Compute the P -value. Check whether the null hypothesis H 0 is not rejected at the α = 0.05 level.

Question

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Answer 1

Question

Chapter 8.2, Problem 72E

a.

To determine

Compute the P-value.

Check whether the null hypothesis $H0$ is not rejected at the $α=0.05$ level.

a.

Expert Solution

Answer to Problem 72E

The P-value is 0.159.

The null hypothesis $H0$ is not rejected at the $α=0.05$ level.

Explanation of Solution

Calculation:

The given test hypotheses are:

Null hypothesis: $H0:μ=20$ .

Alternate hypothesis: $H1:μ>20$ .

It is given that the sample size $n=100$ ,sample mean is $x¯=21$ and the assumed standard deviation is $σ=10$ . It has been stated that the population mean $μ$ will not be of practical significance unless $μ>25$ .

Assumptions for performing a hypothesis test about $μ$ when $σ$ is known:

The samples are taken from the population are simple random samples.
The sample size should be large that is, $n>30$ or the population should be distributed according to normal.

Requirement check:

The sample of 100 is assumed to be simple random samples.
The sample size of 100 is greater than 30. That is, $100>30$ .

Denote $μ$ as the population mean.

It is given that the level of significance $α=0.05$ .

Test statistic:

The z-test statistic is:

$z=x¯−μσn$ ,

Where, $x¯$ be the sample mean, $μ$ be the hypothesized mean, $σ$ be the standard deviation and n be the sample size.

Software procedure:

Step by step procedure to find the test statistic using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data select the Sample size as 100, Sample mean as 21.
Select Known standard deviation as 10.
Enter Hypothesized mean as 20.
Select Options.
Choose $Mean>hypothesized mean$ in Alternate hypothesis.
Click OK in all dialogue boxes.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8.2, Problem 72E , additional homework tip 1

From the MINITAB output, the P-value is 0.159.

Thus, the P-value is 0.159.

Decision based on the P-value method:

If $P−value≤α$ , reject $H0$ .
If $P−value>α$ , fail to reject $H0$ .

Conclusion:

The significance level is, $α=0.05$ and the P-value is 0.159.

Here, the P-value of 0.159 is greater than the significance level 0.05.

That is, $P-value(=0.159)>α(=0.05)$ .

Therefore, the null hypothesis is not rejected.

Hence, it is verified that the null hypothesis is not rejected.

b.

To determine

Compute the P-value.

Check whether the null hypothesis $H0$ is rejected at the $α=0.05$ level.

b.

Expert Solution

Answer to Problem 72E

The P-value is 0.0008.

Yes, the null hypothesis $H0$ is rejected at the $α=0.05$ level.

Explanation of Solution

Calculation:

It is given that the sample size $n=1,000$ ,sample mean is $x¯=21$ and the assumed standard deviation $σ=10$ .

Assumptions for performing a hypothesis test about $μ$ when $σ$ is known:

The samples are taken from the population are simple random samples.
The sample size should be large that is, $n>30$ or the population should be distributed according to normal.

Requirement check:

The sample of 1,000 is assumed to be simple random samples.
The sample size of 1,000 is greater than 30. That is, $1,000>30$ .

Denote $μ$ as the population mean.

It is given that the level of significance $α=0.05$ .

Test statistic:

Software procedure:

Step by step procedure to find the test statistic using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data select the Sample size as 1,000, Sample mean as 21.
Select Known standard deviation as 10.
Enter Hypothesized mean as 20.
Select Options.
Choose $Mean>hypothesized mean$ in Alternate hypothesis.
Click OK in all dialogue boxes.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8.2, Problem 72E , additional homework tip 2

From the MINITAB output, the P-value is 0.0008.

Thus, the P-value is 0.0008.

Conclusion:

The significance level is, $α=0.05$ and the P-value is 0.0008.

Here, the P-value of 0.0008 is less than the significance level 0.05.

That is, $P-value(=0.0008)<α(=0.05)$ .

Therefore, the null hypothesis is rejected.

Hence, it has verified that the null hypothesis is rejected.

c.

To determine

Explain whether the result is likely to be of practical significance.

c.

Expert Solution

Answer to Problem 72E

No, the result may not be of practical significance.

Explanation of Solution

Interpretation:

Practical significance:

The practical significance means the real life applicability of a characteristic under study.

It is often observed that even if the tested mean is significant, in common sense or practically, it may not have enough significanceto be considered as practically significant. That is, it suggests that the difference may not be sufficiently large to be considered of practical significance, even if it is statistically significant.

Here, the test of hypothesis gives statistically significant results. The hypothesized mean valueis 20. In 100 sampled individuals,themean is found assignificantlygreater than20.

However, this is not necessarily a great increase for practical purposes. In other words, meangreater than20 is not useful, unless $μ>25$ . There is no evidence to believe the population mean $μ$ is greater than 25 from this test.

Hence, the result may not be of practical significance.

d.

To determine

Explain the reason the larger sample can be more likely to produce statistically significant results that are not a practical significance.

d.

