The test statistic and the corresponding P- value for the sample of size n = 60 .

Question

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Answer 1

Question

Chapter 8, Problem 93E

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=60.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.25_ and P-value is obtained as P−value=0.025_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=60,n2=60. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×60=39

X2=0.45×60=27

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.25 and the P-value is obtained as P−value=0.025.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=70.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.43_ and P-value is obtained as P−value=0.015_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=70,n2=70. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×70=45.5≈46

X2=0.45×70=31.5≈32

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.43 and the P-value is obtained as P−value=0.015.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=80.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.60_ and P-value is obtained as P−value=0.009_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=80,n2=80. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×80=52

X2=0.45×80=36

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.60 and the P-value is obtained as P−value=0.009.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=100.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.90_ and P-value is obtained as P−value=0.004_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=100,n2=100. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×100=65

X2=0.45×100=45

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.90 and the P-value is obtained as P−value=0.004.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=400.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=5.80_ and P-value is obtained as P−value=0.000_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=400,n2=400. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×400=260

X2=0.45×400=180

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=5.80 and the P-value is obtained as P−value=0.000.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=500.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=6.49_ and P-value is obtained as P−value=0.000_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=500,n2=500. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×500=325

X2=0.45×500=225

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=6.49 and the P-value is obtained as P−value=0.000.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=1000.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=9.18_ and P-value is obtained as P−value=0.000_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=1000,n2=1000. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×1000=650

X2=0.45×1000=450

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=9.18 and the P-value is obtained as P−value=0.000.

To determine

To explain: The obtained results of test statistic and the corresponding P-value.

Expert Solution

Answer to Problem 93E

Solution: The obtained results of test statistic and the corresponding P- value are summarized as:

n	z	P
60	2.25	0.025
70	2.43	0.015
80	2.60	0.009
100	2.90	0.004
400	5.80	0.000
500	6.49	0.000
1000	9.18	0.000

Explanation of Solution

The statistic and the associated P-value is obtained for different sample sizes and tabulated as below.

n	z	P
60	2.25	0.025
70	2.43	0.015
80	2.60	0.009
100	2.90	0.004
400	5.80	0.000
500	6.49	0.000
1000	9.18	0.000

To determine

To explain: The effect of the sample size on the statistical significance for the same sample proportions.

Expert Solution

Answer to Problem 93E

Solution: The difference of proportions becomes more significant as the sample size increases.

Explanation of Solution

The different values for the statistic and the associated P-value are obtained for different sample sizes.

The obtained values show that for small sample sizes, the results are not significant but for larger sample sizes, the result are significant and show that the sample proportions are significantly different.

Therefore, conclude that the difference becomes more significant as the sample size increases.

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Answer 2

Question

Chapter 8, Problem 93E

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=60.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.25_ and P-value is obtained as P−value=0.025_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=60,n2=60. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×60=39

X2=0.45×60=27

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.25 and the P-value is obtained as P−value=0.025.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=70.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.43_ and P-value is obtained as P−value=0.015_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=70,n2=70. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×70=45.5≈46

X2=0.45×70=31.5≈32

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.43 and the P-value is obtained as P−value=0.015.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=80.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.60_ and P-value is obtained as P−value=0.009_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=80,n2=80. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×80=52

X2=0.45×80=36

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.60 and the P-value is obtained as P−value=0.009.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=100.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.90_ and P-value is obtained as P−value=0.004_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=100,n2=100. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×100=65

X2=0.45×100=45

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.90 and the P-value is obtained as P−value=0.004.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=400.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=5.80_ and P-value is obtained as P−value=0.000_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=400,n2=400. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×400=260

X2=0.45×400=180

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=5.80 and the P-value is obtained as P−value=0.000.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=500.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=6.49_ and P-value is obtained as P−value=0.000_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=500,n2=500. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×500=325

X2=0.45×500=225

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=6.49 and the P-value is obtained as P−value=0.000.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=1000.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=9.18_ and P-value is obtained as P−value=0.000_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=1000,n2=1000. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×1000=650

X2=0.45×1000=450

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=9.18 and the P-value is obtained as P−value=0.000.

To determine

To explain: The obtained results of test statistic and the corresponding P-value.

Expert Solution

Answer to Problem 93E

Solution: The obtained results of test statistic and the corresponding P- value are summarized as:

n	z	P
60	2.25	0.025
70	2.43	0.015
80	2.60	0.009
100	2.90	0.004
400	5.80	0.000
500	6.49	0.000
1000	9.18	0.000

Explanation of Solution

The statistic and the associated P-value is obtained for different sample sizes and tabulated as below.

n	z	P
60	2.25	0.025
70	2.43	0.015
80	2.60	0.009
100	2.90	0.004
400	5.80	0.000
500	6.49	0.000
1000	9.18	0.000

To determine

To explain: The effect of the sample size on the statistical significance for the same sample proportions.

Expert Solution

Answer to Problem 93E

Solution: The difference of proportions becomes more significant as the sample size increases.

