State the null and alternate hypotheses.

Question

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Answer 1

Question

Chapter 8, Problem 15RE

a.

To determine

State the null and alternate hypotheses.

a.

Expert Solution

Answer to Problem 15RE

Null hypothesis: $H0:p=0.23$ .

Alternate hypothesis: $H1:p>0.23$ .

Explanation of Solution

The Pew Research Center conducted the survey among the students at University who watch cable news regularly. Of 200 students, 66 were found to watch cable news regularly.

Denote p as the true population proportion of the students who watch cable news regularly.

The given test hypotheses are:

Null hypothesis:

$H0:p=0.23$ .

That is, the proportion of the students who watch cable news regularly is 0.23.

Alternate hypothesis:

$H1:p>0.23$ .

That is, the proportion of the students who watch cable news regularly is greater than 0.23.

b.

To determine

Find the value of test statistic.

b.

Expert Solution

Answer to Problem 15RE

The value of test statistic is 3.36.

Explanation of Solution

Calculation:

Assumptions for performing a hypothesis test for a population proportion:

• The samples taken from the population are simple random samples.
• The population is at least 20 times as large as the sample.
• The samples in the population are divided into two categories.
• The values of $np^0$ and $nq^0$ should be greater than or equal to 10.

Requirement check:

• The sample of 200 students is simple random samples.
• The information about the population is not known. However, the population size is assumed to be more than 20 times as large as the sample.
• The samples in the population seemed to be categorized into two parts. That is, students who watch cable news and students who do not watch cable news.
• Verify the condition: $np^0≥10$ and $nq^0≥10$

Here,

$p^0=66200=0.33$

Substitute n as 200 and $p^0$ as 0.33,

$np^0=200(0.33)=66>10$

Substitute n as 200 and $q^0$ as $0.67(=1−p^0)$ ,

$nq^0=200(0.67)=134>10$

Therefore, all the conditions are satisfied.

Test statistic:

The z-test statistic is:

$z=p^−p0p0(1−p0)n$ ,

Where, $p^$ be the sample proportion, $p0$ be the population proportion and n be the sample size.

Software procedure:

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

• Choose Stat > Basic Statistics > 1 Proportion.
• Choose Summarized data.
• In Number of events, enter 66. In Number of trials, enter 200.
• Check Perform hypothesis test. In Hypothesized proportion, enter 0.23.
• Click Options. Under Alternative, and choose $Proportion>Hypothesized proportion$ .
• Click OK in each dialog box.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8, Problem 15RE

From the MINITAB output, the test statistic, that is, the z-value is 3.36.

Thus, the value of test statistic z is 3.36.

c.

To determine

Decide whether the null hypothesis $H0$ is rejected at $α=0.01$ level.

c.

Expert Solution

Answer to Problem 15RE

The null hypothesis $H0$ is rejected at $α=0.01$ level.

Explanation of Solution

From previous part (b), it has been found that the value of test statistic z is 3.36.

From the given hypotheses, the alternative hypothesis contains the greater $(>)$ symbol. Thus, it is clear that the hypothesis follows the Two-tailed test.

From Table 8.1 “Table of Critical Values”, the critical values for right-tailed test at $α=0.01$ level are 2.326.

Decision based on the critical value method:

For left-tailed test: If $z≤−zα$ , reject $H0$ .

For right-tailed test: If $z≥zα$ , reject $H0$ .

For two-tailed test: If $z≤−zα2 or z≥zα2$ , reject $H0$ .

Conclusion:

The critical value at $α=0.01$ level is 2.326 and the test statistic is 3.36.

Here, the test statistic value of 3.36 lies in the critical region.

That is, $z(=3.36)>zα(=2.326)$ .

Therefore, the null hypothesis is rejected.

d.

To determine

State a conclusion.

d.

Expert Solution

Answer to Problem 15RE

The conclusion is that, there is evidence that the proportion of the students who watch cable news regularly is greater than 0.23.

Explanation of Solution

From previous part (c), it has been found that the null hypothesis is rejected.

Hence, there is evidence that the true proportion of the students who watch cable news regularly is greater than 0.23.

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Answer 2

Question

Chapter 8, Problem 15RE

a.

To determine

State the null and alternate hypotheses.

a.