Expert Solution

Explanation of Solution

Increasing in the sample size increases the power of a statistical test. Now, power is the probability of Type II error subtracted from 1. Thus, a larger power suggests smaller probability of Type II error. Again, decreasing the Type II error increases the probability of Type I error, leading to a greater chance of rejecting the null hypothesis when it is true.

As a result, the increase in the sample size increases the chance of rejecting the null hypothesis when it is true. This produces a statistically significant result for larger sample sizes.

Hence, the larger sample can be more likely to produce statistically significant results that are not a practical significance.

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Answer 2

Question

Chapter 8.2, Problem 72E

a.

To determine

Compute the P-value.

Check whether the null hypothesis $H0$ is not rejected at the $α=0.05$ level.

a.

Expert Solution

Answer to Problem 72E

The P-value is 0.159.

The null hypothesis $H0$ is not rejected at the $α=0.05$ level.

Explanation of Solution

Calculation:

The given test hypotheses are:

Null hypothesis: $H0:μ=20$ .

Alternate hypothesis: $H1:μ>20$ .

It is given that the sample size $n=100$ ,sample mean is $x¯=21$ and the assumed standard deviation is $σ=10$ . It has been stated that the population mean $μ$ will not be of practical significance unless $μ>25$ .

Assumptions for performing a hypothesis test about $μ$ when $σ$ is known:

The samples are taken from the population are simple random samples.
The sample size should be large that is, $n>30$ or the population should be distributed according to normal.

Requirement check:

The sample of 100 is assumed to be simple random samples.
The sample size of 100 is greater than 30. That is, $100>30$ .

Denote $μ$ as the population mean.

It is given that the level of significance $α=0.05$ .

Test statistic:

The z-test statistic is:

$z=x¯−μσn$ ,

Where, $x¯$ be the sample mean, $μ$ be the hypothesized mean, $σ$ be the standard deviation and n be the sample size.

Software procedure:

Step by step procedure to find the test statistic using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data select the Sample size as 100, Sample mean as 21.
Select Known standard deviation as 10.
Enter Hypothesized mean as 20.
Select Options.
Choose $Mean>hypothesized mean$ in Alternate hypothesis.
Click OK in all dialogue boxes.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8.2, Problem 72E , additional homework tip 1

From the MINITAB output, the P-value is 0.159.

Thus, the P-value is 0.159.

Decision based on the P-value method:

If $P−value≤α$ , reject $H0$ .
If $P−value>α$ , fail to reject $H0$ .

Conclusion:

The significance level is, $α=0.05$ and the P-value is 0.159.

Here, the P-value of 0.159 is greater than the significance level 0.05.

That is, $P-value(=0.159)>α(=0.05)$ .

Therefore, the null hypothesis is not rejected.

Hence, it is verified that the null hypothesis is not rejected.

b.

To determine

Compute the P-value.

Check whether the null hypothesis $H0$ is rejected at the $α=0.05$ level.

b.

Expert Solution

Answer to Problem 72E

The P-value is 0.0008.

Yes, the null hypothesis $H0$ is rejected at the $α=0.05$ level.

Explanation of Solution

Calculation:

It is given that the sample size $n=1,000$ ,sample mean is $x¯=21$ and the assumed standard deviation $σ=10$ .

Assumptions for performing a hypothesis test about $μ$ when $σ$ is known:

The samples are taken from the population are simple random samples.
The sample size should be large that is, $n>30$ or the population should be distributed according to normal.

Requirement check:

The sample of 1,000 is assumed to be simple random samples.
The sample size of 1,000 is greater than 30. That is, $1,000>30$ .

Denote $μ$ as the population mean.

It is given that the level of significance $α=0.05$ .

Test statistic:

Software procedure:

Step by step procedure to find the test statistic using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data select the Sample size as 1,000, Sample mean as 21.
Select Known standard deviation as 10.
Enter Hypothesized mean as 20.
Select Options.
Choose $Mean>hypothesized mean$ in Alternate hypothesis.
Click OK in all dialogue boxes.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8.2, Problem 72E , additional homework tip 2

From the MINITAB output, the P-value is 0.0008.

Thus, the P-value is 0.0008.

Conclusion:

The significance level is, $α=0.05$ and the P-value is 0.0008.

Here, the P-value of 0.0008 is less than the significance level 0.05.

That is, $P-value(=0.0008)<α(=0.05)$ .

Therefore, the null hypothesis is rejected.

Hence, it has verified that the null hypothesis is rejected.

c.

To determine

Explain whether the result is likely to be of practical significance.

c.

Expert Solution

Answer to Problem 72E

No, the result may not be of practical significance.

Explanation of Solution

Interpretation:

Practical significance:

The practical significance means the real life applicability of a characteristic under study.

It is often observed that even if the tested mean is significant, in common sense or practically, it may not have enough significanceto be considered as practically significant. That is, it suggests that the difference may not be sufficiently large to be considered of practical significance, even if it is statistically significant.

Here, the test of hypothesis gives statistically significant results. The hypothesized mean valueis 20. In 100 sampled individuals,themean is found assignificantlygreater than20.

However, this is not necessarily a great increase for practical purposes. In other words, meangreater than20 is not useful, unless $μ>25$ . There is no evidence to believe the population mean $μ$ is greater than 25 from this test.