Explanation of Solution

The different values for the statistic and the associated P-value are obtained for different sample sizes.

The obtained values show that for small sample sizes, the results are not significant but for larger sample sizes, the result are significant and show that the sample proportions are significantly different.

Therefore, conclude that the difference becomes more significant as the sample size increases.

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Answer 3

Question

Chapter 8, Problem 93E

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=60.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.25_ and P-value is obtained as P−value=0.025_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=60,n2=60. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×60=39

X2=0.45×60=27

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.25 and the P-value is obtained as P−value=0.025.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=70.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.43_ and P-value is obtained as P−value=0.015_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=70,n2=70. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×70=45.5≈46

X2=0.45×70=31.5≈32

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.43 and the P-value is obtained as P−value=0.015.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=80.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.60_ and P-value is obtained as P−value=0.009_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=80,n2=80. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×80=52

X2=0.45×80=36

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.60 and the P-value is obtained as P−value=0.009.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=100.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.90_ and P-value is obtained as P−value=0.004_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=100,n2=100. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×100=65

X2=0.45×100=45

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.90 and the P-value is obtained as P−value=0.004.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=400.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=5.80_ and P-value is obtained as P−value=0.000_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=400,n2=400. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×400=260

X2=0.45×400=180

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=5.80 and the P-value is obtained as P−value=0.000.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=500.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=6.49_ and P-value is obtained as P−value=0.000_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=500,n2=500. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×500=325

X2=0.45×500=225

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=6.49 and the P-value is obtained as P−value=0.000.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=1000.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=9.18_ and P-value is obtained as P−value=0.000_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=1000,n2=1000. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×1000=650

X2=0.45×1000=450

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=9.18 and the P-value is obtained as P−value=0.000.

To determine

To explain: The obtained results of test statistic and the corresponding P-value.

Expert Solution

Answer to Problem 93E

Solution: The obtained results of test statistic and the corresponding P- value are summarized as:

n	z	P
60	2.25	0.025
70	2.43	0.015
80	2.60	0.009
100	2.90	0.004
400	5.80	0.000
500	6.49	0.000
1000	9.18	0.000

Explanation of Solution

The statistic and the associated P-value is obtained for different sample sizes and tabulated as below.

n	z	P
60	2.25	0.025
70	2.43	0.015
80	2.60	0.009
100	2.90	0.004
400	5.80	0.000
500	6.49	0.000
1000	9.18	0.000

To determine

To explain: The effect of the sample size on the statistical significance for the same sample proportions.

Expert Solution

Answer to Problem 93E

Solution: The difference of proportions becomes more significant as the sample size increases.

Explanation of Solution

The different values for the statistic and the associated P-value are obtained for different sample sizes.

The obtained values show that for small sample sizes, the results are not significant but for larger sample sizes, the result are significant and show that the sample proportions are significantly different.

Therefore, conclude that the difference becomes more significant as the sample size increases.

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Answer 4

Question

Chapter 8, Problem 93E

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=60.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.25_ and P-value is obtained as P−value=0.025_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=60,n2=60. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×60=39

X2=0.45×60=27

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.25 and the P-value is obtained as P−value=0.025.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=70.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.43_ and P-value is obtained as P−value=0.015_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=70,n2=70. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×70=45.5≈46

X2=0.45×70=31.5≈32

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.43 and the P-value is obtained as P−value=0.015.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=80.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.60_ and P-value is obtained as P−value=0.009_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=80,n2=80. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×80=52

X2=0.45×80=36

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.60 and the P-value is obtained as P−value=0.009.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=100.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.90_ and P-value is obtained as P−value=0.004_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=100,n2=100. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×100=65

X2=0.45×100=45

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.90 and the P-value is obtained as P−value=0.004.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=400.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=5.80_ and P-value is obtained as P−value=0.000_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=400,n2=400. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×400=260

X2=0.45×400=180

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=5.80 and the P-value is obtained as P−value=0.000.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=500.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=6.49_ and P-value is obtained as P−value=0.000_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=500,n2=500. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×500=325

X2=0.45×500=225

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=6.49 and the P-value is obtained as P−value=0.000.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=1000.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=9.18_ and P-value is obtained as P−value=0.000_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=1000,n2=1000. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×1000=650

X2=0.45×1000=450

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=9.18 and the P-value is obtained as P−value=0.000.

To determine

To explain: The obtained results of test statistic and the corresponding P-value.

Expert Solution

Answer to Problem 93E

Solution: The obtained results of test statistic and the corresponding P- value are summarized as:

n	z	P
60	2.25	0.025
70	2.43	0.015
80	2.60	0.009
100	2.90	0.004
400	5.80	0.000
500	6.49	0.000
1000	9.18	0.000

Explanation of Solution

The statistic and the associated P-value is obtained for different sample sizes and tabulated as below.

n	z	P
60	2.25	0.025
70	2.43	0.015
80	2.60	0.009
100	2.90	0.004
400	5.80	0.000
500	6.49	0.000
1000	9.18	0.000

To determine

To explain: The effect of the sample size on the statistical significance for the same sample proportions.

