficient evidence in the findings is considered an ethical practice.

Accepted Answer

This practice is unethical as it can give a biased or incomplete view of the results. It is important to report all hypothesis tests conducted, regardless of the outcome, to ensure transparency and accuracy in the research.

Question 7

Suppose, in testing a hypothesis about a proportion, 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 p-value is less than or equal to the chosen level of significance (0.05 in this case), then we reject the null hypothesis. Since the p-value (0.043) is less than 0.05, we can reject the null hypothesis.

Question 8

TABLE 9-5
 A bank tests the null hypothesis that the mean age of the bank's mortgage holders is less than or equal to 45, versus an alternative that the mean age is greater than 45. They take a sample and calculate a p-value of 0.0202.
-Referring to Table 9-5, the null hypothesis would be rejected at a significance level of ? = 0.01.

Accepted Answer

The answer of TABLE 9-5
 A bank tests the null...

Question 9

$\text { TABLE 9-1 }$
Microsoft Excel was used on a set of data involving the number of parasites found on 46 Monarch butterflies captured in Pismo Beach State Park. A biologist wantsto know if the mean number of parasites per butterfly is over 20 . She will make her decision usinga test witha level of significance of $0.10$. The following informationwas extracted from the Microsoft Excel outputfor the sample of 46 Monarchbutterflies:

$\begin{array}{llcc} \hline n=46 ; \text { Arithmetic Mean }=28.00 ; \text { Standard Deviation }=25.92 ; \text { Standard Error }=3.82 ; \
\text { Null Hypothesis:} H_{0}: \mu \leq 20.000 ; \alpha=0.10 ; d f=45 ; T \text {  Test Statistic} = 2.09;\
\text { One-Tailed Test Upper Critical Value} =1.3006 ;\text {  p-value}  =0.021 ; \text {  Decision = Reject.}\
\hline
\end{array}$

-Referring to Table 9-1, the null hypothesis would be rejected.

Accepted Answer

Referring to Table 9-1, if the calculated F-statistic is greater than the critical F-value at a given level of significance, the null hypothesis would be rejected.

Question 10

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 for the mean of the population does not contain the value of 20, it suggests that the mean of the population is unlikely to be 20. Therefore, the null hypothesis can be rejected at a level of significance of 0.10.

Question 11

The larger is the p-value, the more likely one is to reject the null hypothesis.

Accepted Answer

The opposite is true. The larger the p-value, the less likely one is to reject the null hypothesis. A small p-value indicates strong evidence against the null hypothesis, while a large p-value indicates weak evidence against the null hypothesis.

Question 12

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

Accepted Answer

A smaller p-value indicates a lower probability of obtaining the observed results by chance alone, providing stronger evidence against 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

TABLE 9-8
One of the biggest issues facing e-retailers is the ability to turn browsers into buyers. This is measured by the conversion rate, the percentage of browsers who buy something in their visit to a site. The conversion rate for a company's web site was 10.1% The web site at the company was redesigned in an attempt to increase its conversion rates. Samples of 200 browsers at the redesigned site were selected. Suppose that 24 browsers made a purchase. The company officials would like to know if there is evidence of an increase in conversion rate at the 5% level of significance.
-Referring to Table 9-8, the company officials can conclude that there is sufficient evidence that the conversion rate at the company's web site has increased using a level of significance of 0.05.

Accepted Answer

The answer of TABLE 9-8
One of the biggest issues facing...

Question 15

TABLE 9-3
An appliance manufacturer claims to have developed a compact microwave oven that consumes an average of no more than 250 W. From previous studies, it is believed that power consumption for microwave ovens is normally distributed with a standard deviation of 15 W. A consumer group has decided to try to discover if the claim appears true. They take a sample of 20 microwave ovens and find that they consume an average of 257.3 W.
-Referring to Table 9-3, the consumer group can conclude that there is enough evidence to prove that the manufacturer's claim is not true when allowing for a 5% probability of committing a type I error.