Hence, the result may not be of practical significance.

d.

To determine

Explain the reason the larger sample can be more likely to produce statistically significant results that are not a practical significance.

d.

Expert Solution

Explanation of Solution

Increasing in the sample size increases the power of a statistical test. Now, power is the probability of Type II error subtracted from 1. Thus, a larger power suggests smaller probability of Type II error. Again, decreasing the Type II error increases the probability of Type I error, leading to a greater chance of rejecting the null hypothesis when it is true.

As a result, the increase in the sample size increases the chance of rejecting the null hypothesis when it is true. This produces a statistically significant result for larger sample sizes.

Hence, the larger sample can be more likely to produce statistically significant results that are not a practical significance.

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Answer 3

Question

Chapter 8.2, Problem 72E

a.

To determine

Compute the P-value.

Check whether the null hypothesis $H0$ is not rejected at the $α=0.05$ level.

a.

Expert Solution

Answer to Problem 72E

The P-value is 0.159.

The null hypothesis $H0$ is not rejected at the $α=0.05$ level.

Explanation of Solution

Calculation:

The given test hypotheses are:

Null hypothesis: $H0:μ=20$ .

Alternate hypothesis: $H1:μ>20$ .

It is given that the sample size $n=100$ ,sample mean is $x¯=21$ and the assumed standard deviation is $σ=10$ . It has been stated that the population mean $μ$ will not be of practical significance unless $μ>25$ .

Assumptions for performing a hypothesis test about $μ$ when $σ$ is known:

The samples are taken from the population are simple random samples.
The sample size should be large that is, $n>30$ or the population should be distributed according to normal.

Requirement check:

The sample of 100 is assumed to be simple random samples.
The sample size of 100 is greater than 30. That is, $100>30$ .

Denote $μ$ as the population mean.

It is given that the level of significance $α=0.05$ .

Test statistic:

The z-test statistic is:

$z=x¯−μσn$ ,

Where, $x¯$ be the sample mean, $μ$ be the hypothesized mean, $σ$ be the standard deviation and n be the sample size.

Software procedure:

Step by step procedure to find the test statistic using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data select the Sample size as 100, Sample mean as 21.
Select Known standard deviation as 10.
Enter Hypothesized mean as 20.
Select Options.
Choose $Mean>hypothesized mean$ in Alternate hypothesis.
Click OK in all dialogue boxes.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8.2, Problem 72E , additional homework tip 1

From the MINITAB output, the P-value is 0.159.

Thus, the P-value is 0.159.

Decision based on the P-value method:

If $P−value≤α$ , reject $H0$ .
If $P−value>α$ , fail to reject $H0$ .

Conclusion:

The significance level is, $α=0.05$ and the P-value is 0.159.

Here, the P-value of 0.159 is greater than the significance level 0.05.

That is, $P-value(=0.159)>α(=0.05)$ .

Therefore, the null hypothesis is not rejected.

Hence, it is verified that the null hypothesis is not rejected.

b.

To determine

Compute the P-value.

Check whether the null hypothesis $H0$ is rejected at the $α=0.05$ level.

b.

Expert Solution

Answer to Problem 72E

The P-value is 0.0008.

Yes, the null hypothesis $H0$ is rejected at the $α=0.05$ level.

Explanation of Solution

Calculation:

It is given that the sample size $n=1,000$ ,sample mean is $x¯=21$ and the assumed standard deviation $σ=10$ .

Assumptions for performing a hypothesis test about $μ$ when $σ$ is known:

The samples are taken from the population are simple random samples.
The sample size should be large that is, $n>30$ or the population should be distributed according to normal.

Requirement check:

The sample of 1,000 is assumed to be simple random samples.
The sample size of 1,000 is greater than 30. That is, $1,000>30$ .

Denote $μ$ as the population mean.

It is given that the level of significance $α=0.05$ .

Test statistic:

Software procedure:

Step by step procedure to find the test statistic using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data select the Sample size as 1,000, Sample mean as 21.
Select Known standard deviation as 10.
Enter Hypothesized mean as 20.
Select Options.
Choose $Mean>hypothesized mean$ in Alternate hypothesis.
Click OK in all dialogue boxes.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8.2, Problem 72E , additional homework tip 2

From the MINITAB output, the P-value is 0.0008.

Thus, the P-value is 0.0008.

Conclusion:

The significance level is, $α=0.05$ and the P-value is 0.0008.

Here, the P-value of 0.0008 is less than the significance level 0.05.

That is, $P-value(=0.0008)<α(=0.05)$ .

Therefore, the null hypothesis is rejected.

Hence, it has verified that the null hypothesis is rejected.

c.

To determine

Explain whether the result is likely to be of practical significance.

c.

Expert Solution

Answer to Problem 72E

No, the result may not be of practical significance.

Explanation of Solution

Interpretation:

Practical significance:

The practical significance means the real life applicability of a characteristic under study.

It is often observed that even if the tested mean is significant, in common sense or practically, it may not have enough significanceto be considered as practically significant. That is, it suggests that the difference may not be sufficiently large to be considered of practical significance, even if it is statistically significant.