Expert Solution

Answer to Problem 15RE

Null hypothesis: $H0:p=0.23$ .

Alternate hypothesis: $H1:p>0.23$ .

Explanation of Solution

The Pew Research Center conducted the survey among the students at University who watch cable news regularly. Of 200 students, 66 were found to watch cable news regularly.

Denote p as the true population proportion of the students who watch cable news regularly.

The given test hypotheses are:

Null hypothesis:

$H0:p=0.23$ .

That is, the proportion of the students who watch cable news regularly is 0.23.

Alternate hypothesis:

$H1:p>0.23$ .

That is, the proportion of the students who watch cable news regularly is greater than 0.23.

b.

To determine

Find the value of test statistic.

b.

Expert Solution

Answer to Problem 15RE

The value of test statistic is 3.36.

Explanation of Solution

Calculation:

Assumptions for performing a hypothesis test for a population proportion:

• The samples taken from the population are simple random samples.
• The population is at least 20 times as large as the sample.
• The samples in the population are divided into two categories.
• The values of $np^0$ and $nq^0$ should be greater than or equal to 10.

Requirement check:

• The sample of 200 students is simple random samples.
• The information about the population is not known. However, the population size is assumed to be more than 20 times as large as the sample.
• The samples in the population seemed to be categorized into two parts. That is, students who watch cable news and students who do not watch cable news.
• Verify the condition: $np^0≥10$ and $nq^0≥10$

Here,

$p^0=66200=0.33$

Substitute n as 200 and $p^0$ as 0.33,

$np^0=200(0.33)=66>10$

Substitute n as 200 and $q^0$ as $0.67(=1−p^0)$ ,

$nq^0=200(0.67)=134>10$

Therefore, all the conditions are satisfied.

Test statistic:

The z-test statistic is:

$z=p^−p0p0(1−p0)n$ ,

Where, $p^$ be the sample proportion, $p0$ be the population proportion and n be the sample size.

Software procedure:

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

• Choose Stat > Basic Statistics > 1 Proportion.
• Choose Summarized data.
• In Number of events, enter 66. In Number of trials, enter 200.
• Check Perform hypothesis test. In Hypothesized proportion, enter 0.23.
• Click Options. Under Alternative, and choose $Proportion>Hypothesized proportion$ .
• Click OK in each dialog box.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8, Problem 15RE

From the MINITAB output, the test statistic, that is, the z-value is 3.36.

Thus, the value of test statistic z is 3.36.

c.

To determine

Decide whether the null hypothesis $H0$ is rejected at $α=0.01$ level.

c.

Expert Solution

Answer to Problem 15RE

The null hypothesis $H0$ is rejected at $α=0.01$ level.

Explanation of Solution

From previous part (b), it has been found that the value of test statistic z is 3.36.

From the given hypotheses, the alternative hypothesis contains the greater $(>)$ symbol. Thus, it is clear that the hypothesis follows the Two-tailed test.

From Table 8.1 “Table of Critical Values”, the critical values for right-tailed test at $α=0.01$ level are 2.326.

Decision based on the critical value method:

For left-tailed test: If $z≤−zα$ , reject $H0$ .

For right-tailed test: If $z≥zα$ , reject $H0$ .

For two-tailed test: If $z≤−zα2 or z≥zα2$ , reject $H0$ .

Conclusion:

The critical value at $α=0.01$ level is 2.326 and the test statistic is 3.36.

Here, the test statistic value of 3.36 lies in the critical region.

That is, $z(=3.36)>zα(=2.326)$ .

Therefore, the null hypothesis is rejected.

d.

To determine

State a conclusion.

d.

Expert Solution

Answer to Problem 15RE

The conclusion is that, there is evidence that the proportion of the students who watch cable news regularly is greater than 0.23.

Explanation of Solution

From previous part (c), it has been found that the null hypothesis is rejected.

Hence, there is evidence that the true proportion of the students who watch cable news regularly is greater than 0.23.

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Answer 3

Question

Chapter 8, Problem 15RE

a.

To determine

State the null and alternate hypotheses.

a.

Expert Solution

Answer to Problem 15RE

Null hypothesis: $H0:p=0.23$ .

Alternate hypothesis: $H1:p>0.23$ .