Accepted Answer

The consumer group can perform a hypothesis test with the null hypothesis that the mean power consumption is 250 W. Given the sample mean of 257.3 W, a standard deviation of 15 W, and a sample size of 20, the calculated test statistic would likely exceed the critical value for a 5% significance level, indicating enough evidence to reject the null hypothesis and conclude the manufacturer's claim is not true.

Question 16

In instances in which there is insufficient evidence to reject the null hypothesis, making it clear that this does not prove that the null hypothesis is true is considered an ethical practice.

Accepted Answer

It is ethical to acknowledge that insufficient evidence to reject the null hypothesis does not prove that the null hypothesis is true. This is because not finding evidence against the null hypothesis does not necessarily mean that there is truly no effect or difference. It could be that the study did not have enough power or that there were other limitations. Therefore, it is important to be transparent and avoid making any unwarranted conclusions.

Question 17

TABLE 9-4
A drug company is considering marketing a new local anesthetic. The effective time of the anesthetic the drug company is currently producing has a normal distribution with an average of 7.4 minutes with a standard deviation of 1.2 minutes. The chemistry of the new anesthetic is such that the effective time should be normal with the same standard deviation, but the mean effective time may be lower. If it is lower, the drug company will market the new anesthetic; otherwise, they will continue to produce the older one. A sample of size 36 results in a sample mean of 7.1. A hypothesis test will be done to help make the decision.
-Referring to Table 9-4, the null hypothesis will be rejected with a level of significance of 0.10.

Accepted Answer

The answer of TABLE 9-4
A drug company is considering marketing...

Question 18

TABLE 9-1
Microsoft Excel was used on a set of data involving the number of parasites found on 46 Monarch butterflies captured in Pismo Beach State Park. A biologist wants to know if the mean number of parasites per butterfly is over 20. 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 Monarch butterflies:

$\begin{array}{llcc} \hline n=46 ; \text { Arithmetic Mean }=28.00 ; \text { Standard Deviation }=25.92 ; \text { Standard Error }=3.82 ; \
\text { Null Hypothesis:} H_{0}: \mu \leq 20.000 ; \alpha=0.10 ; d f=45 ; T \text {  Test Statistic} = 2.09;\
\text { One-Tailed Test Upper Critical Value} =1.3006 ;\text {  p-value}  =0.021 ; \text {  Decision = Reject.}\
\hline
\end{array}$

-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 19

$\text { TABLE 9-1 }$
Microsoft Excel was used on a set of data involving the number of parasites found on 46 Monarch butterflies captured in Pismo Beach State Park. A biologist wantsto know if the mean number of parasites per butterfly is over 20 . She will make her decision usinga test witha level of significance of $0.10$. The following informationwas extracted from the Microsoft Excel outputfor the sample of 46 Monarchbutterflies:

$\begin{array}{llcc} \hline n=46 ; \text { Arithmetic Mean }=28.00 ; \text { Standard Deviation }=25.92 ; \text { Standard Error }=3.82 ; \
\text { Null Hypothesis:} H_{0}: \mu \leq 20.000 ; \alpha=0.10 ; d f=45 ; T \text {  Test Statistic} = 2.09;\
\text { One-Tailed Test Upper Critical Value} =1.3006 ;\text {  p-value}  =0.021 ; \text {  Decision = Reject.}\
\hline
\end{array}$

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

Accepted Answer

The p-value for a two-tailed test is twice the p-value for a one-tailed test.

Question 20

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 accepted at a level of significance of 0.02.

Accepted Answer

Since the confidence interval does not include 20 (the value stated in the null hypothesis), it would be rejected at a significance level of 0.05 (which corresponds to the 95% confidence level). Therefore, it cannot be accepted at a lower significance level of 0.02.

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

Just reporting the results of hypothesis tests that show statistical significance but omitting those for which there is insufficient evidence in the findings is considered an ethical practice.

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

The larger is the p-value, the more likely one is to reject the null hypothesis.

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

In instances in which there is insufficient evidence to reject the null hypothesis, making it clear that this does not prove that the null hypothesis is true is considered an ethical practice.