Here, the test of hypothesis gives statistically significant results. The hypothesized mean valueis 20. In 100 sampled individuals,themean is found assignificantlygreater than20.

However, this is not necessarily a great increase for practical purposes. In other words, meangreater than20 is not useful, unless $μ>25$ . There is no evidence to believe the population mean $μ$ is greater than 25 from this test.

Hence, the result may not be of practical significance.

d.

To determine

Explain the reason the larger sample can be more likely to produce statistically significant results that are not a practical significance.

d.

Expert Solution

Explanation of Solution

Increasing in the sample size increases the power of a statistical test. Now, power is the probability of Type II error subtracted from 1. Thus, a larger power suggests smaller probability of Type II error. Again, decreasing the Type II error increases the probability of Type I error, leading to a greater chance of rejecting the null hypothesis when it is true.

As a result, the increase in the sample size increases the chance of rejecting the null hypothesis when it is true. This produces a statistically significant result for larger sample sizes.

Hence, the larger sample can be more likely to produce statistically significant results that are not a practical significance.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 8.2, Problem 72E

a.

To determine

Compute the P-value.

Check whether the null hypothesis $H0$ is not rejected at the $α=0.05$ level.

a.

Expert Solution

Answer to Problem 72E

The P-value is 0.159.

The null hypothesis $H0$ is not rejected at the $α=0.05$ level.

Explanation of Solution

Calculation:

The given test hypotheses are:

Null hypothesis: $H0:μ=20$ .

Alternate hypothesis: $H1:μ>20$ .

It is given that the sample size $n=100$ ,sample mean is $x¯=21$ and the assumed standard deviation is $σ=10$ . It has been stated that the population mean $μ$ will not be of practical significance unless $μ>25$ .

Assumptions for performing a hypothesis test about $μ$ when $σ$ is known:

The samples are taken from the population are simple random samples.
The sample size should be large that is, $n>30$ or the population should be distributed according to normal.

Requirement check:

The sample of 100 is assumed to be simple random samples.
The sample size of 100 is greater than 30. That is, $100>30$ .

Denote $μ$ as the population mean.

It is given that the level of significance $α=0.05$ .

Test statistic:

The z-test statistic is:

$z=x¯−μσn$ ,

Where, $x¯$ be the sample mean, $μ$ be the hypothesized mean, $σ$ be the standard deviation and n be the sample size.

Software procedure:

Step by step procedure to find the test statistic using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data select the Sample size as 100, Sample mean as 21.
Select Known standard deviation as 10.
Enter Hypothesized mean as 20.
Select Options.
Choose $Mean>hypothesized mean$ in Alternate hypothesis.
Click OK in all dialogue boxes.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8.2, Problem 72E , additional homework tip 1

From the MINITAB output, the P-value is 0.159.

Thus, the P-value is 0.159.

Decision based on the P-value method:

If $P−value≤α$ , reject $H0$ .
If $P−value>α$ , fail to reject $H0$ .

Conclusion:

The significance level is, $α=0.05$ and the P-value is 0.159.

Here, the P-value of 0.159 is greater than the significance level 0.05.

That is, $P-value(=0.159)>α(=0.05)$ .

Therefore, the null hypothesis is not rejected.

Hence, it is verified that the null hypothesis is not rejected.

b.

To determine

Compute the P-value.

Check whether the null hypothesis $H0$ is rejected at the $α=0.05$ level.

b.

Expert Solution

Answer to Problem 72E

The P-value is 0.0008.

Yes, the null hypothesis $H0$ is rejected at the $α=0.05$ level.

Explanation of Solution

Calculation:

It is given that the sample size $n=1,000$ ,sample mean is $x¯=21$ and the assumed standard deviation $σ=10$ .

Assumptions for performing a hypothesis test about $μ$ when $σ$ is known:

The samples are taken from the population are simple random samples.
The sample size should be large that is, $n>30$ or the population should be distributed according to normal.

Requirement check:

The sample of 1,000 is assumed to be simple random samples.
The sample size of 1,000 is greater than 30. That is, $1,000>30$ .

Denote $μ$ as the population mean.

It is given that the level of significance $α=0.05$ .

Test statistic:

Software procedure:

Step by step procedure to find the test statistic using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data select the Sample size as 1,000, Sample mean as 21.
Select Known standard deviation as 10.
Enter Hypothesized mean as 20.
Select Options.
Choose $Mean>hypothesized mean$ in Alternate hypothesis.
Click OK in all dialogue boxes.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8.2, Problem 72E , additional homework tip 2

From the MINITAB output, the P-value is 0.0008.

Thus, the P-value is 0.0008.

Conclusion:

The significance level is, $α=0.05$ and the P-value is 0.0008.

Here, the P-value of 0.0008 is less than the significance level 0.05.

That is, $P-value(=0.0008)<α(=0.05)$ .

Therefore, the null hypothesis is rejected.

Hence, it has verified that the null hypothesis is rejected.

c.

To determine

Explain whether the result is likely to be of practical significance.

c.

Expert Solution

Answer to Problem 72E

No, the result may not be of practical significance.

Explanation of Solution

Interpretation:

Practical significance:

The practical significance means the real life applicability of a characteristic under study.