Explanation of Solution

The Pew Research Center conducted the survey among the students at University who watch cable news regularly. Of 200 students, 66 were found to watch cable news regularly.

Denote p as the true population proportion of the students who watch cable news regularly.

The given test hypotheses are:

Null hypothesis:

$H0:p=0.23$ .

That is, the proportion of the students who watch cable news regularly is 0.23.

Alternate hypothesis:

$H1:p>0.23$ .

That is, the proportion of the students who watch cable news regularly is greater than 0.23.

b.

To determine

Find the value of test statistic.

b.

Expert Solution

Answer to Problem 15RE

The value of test statistic is 3.36.

Explanation of Solution

Calculation:

Assumptions for performing a hypothesis test for a population proportion:

• The samples taken from the population are simple random samples.
• The population is at least 20 times as large as the sample.
• The samples in the population are divided into two categories.
• The values of $np^0$ and $nq^0$ should be greater than or equal to 10.

Requirement check:

• The sample of 200 students is simple random samples.
• The information about the population is not known. However, the population size is assumed to be more than 20 times as large as the sample.
• The samples in the population seemed to be categorized into two parts. That is, students who watch cable news and students who do not watch cable news.
• Verify the condition: $np^0≥10$ and $nq^0≥10$

Here,

$p^0=66200=0.33$

Substitute n as 200 and $p^0$ as 0.33,

$np^0=200(0.33)=66>10$

Substitute n as 200 and $q^0$ as $0.67(=1−p^0)$ ,

$nq^0=200(0.67)=134>10$

Therefore, all the conditions are satisfied.

Test statistic:

The z-test statistic is:

$z=p^−p0p0(1−p0)n$ ,

Where, $p^$ be the sample proportion, $p0$ be the population proportion and n be the sample size.

Software procedure:

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

• Choose Stat > Basic Statistics > 1 Proportion.
• Choose Summarized data.
• In Number of events, enter 66. In Number of trials, enter 200.
• Check Perform hypothesis test. In Hypothesized proportion, enter 0.23.
• Click Options. Under Alternative, and choose $Proportion>Hypothesized proportion$ .
• Click OK in each dialog box.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8, Problem 15RE

From the MINITAB output, the test statistic, that is, the z-value is 3.36.

Thus, the value of test statistic z is 3.36.

c.

To determine

Decide whether the null hypothesis $H0$ is rejected at $α=0.01$ level.

c.

Expert Solution

Answer to Problem 15RE

The null hypothesis $H0$ is rejected at $α=0.01$ level.

Explanation of Solution

From previous part (b), it has been found that the value of test statistic z is 3.36.

From the given hypotheses, the alternative hypothesis contains the greater $(>)$ symbol. Thus, it is clear that the hypothesis follows the Two-tailed test.

From Table 8.1 “Table of Critical Values”, the critical values for right-tailed test at $α=0.01$ level are 2.326.

Decision based on the critical value method:

For left-tailed test: If $z≤−zα$ , reject $H0$ .

For right-tailed test: If $z≥zα$ , reject $H0$ .

For two-tailed test: If $z≤−zα2 or z≥zα2$ , reject $H0$ .

Conclusion:

The critical value at $α=0.01$ level is 2.326 and the test statistic is 3.36.

Here, the test statistic value of 3.36 lies in the critical region.

That is, $z(=3.36)>zα(=2.326)$ .

Therefore, the null hypothesis is rejected.

d.

To determine

State a conclusion.

d.

Expert Solution

Answer to Problem 15RE

The conclusion is that, there is evidence that the proportion of the students who watch cable news regularly is greater than 0.23.

Explanation of Solution

From previous part (c), it has been found that the null hypothesis is rejected.

Hence, there is evidence that the true proportion of the students who watch cable news regularly is greater than 0.23.

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Answer 4

Question

Chapter 8, Problem 15RE

a.

To determine

State the null and alternate hypotheses.

a.

Expert Solution

Answer to Problem 15RE

Null hypothesis: $H0:p=0.23$ .

Alternate hypothesis: $H1:p>0.23$ .

Explanation of Solution

The Pew Research Center conducted the survey among the students at University who watch cable news regularly. Of 200 students, 66 were found to watch cable news regularly.

Denote p as the true population proportion of the students who watch cable news regularly.

The given test hypotheses are:

Null hypothesis:

$H0:p=0.23$ .