It is often observed that even if the tested mean is significant, in common sense or practically, it may not have enough significanceto be considered as practically significant. That is, it suggests that the difference may not be sufficiently large to be considered of practical significance, even if it is statistically significant.

Here, the test of hypothesis gives statistically significant results. The hypothesized mean valueis 20. In 100 sampled individuals,themean is found assignificantlygreater than20.

However, this is not necessarily a great increase for practical purposes. In other words, meangreater than20 is not useful, unless $μ>25$ . There is no evidence to believe the population mean $μ$ is greater than 25 from this test.

Hence, the result may not be of practical significance.

d.

To determine

Explain the reason the larger sample can be more likely to produce statistically significant results that are not a practical significance.

d.

Expert Solution

Explanation of Solution

Increasing in the sample size increases the power of a statistical test. Now, power is the probability of Type II error subtracted from 1. Thus, a larger power suggests smaller probability of Type II error. Again, decreasing the Type II error increases the probability of Type I error, leading to a greater chance of rejecting the null hypothesis when it is true.

As a result, the increase in the sample size increases the chance of rejecting the null hypothesis when it is true. This produces a statistically significant result for larger sample sizes.

Hence, the larger sample can be more likely to produce statistically significant results that are not a practical significance.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 8.2, Problem 72E

a.

To determine

Compute the P-value.

Check whether the null hypothesis $H0$ is not rejected at the $α=0.05$ level.

a.

Expert Solution

Answer to Problem 72E

The P-value is 0.159.

The null hypothesis $H0$ is not rejected at the $α=0.05$ level.

Explanation of Solution

Calculation:

The given test hypotheses are:

Null hypothesis: $H0:μ=20$ .

Alternate hypothesis: $H1:μ>20$ .

It is given that the sample size $n=100$ ,sample mean is $x¯=21$ and the assumed standard deviation is $σ=10$ . It has been stated that the population mean $μ$ will not be of practical significance unless $μ>25$ .

Assumptions for performing a hypothesis test about $μ$ when $σ$ is known:

The samples are taken from the population are simple random samples.
The sample size should be large that is, $n>30$ or the population should be distributed according to normal.

Requirement check:

The sample of 100 is assumed to be simple random samples.
The sample size of 100 is greater than 30. That is, $100>30$ .

Denote $μ$ as the population mean.

It is given that the level of significance $α=0.05$ .

Test statistic:

The z-test statistic is:

$z=x¯−μσn$ ,

Where, $x¯$ be the sample mean, $μ$ be the hypothesized mean, $σ$ be the standard deviation and n be the sample size.

Software procedure:

Step by step procedure to find the test statistic using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data select the Sample size as 100, Sample mean as 21.
Select Known standard deviation as 10.
Enter Hypothesized mean as 20.
Select Options.
Choose $Mean>hypothesized mean$ in Alternate hypothesis.
Click OK in all dialogue boxes.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8.2, Problem 72E , additional homework tip 1

From the MINITAB output, the P-value is 0.159.

Thus, the P-value is 0.159.

Decision based on the P-value method:

If $P−value≤α$ , reject $H0$ .
If $P−value>α$ , fail to reject $H0$ .

Conclusion:

The significance level is, $α=0.05$ and the P-value is 0.159.

Here, the P-value of 0.159 is greater than the significance level 0.05.

That is, $P-value(=0.159)>α(=0.05)$ .

Therefore, the null hypothesis is not rejected.

Hence, it is verified that the null hypothesis is not rejected.

b.

To determine

Compute the P-value.

Check whether the null hypothesis $H0$ is rejected at the $α=0.05$ level.

b.

Expert Solution

Answer to Problem 72E

The P-value is 0.0008.

Yes, the null hypothesis $H0$ is rejected at the $α=0.05$ level.

Explanation of Solution

Calculation:

It is given that the sample size $n=1,000$ ,sample mean is $x¯=21$ and the assumed standard deviation $σ=10$ .

Assumptions for performing a hypothesis test about $μ$ when $σ$ is known:

The samples are taken from the population are simple random samples.
The sample size should be large that is, $n>30$ or the population should be distributed according to normal.

Requirement check:

The sample of 1,000 is assumed to be simple random samples.
The sample size of 1,000 is greater than 30. That is, $1,000>30$ .

Denote $μ$ as the population mean.

It is given that the level of significance $α=0.05$ .

Test statistic:

Software procedure:

Step by step procedure to find the test statistic using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data select the Sample size as 1,000, Sample mean as 21.
Select Known standard deviation as 10.
Enter Hypothesized mean as 20.
Select Options.
Choose $Mean>hypothesized mean$ in Alternate hypothesis.
Click OK in all dialogue boxes.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8.2, Problem 72E , additional homework tip 2

From the MINITAB output, the P-value is 0.0008.

Thus, the P-value is 0.0008.

Conclusion:

The significance level is, $α=0.05$ and the P-value is 0.0008.

Here, the P-value of 0.0008 is less than the significance level 0.05.

That is, $P-value(=0.0008)<α(=0.05)$ .

Therefore, the null hypothesis is rejected.

Hence, it has verified that the null hypothesis is rejected.

c.

To determine

Explain whether the result is likely to be of practical significance.

c.

Expert Solution

Answer to Problem 72E

No, the result may not be of practical significance.

Explanation of Solution

Interpretation:

Practical significance:

The practical significance means the real life applicability of a characteristic under study.

It is often observed that even if the tested mean is significant, in common sense or practically, it may not have enough significanceto be considered as practically significant. That is, it suggests that the difference may not be sufficiently large to be considered of practical significance, even if it is statistically significant.

Here, the test of hypothesis gives statistically significant results. The hypothesized mean valueis 20. In 100 sampled individuals,themean is found assignificantlygreater than20.

However, this is not necessarily a great increase for practical purposes. In other words, meangreater than20 is not useful, unless $μ>25$ . There is no evidence to believe the population mean $μ$ is greater than 25 from this test.

Hence, the result may not be of practical significance.

d.

To determine

Explain the reason the larger sample can be more likely to produce statistically significant results that are not a practical significance.

d.

Expert Solution

Explanation of Solution

Increasing in the sample size increases the power of a statistical test. Now, power is the probability of Type II error subtracted from 1. Thus, a larger power suggests smaller probability of Type II error. Again, decreasing the Type II error increases the probability of Type I error, leading to a greater chance of rejecting the null hypothesis when it is true.

As a result, the increase in the sample size increases the chance of rejecting the null hypothesis when it is true. This produces a statistically significant result for larger sample sizes.

Hence, the larger sample can be more likely to produce statistically significant results that are not a practical significance.

Answer 6

Chapter 8.2, Problem 72E

a.

To determine

Compute the P-value.

Check whether the null hypothesis $H0$ is not rejected at the $α=0.05$ level.

a.

Expert Solution

Answer to Problem 72E

The P-value is 0.159.

The null hypothesis $H0$ is not rejected at the $α=0.05$ level.

Explanation of Solution

Calculation:

The given test hypotheses are:

Null hypothesis: $H0:μ=20$ .

Alternate hypothesis: $H1:μ>20$ .

It is given that the sample size $n=100$ ,sample mean is $x¯=21$ and the assumed standard deviation is $σ=10$ . It has been stated that the population mean $μ$ will not be of practical significance unless $μ>25$ .

Assumptions for performing a hypothesis test about $μ$ when $σ$ is known:

The samples are taken from the population are simple random samples.
The sample size should be large that is, $n>30$ or the population should be distributed according to normal.

Requirement check:

The sample of 100 is assumed to be simple random samples.
The sample size of 100 is greater than 30. That is, $100>30$ .

Denote $μ$ as the population mean.

It is given that the level of significance $α=0.05$ .

Test statistic:

The z-test statistic is:

$z=x¯−μσn$ ,

Where, $x¯$ be the sample mean, $μ$ be the hypothesized mean, $σ$ be the standard deviation and n be the sample size.

Software procedure:

Step by step procedure to find the test statistic using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data select the Sample size as 100, Sample mean as 21.
Select Known standard deviation as 10.
Enter Hypothesized mean as 20.
Select Options.
Choose $Mean>hypothesized mean$ in Alternate hypothesis.
Click OK in all dialogue boxes.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8.2, Problem 72E , additional homework tip 1

From the MINITAB output, the P-value is 0.159.

Thus, the P-value is 0.159.

Decision based on the P-value method:

If $P−value≤α$ , reject $H0$ .
If $P−value>α$ , fail to reject $H0$ .

Conclusion:

The significance level is, $α=0.05$ and the P-value is 0.159.

Here, the P-value of 0.159 is greater than the significance level 0.05.

That is, $P-value(=0.159)>α(=0.05)$ .

Therefore, the null hypothesis is not rejected.

Hence, it is verified that the null hypothesis is not rejected.

Answer 7

Chapter 8.2, Problem 72E

a.

To determine

Compute the P-value.

Check whether the null hypothesis $H0$ is not rejected at the $α=0.05$ level.

Answer 8

a.

Expert Solution

Answer to Problem 72E

The P-value is 0.159.

The null hypothesis $H0$ is not rejected at the $α=0.05$ level.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation:

The given test hypotheses are:

Null hypothesis: $H0:μ=20$ .

Alternate hypothesis: $H1:μ>20$ .

It is given that the sample size $n=100$ ,sample mean is $x¯=21$ and the assumed standard deviation is $σ=10$ . It has been stated that the population mean $μ$ will not be of practical significance unless $μ>25$ .

Assumptions for performing a hypothesis test about $μ$ when $σ$ is known:

The samples are taken from the population are simple random samples.
The sample size should be large that is, $n>30$ or the population should be distributed according to normal.

Requirement check:

The sample of 100 is assumed to be simple random samples.
The sample size of 100 is greater than 30. That is, $100>30$ .

Denote $μ$ as the population mean.

It is given that the level of significance $α=0.05$ .

Test statistic:

The z-test statistic is:

$z=x¯−μσn$ ,

Where, $x¯$ be the sample mean, $μ$ be the hypothesized mean, $σ$ be the standard deviation and n be the sample size.