That is, the proportion of the students who watch cable news regularly is 0.23.

Alternate hypothesis:

$H1:p>0.23$ .

That is, the proportion of the students who watch cable news regularly is greater than 0.23.

b.

To determine

Find the value of test statistic.

b.

Expert Solution

Answer to Problem 15RE

The value of test statistic is 3.36.

Explanation of Solution

Calculation:

Assumptions for performing a hypothesis test for a population proportion:

• The samples taken from the population are simple random samples.
• The population is at least 20 times as large as the sample.
• The samples in the population are divided into two categories.
• The values of $np^0$ and $nq^0$ should be greater than or equal to 10.

Requirement check:

• The sample of 200 students is simple random samples.
• The information about the population is not known. However, the population size is assumed to be more than 20 times as large as the sample.
• The samples in the population seemed to be categorized into two parts. That is, students who watch cable news and students who do not watch cable news.
• Verify the condition: $np^0≥10$ and $nq^0≥10$

Here,

$p^0=66200=0.33$

Substitute n as 200 and $p^0$ as 0.33,

$np^0=200(0.33)=66>10$

Substitute n as 200 and $q^0$ as $0.67(=1−p^0)$ ,

$nq^0=200(0.67)=134>10$

Therefore, all the conditions are satisfied.

Test statistic:

The z-test statistic is:

$z=p^−p0p0(1−p0)n$ ,

Where, $p^$ be the sample proportion, $p0$ be the population proportion and n be the sample size.

Software procedure:

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

• Choose Stat > Basic Statistics > 1 Proportion.
• Choose Summarized data.
• In Number of events, enter 66. In Number of trials, enter 200.
• Check Perform hypothesis test. In Hypothesized proportion, enter 0.23.
• Click Options. Under Alternative, and choose $Proportion>Hypothesized proportion$ .
• Click OK in each dialog box.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8, Problem 15RE

From the MINITAB output, the test statistic, that is, the z-value is 3.36.

Thus, the value of test statistic z is 3.36.

c.

To determine

Decide whether the null hypothesis $H0$ is rejected at $α=0.01$ level.

c.

Expert Solution

Answer to Problem 15RE

The null hypothesis $H0$ is rejected at $α=0.01$ level.

Explanation of Solution

From previous part (b), it has been found that the value of test statistic z is 3.36.

From the given hypotheses, the alternative hypothesis contains the greater $(>)$ symbol. Thus, it is clear that the hypothesis follows the Two-tailed test.

From Table 8.1 “Table of Critical Values”, the critical values for right-tailed test at $α=0.01$ level are 2.326.

Decision based on the critical value method:

For left-tailed test: If $z≤−zα$ , reject $H0$ .

For right-tailed test: If $z≥zα$ , reject $H0$ .

For two-tailed test: If $z≤−zα2 or z≥zα2$ , reject $H0$ .

Conclusion:

The critical value at $α=0.01$ level is 2.326 and the test statistic is 3.36.

Here, the test statistic value of 3.36 lies in the critical region.

That is, $z(=3.36)>zα(=2.326)$ .

Therefore, the null hypothesis is rejected.

d.

To determine

State a conclusion.

d.

Expert Solution

Answer to Problem 15RE

The conclusion is that, there is evidence that the proportion of the students who watch cable news regularly is greater than 0.23.

Explanation of Solution

From previous part (c), it has been found that the null hypothesis is rejected.

Hence, there is evidence that the true proportion of the students who watch cable news regularly is greater than 0.23.

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Answer 5

Chapter 8, Problem 15RE

a.

To determine

State the null and alternate hypotheses.

a.

Expert Solution

Answer to Problem 15RE

Null hypothesis: $H0:p=0.23$ .

Alternate hypothesis: $H1:p>0.23$ .

Explanation of Solution

The Pew Research Center conducted the survey among the students at University who watch cable news regularly. Of 200 students, 66 were found to watch cable news regularly.

Denote p as the true population proportion of the students who watch cable news regularly.

The given test hypotheses are:

Null hypothesis:

$H0:p=0.23$ .

That is, the proportion of the students who watch cable news regularly is 0.23.

Alternate hypothesis:

$H1:p>0.23$ .