Software procedure:

Step by step procedure to find the test statistic using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data select the Sample size as 100, Sample mean as 21.
Select Known standard deviation as 10.
Enter Hypothesized mean as 20.
Select Options.
Choose $Mean>hypothesized mean$ in Alternate hypothesis.
Click OK in all dialogue boxes.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8.2, Problem 72E , additional homework tip 1

From the MINITAB output, the P-value is 0.159.

Thus, the P-value is 0.159.

Decision based on the P-value method:

If $P−value≤α$ , reject $H0$ .
If $P−value>α$ , fail to reject $H0$ .

Conclusion:

The significance level is, $α=0.05$ and the P-value is 0.159.

Here, the P-value of 0.159 is greater than the significance level 0.05.

That is, $P-value(=0.159)>α(=0.05)$ .

Therefore, the null hypothesis is not rejected.

Hence, it is verified that the null hypothesis is not rejected.

Answer 13

b.

To determine

Compute the P-value.

Check whether the null hypothesis $H0$ is rejected at the $α=0.05$ level.

b.

Expert Solution

Answer to Problem 72E

The P-value is 0.0008.

Yes, the null hypothesis $H0$ is rejected at the $α=0.05$ level.

Explanation of Solution

Calculation:

It is given that the sample size $n=1,000$ ,sample mean is $x¯=21$ and the assumed standard deviation $σ=10$ .

Assumptions for performing a hypothesis test about $μ$ when $σ$ is known:

The samples are taken from the population are simple random samples.
The sample size should be large that is, $n>30$ or the population should be distributed according to normal.

Requirement check:

The sample of 1,000 is assumed to be simple random samples.
The sample size of 1,000 is greater than 30. That is, $1,000>30$ .

Denote $μ$ as the population mean.

It is given that the level of significance $α=0.05$ .

Test statistic:

Software procedure:

Step by step procedure to find the test statistic using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data select the Sample size as 1,000, Sample mean as 21.
Select Known standard deviation as 10.
Enter Hypothesized mean as 20.
Select Options.
Choose $Mean>hypothesized mean$ in Alternate hypothesis.
Click OK in all dialogue boxes.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8.2, Problem 72E , additional homework tip 2

From the MINITAB output, the P-value is 0.0008.

Thus, the P-value is 0.0008.

Conclusion:

The significance level is, $α=0.05$ and the P-value is 0.0008.

Here, the P-value of 0.0008 is less than the significance level 0.05.

That is, $P-value(=0.0008)<α(=0.05)$ .

Therefore, the null hypothesis is rejected.

Hence, it has verified that the null hypothesis is rejected.

Answer 14

b.

To determine

Compute the P-value.

Check whether the null hypothesis $H0$ is rejected at the $α=0.05$ level.

Answer 15

b.

Expert Solution

Answer to Problem 72E

The P-value is 0.0008.

Yes, the null hypothesis $H0$ is rejected at the $α=0.05$ level.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

It is given that the sample size $n=1,000$ ,sample mean is $x¯=21$ and the assumed standard deviation $σ=10$ .

Assumptions for performing a hypothesis test about $μ$ when $σ$ is known:

The samples are taken from the population are simple random samples.
The sample size should be large that is, $n>30$ or the population should be distributed according to normal.

Requirement check:

The sample of 1,000 is assumed to be simple random samples.
The sample size of 1,000 is greater than 30. That is, $1,000>30$ .

Denote $μ$ as the population mean.

It is given that the level of significance $α=0.05$ .

Test statistic:

Software procedure:

Step by step procedure to find the test statistic using the MINITAB software:

Choose Stat > Basic Statistics > 1-Sample Z.
In Summarized data select the Sample size as 1,000, Sample mean as 21.
Select Known standard deviation as 10.
Enter Hypothesized mean as 20.
Select Options.
Choose $Mean>hypothesized mean$ in Alternate hypothesis.
Click OK in all dialogue boxes.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8.2, Problem 72E , additional homework tip 2

From the MINITAB output, the P-value is 0.0008.

Thus, the P-value is 0.0008.

Conclusion:

The significance level is, $α=0.05$ and the P-value is 0.0008.

Here, the P-value of 0.0008 is less than the significance level 0.05.

That is, $P-value(=0.0008)<α(=0.05)$ .

Therefore, the null hypothesis is rejected.

Hence, it has verified that the null hypothesis is rejected.

Answer 20

c.

To determine

Explain whether the result is likely to be of practical significance.

c.

Expert Solution

Answer to Problem 72E

No, the result may not be of practical significance.

Explanation of Solution

Interpretation:

Practical significance:

The practical significance means the real life applicability of a characteristic under study.

It is often observed that even if the tested mean is significant, in common sense or practically, it may not have enough significanceto be considered as practically significant. That is, it suggests that the difference may not be sufficiently large to be considered of practical significance, even if it is statistically significant.

Here, the test of hypothesis gives statistically significant results. The hypothesized mean valueis 20. In 100 sampled individuals,themean is found assignificantlygreater than20.

However, this is not necessarily a great increase for practical purposes. In other words, meangreater than20 is not useful, unless $μ>25$ . There is no evidence to believe the population mean $μ$ is greater than 25 from this test.