That is, the proportion of the students who watch cable news regularly is greater than 0.23.

b.

To determine

Find the value of test statistic.

b.

Expert Solution

Answer to Problem 15RE

The value of test statistic is 3.36.

Explanation of Solution

Calculation:

Assumptions for performing a hypothesis test for a population proportion:

• The samples taken from the population are simple random samples.
• The population is at least 20 times as large as the sample.
• The samples in the population are divided into two categories.
• The values of $np^0$ and $nq^0$ should be greater than or equal to 10.

Requirement check:

• The sample of 200 students is simple random samples.
• The information about the population is not known. However, the population size is assumed to be more than 20 times as large as the sample.
• The samples in the population seemed to be categorized into two parts. That is, students who watch cable news and students who do not watch cable news.
• Verify the condition: $np^0≥10$ and $nq^0≥10$

Here,

$p^0=66200=0.33$

Substitute n as 200 and $p^0$ as 0.33,

$np^0=200(0.33)=66>10$

Substitute n as 200 and $q^0$ as $0.67(=1−p^0)$ ,

$nq^0=200(0.67)=134>10$

Therefore, all the conditions are satisfied.

Test statistic:

The z-test statistic is:

$z=p^−p0p0(1−p0)n$ ,

Where, $p^$ be the sample proportion, $p0$ be the population proportion and n be the sample size.

Software procedure:

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

• Choose Stat > Basic Statistics > 1 Proportion.
• Choose Summarized data.
• In Number of events, enter 66. In Number of trials, enter 200.
• Check Perform hypothesis test. In Hypothesized proportion, enter 0.23.
• Click Options. Under Alternative, and choose $Proportion>Hypothesized proportion$ .
• Click OK in each dialog box.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8, Problem 15RE

From the MINITAB output, the test statistic, that is, the z-value is 3.36.

Thus, the value of test statistic z is 3.36.

c.

To determine

Decide whether the null hypothesis $H0$ is rejected at $α=0.01$ level.

c.

Expert Solution

Answer to Problem 15RE

The null hypothesis $H0$ is rejected at $α=0.01$ level.

Explanation of Solution

From previous part (b), it has been found that the value of test statistic z is 3.36.

From the given hypotheses, the alternative hypothesis contains the greater $(>)$ symbol. Thus, it is clear that the hypothesis follows the Two-tailed test.

From Table 8.1 “Table of Critical Values”, the critical values for right-tailed test at $α=0.01$ level are 2.326.

Decision based on the critical value method:

For left-tailed test: If $z≤−zα$ , reject $H0$ .

For right-tailed test: If $z≥zα$ , reject $H0$ .

For two-tailed test: If $z≤−zα2 or z≥zα2$ , reject $H0$ .

Conclusion:

The critical value at $α=0.01$ level is 2.326 and the test statistic is 3.36.

Here, the test statistic value of 3.36 lies in the critical region.

That is, $z(=3.36)>zα(=2.326)$ .

Therefore, the null hypothesis is rejected.

d.

To determine

State a conclusion.

d.

Expert Solution

Answer to Problem 15RE

The conclusion is that, there is evidence that the proportion of the students who watch cable news regularly is greater than 0.23.

Explanation of Solution

From previous part (c), it has been found that the null hypothesis is rejected.

Hence, there is evidence that the true proportion of the students who watch cable news regularly is greater than 0.23.

Answer 6

Chapter 8, Problem 15RE

a.

To determine

State the null and alternate hypotheses.

a.

Expert Solution

Answer to Problem 15RE

Null hypothesis: $H0:p=0.23$ .

Alternate hypothesis: $H1:p>0.23$ .

Explanation of Solution

The Pew Research Center conducted the survey among the students at University who watch cable news regularly. Of 200 students, 66 were found to watch cable news regularly.

Denote p as the true population proportion of the students who watch cable news regularly.

The given test hypotheses are:

Null hypothesis:

$H0:p=0.23$ .

That is, the proportion of the students who watch cable news regularly is 0.23.

Alternate hypothesis:

$H1:p>0.23$ .

That is, the proportion of the students who watch cable news regularly is greater than 0.23.

Answer 7

Chapter 8, Problem 15RE

a.

To determine

State the null and alternate hypotheses.

Answer 8

a.

Expert Solution

Answer to Problem 15RE

Null hypothesis: $H0:p=0.23$ .

Alternate hypothesis: $H1:p>0.23$ .

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The Pew Research Center conducted the survey among the students at University who watch cable news regularly. Of 200 students, 66 were found to watch cable news regularly.

Denote p as the true population proportion of the students who watch cable news regularly.

The given test hypotheses are:

Null hypothesis:

$H0:p=0.23$ .

That is, the proportion of the students who watch cable news regularly is 0.23.

Alternate hypothesis:

$H1:p>0.23$ .

That is, the proportion of the students who watch cable news regularly is greater than 0.23.

Answer 13

b.

To determine

Find the value of test statistic.

b.

Expert Solution

Answer to Problem 15RE

The value of test statistic is 3.36.

Explanation of Solution

Calculation:

Assumptions for performing a hypothesis test for a population proportion:

• The samples taken from the population are simple random samples.
• The population is at least 20 times as large as the sample.
• The samples in the population are divided into two categories.
• The values of $np^0$ and $nq^0$ should be greater than or equal to 10.

Requirement check:

• The sample of 200 students is simple random samples.
• The information about the population is not known. However, the population size is assumed to be more than 20 times as large as the sample.
• The samples in the population seemed to be categorized into two parts. That is, students who watch cable news and students who do not watch cable news.
• Verify the condition: $np^0≥10$ and $nq^0≥10$

Here,

$p^0=66200=0.33$

Substitute n as 200 and $p^0$ as 0.33,

$np^0=200(0.33)=66>10$

Substitute n as 200 and $q^0$ as $0.67(=1−p^0)$ ,

$nq^0=200(0.67)=134>10$

Therefore, all the conditions are satisfied.

Test statistic:

The z-test statistic is:

$z=p^−p0p0(1−p0)n$ ,

Where, $p^$ be the sample proportion, $p0$ be the population proportion and n be the sample size.

Software procedure:

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

• Choose Stat > Basic Statistics > 1 Proportion.
• Choose Summarized data.
• In Number of events, enter 66. In Number of trials, enter 200.
• Check Perform hypothesis test. In Hypothesized proportion, enter 0.23.
• Click Options. Under Alternative, and choose $Proportion>Hypothesized proportion$ .
• Click OK in each dialog box.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8, Problem 15RE

From the MINITAB output, the test statistic, that is, the z-value is 3.36.

Thus, the value of test statistic z is 3.36.

Answer 14

b.

To determine

Find the value of test statistic.

Answer 15

b.

Expert Solution

Answer to Problem 15RE

The value of test statistic is 3.36.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

Assumptions for performing a hypothesis test for a population proportion:

• The samples taken from the population are simple random samples.
• The population is at least 20 times as large as the sample.
• The samples in the population are divided into two categories.
• The values of $np^0$ and $nq^0$ should be greater than or equal to 10.

Requirement check:

• The sample of 200 students is simple random samples.
• The information about the population is not known. However, the population size is assumed to be more than 20 times as large as the sample.
• The samples in the population seemed to be categorized into two parts. That is, students who watch cable news and students who do not watch cable news.
• Verify the condition: $np^0≥10$ and $nq^0≥10$

Here,

$p^0=66200=0.33$

Substitute n as 200 and $p^0$ as 0.33,

$np^0=200(0.33)=66>10$

Substitute n as 200 and $q^0$ as $0.67(=1−p^0)$ ,

$nq^0=200(0.67)=134>10$

Therefore, all the conditions are satisfied.

Test statistic:

The z-test statistic is:

$z=p^−p0p0(1−p0)n$ ,

Where, $p^$ be the sample proportion, $p0$ be the population proportion and n be the sample size.

Software procedure:

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

• Choose Stat > Basic Statistics > 1 Proportion.
• Choose Summarized data.
• In Number of events, enter 66. In Number of trials, enter 200.
• Check Perform hypothesis test. In Hypothesized proportion, enter 0.23.
• Click Options. Under Alternative, and choose $Proportion>Hypothesized proportion$ .
• Click OK in each dialog box.

The output using Minitab is given below:

ALEKS 360 ESSENT. STAT ACCESS CARD, Chapter 8, Problem 15RE

From the MINITAB output, the test statistic, that is, the z-value is 3.36.