Hence, the result may not be of practical significance.

Answer 21

c.

To determine

Explain whether the result is likely to be of practical significance.

Answer 22

c.

Expert Solution

Answer to Problem 72E

No, the result may not be of practical significance.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Interpretation:

Practical significance:

The practical significance means the real life applicability of a characteristic under study.

It is often observed that even if the tested mean is significant, in common sense or practically, it may not have enough significanceto be considered as practically significant. That is, it suggests that the difference may not be sufficiently large to be considered of practical significance, even if it is statistically significant.

Here, the test of hypothesis gives statistically significant results. The hypothesized mean valueis 20. In 100 sampled individuals,themean is found assignificantlygreater than20.

However, this is not necessarily a great increase for practical purposes. In other words, meangreater than20 is not useful, unless $μ>25$ . There is no evidence to believe the population mean $μ$ is greater than 25 from this test.

Hence, the result may not be of practical significance.

Answer 27

d.

To determine

Explain the reason the larger sample can be more likely to produce statistically significant results that are not a practical significance.

d.

Expert Solution

Explanation of Solution

Increasing in the sample size increases the power of a statistical test. Now, power is the probability of Type II error subtracted from 1. Thus, a larger power suggests smaller probability of Type II error. Again, decreasing the Type II error increases the probability of Type I error, leading to a greater chance of rejecting the null hypothesis when it is true.

As a result, the increase in the sample size increases the chance of rejecting the null hypothesis when it is true. This produces a statistically significant result for larger sample sizes.

Hence, the larger sample can be more likely to produce statistically significant results that are not a practical significance.

Answer 28

d.

To determine

Explain the reason the larger sample can be more likely to produce statistically significant results that are not a practical significance.

Answer 29

d.

Expert Solution

Explanation of Solution

Increasing in the sample size increases the power of a statistical test. Now, power is the probability of Type II error subtracted from 1. Thus, a larger power suggests smaller probability of Type II error. Again, decreasing the Type II error increases the probability of Type I error, leading to a greater chance of rejecting the null hypothesis when it is true.

As a result, the increase in the sample size increases the chance of rejecting the null hypothesis when it is true. This produces a statistically significant result for larger sample sizes.

Hence, the larger sample can be more likely to produce statistically significant results that are not a practical significance.

Answer 30

d.

Expert Solution

Answer 31

d.

Expert Solution

Answer 32

Expert Solution

Answer 33

Ch. 8.1 - Check Your Understanding 1. Last year, the mean...Ch. 8.1 - Prob. 2CYU Ch. 8.1 - Prob. 3CYU Ch. 8.1 - Prob. 4CYU Ch. 8.1 - Prob. 5CYU Ch. 8.1 - Prob. 6CYU Ch. 8.1 - Prob. 7E Ch. 8.1 - Prob. 8E Ch. 8.1 - Prob. 9E Ch. 8.1 - In Exercises 9–12, determine whether the statement...

Compute the P -value. Check whether the null hypothesis H 0 is not rejected at the α = 0.05 level.

Concept explainers

Videos

Compute the P-value.

Check whether the null hypothesis $H0$ is not rejected at the $α=0.05$ level.

Answer to Problem 72E

Explanation of Solution

Compute the P-value.

Check whether the null hypothesis $H0$ is rejected at the $α=0.05$ level.

Answer to Problem 72E

Explanation of Solution

Explain whether the result is likely to be of practical significance.

Answer to Problem 72E

Explanation of Solution

Explain the reason the larger sample can be more likely to produce statistically significant results that are not a practical significance.

Explanation of Solution

Want to see more full solutions like this?

Chapter 8 Solutions

Compute the P -value. Check whether the null hypothesis H 0 is not rejected at the α = 0.05 level.

Concept explainers

Videos

Compute the P-value. Check whether the null hypothesis H0<m:math><m:mrow><m:msub><m:mi>H</m:mi><m:mn>0</m:mn></m:msub></m:mrow></m:math> is not rejected at the α=0.05<m:math><m:mrow><m:mi>α</m:mi><m:mo>=</m:mo><m:mn>0.05</m:mn></m:mrow></m:math> level.

Answer to Problem 72E

Explanation of Solution

Compute the P-value. Check whether the null hypothesis H0<m:math><m:mrow><m:msub><m:mi>H</m:mi><m:mn>0</m:mn></m:msub></m:mrow></m:math> is rejected at the α=0.05<m:math><m:mrow><m:mi>α</m:mi><m:mo>=</m:mo><m:mn>0.05</m:mn></m:mrow></m:math> level.

Answer to Problem 72E

Explanation of Solution

Explain whether the result is likely to be of practical significance.

Answer to Problem 72E

Explanation of Solution

Explain the reason the larger sample can be more likely to produce statistically significant results that are not a practical significance.

Explanation of Solution

Want to see more full solutions like this?

Chapter 8 Solutions

Compute the P-value.

Check whether the null hypothesis $H0$ is not rejected at the $α=0.05$ level.

Compute the P-value.

Check whether the null hypothesis $H0$ is rejected at the $α=0.05$ level.