Thus, the value of test statistic z is 3.36.

Answer 20

c.

To determine

Decide whether the null hypothesis $H0$ is rejected at $α=0.01$ level.

c.

Expert Solution

Answer to Problem 15RE

The null hypothesis $H0$ is rejected at $α=0.01$ level.

Explanation of Solution

From previous part (b), it has been found that the value of test statistic z is 3.36.

From the given hypotheses, the alternative hypothesis contains the greater $(>)$ symbol. Thus, it is clear that the hypothesis follows the Two-tailed test.

From Table 8.1 “Table of Critical Values”, the critical values for right-tailed test at $α=0.01$ level are 2.326.

Decision based on the critical value method:

For left-tailed test: If $z≤−zα$ , reject $H0$ .

For right-tailed test: If $z≥zα$ , reject $H0$ .

For two-tailed test: If $z≤−zα2 or z≥zα2$ , reject $H0$ .

Conclusion:

The critical value at $α=0.01$ level is 2.326 and the test statistic is 3.36.

Here, the test statistic value of 3.36 lies in the critical region.

That is, $z(=3.36)>zα(=2.326)$ .

Therefore, the null hypothesis is rejected.

Answer 21

c.

To determine

Decide whether the null hypothesis $H0$ is rejected at $α=0.01$ level.

Answer 22

c.

Expert Solution

Answer to Problem 15RE

The null hypothesis $H0$ is rejected at $α=0.01$ level.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

From previous part (b), it has been found that the value of test statistic z is 3.36.

From the given hypotheses, the alternative hypothesis contains the greater $(>)$ symbol. Thus, it is clear that the hypothesis follows the Two-tailed test.

From Table 8.1 “Table of Critical Values”, the critical values for right-tailed test at $α=0.01$ level are 2.326.

Decision based on the critical value method:

For left-tailed test: If $z≤−zα$ , reject $H0$ .

For right-tailed test: If $z≥zα$ , reject $H0$ .

For two-tailed test: If $z≤−zα2 or z≥zα2$ , reject $H0$ .

Conclusion:

The critical value at $α=0.01$ level is 2.326 and the test statistic is 3.36.

Here, the test statistic value of 3.36 lies in the critical region.

That is, $z(=3.36)>zα(=2.326)$ .

Therefore, the null hypothesis is rejected.

Answer 27

d.

To determine

State a conclusion.

d.

Expert Solution

Answer to Problem 15RE

The conclusion is that, there is evidence that the proportion of the students who watch cable news regularly is greater than 0.23.

Explanation of Solution

From previous part (c), it has been found that the null hypothesis is rejected.

Hence, there is evidence that the true proportion of the students who watch cable news regularly is greater than 0.23.

Answer 28

d.

To determine

State a conclusion.

Answer 29

d.

Expert Solution

Answer to Problem 15RE

The conclusion is that, there is evidence that the proportion of the students who watch cable news regularly is greater than 0.23.

Answer 30

d.

Expert Solution

Answer 31

d.

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

From previous part (c), it has been found that the null hypothesis is rejected.

Hence, there is evidence that the true proportion of the students who watch cable news regularly is greater than 0.23.

Answer 34

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...

State the null and alternate hypotheses.

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State the null and alternate hypotheses.

Answer to Problem 15RE

Explanation of Solution

Find the value of test statistic.

Answer to Problem 15RE

Explanation of Solution

Decide whether the null hypothesis $H0$ is rejected at $α=0.01$ level.

Answer to Problem 15RE

Explanation of Solution

State a conclusion.

Answer to Problem 15RE

Explanation of Solution

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Chapter 8 Solutions

State the null and alternate hypotheses.

Concept explainers

Videos

State the null and alternate hypotheses.

Answer to Problem 15RE

Explanation of Solution

Find the value of test statistic.

Answer to Problem 15RE

Explanation of Solution

Decide 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 α=0.01<m:math><m:mrow><m:mi>α</m:mi><m:mo>=</m:mo><m:mn>0.01</m:mn></m:mrow></m:math> level.

Answer to Problem 15RE

Explanation of Solution

State a conclusion.

Answer to Problem 15RE

Explanation of Solution

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Chapter 8 Solutions

Decide whether the null hypothesis $H0$ is rejected at $α=0.01$ level.