Expert Solution

Answer to Problem 93E

Solution: The difference of proportions becomes more significant as the sample size increases.

Explanation of Solution

The different values for the statistic and the associated P-value are obtained for different sample sizes.

The obtained values show that for small sample sizes, the results are not significant but for larger sample sizes, the result are significant and show that the sample proportions are significantly different.

Therefore, conclude that the difference becomes more significant as the sample size increases.

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Answer 5

Chapter 8, Problem 93E

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=60.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.25_ and P-value is obtained as P−value=0.025_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=60,n2=60. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×60=39

X2=0.45×60=27

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.25 and the P-value is obtained as P−value=0.025.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=70.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.43_ and P-value is obtained as P−value=0.015_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=70,n2=70. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×70=45.5≈46

X2=0.45×70=31.5≈32

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.43 and the P-value is obtained as P−value=0.015.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=80.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.60_ and P-value is obtained as P−value=0.009_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=80,n2=80. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×80=52

X2=0.45×80=36

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.60 and the P-value is obtained as P−value=0.009.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=100.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.90_ and P-value is obtained as P−value=0.004_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=100,n2=100. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×100=65

X2=0.45×100=45

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.90 and the P-value is obtained as P−value=0.004.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=400.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=5.80_ and P-value is obtained as P−value=0.000_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=400,n2=400. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×400=260

X2=0.45×400=180

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=5.80 and the P-value is obtained as P−value=0.000.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=500.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=6.49_ and P-value is obtained as P−value=0.000_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=500,n2=500. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×500=325

X2=0.45×500=225

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=6.49 and the P-value is obtained as P−value=0.000.

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=1000.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=9.18_ and P-value is obtained as P−value=0.000_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=1000,n2=1000. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×1000=650

X2=0.45×1000=450

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=9.18 and the P-value is obtained as P−value=0.000.

To determine

To explain: The obtained results of test statistic and the corresponding P-value.

Expert Solution

Answer to Problem 93E

Solution: The obtained results of test statistic and the corresponding P- value are summarized as:

n	z	P
60	2.25	0.025
70	2.43	0.015
80	2.60	0.009
100	2.90	0.004
400	5.80	0.000
500	6.49	0.000
1000	9.18	0.000

Explanation of Solution

The statistic and the associated P-value is obtained for different sample sizes and tabulated as below.

n	z	P
60	2.25	0.025
70	2.43	0.015
80	2.60	0.009
100	2.90	0.004
400	5.80	0.000
500	6.49	0.000
1000	9.18	0.000

To determine

To explain: The effect of the sample size on the statistical significance for the same sample proportions.

Expert Solution

Answer to Problem 93E

Solution: The difference of proportions becomes more significant as the sample size increases.

Explanation of Solution

The different values for the statistic and the associated P-value are obtained for different sample sizes.

The obtained values show that for small sample sizes, the results are not significant but for larger sample sizes, the result are significant and show that the sample proportions are significantly different.

Therefore, conclude that the difference becomes more significant as the sample size increases.

Answer 6

Chapter 8, Problem 93E

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=60.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.25_ and P-value is obtained as P−value=0.025_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=60,n2=60. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×60=39

X2=0.45×60=27

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.25 and the P-value is obtained as P−value=0.025.

Answer 7

Chapter 8, Problem 93E

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=60.

Answer 8

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.25_ and P-value is obtained as P−value=0.025_.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=60,n2=60. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×60=39

X2=0.45×60=27

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.25 and the P-value is obtained as P−value=0.025.

Answer 13

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=70.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.43_ and P-value is obtained as P−value=0.015_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=70,n2=70. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×70=45.5≈46

X2=0.45×70=31.5≈32

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.43 and the P-value is obtained as P−value=0.015.

Answer 14

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=70.

Answer 15

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.43_ and P-value is obtained as P−value=0.015_.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=70,n2=70. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×70=45.5≈46

X2=0.45×70=31.5≈32

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.43 and the P-value is obtained as P−value=0.015.

Answer 20

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=80.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.60_ and P-value is obtained as P−value=0.009_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=80,n2=80. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×80=52

X2=0.45×80=36

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.60 and the P-value is obtained as P−value=0.009.

Answer 21

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=80.

Answer 22

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.60_ and P-value is obtained as P−value=0.009_.

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=80,n2=80. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×80=52

X2=0.45×80=36

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.60 and the P-value is obtained as P−value=0.009.

Answer 27

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=100.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.90_ and P-value is obtained as P−value=0.004_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=100,n2=100. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×100=65

X2=0.45×100=45

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.90 and the P-value is obtained as P−value=0.004.

Answer 28

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=100.

Answer 29

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=2.90_ and P-value is obtained as P−value=0.004_.

Answer 30

Expert Solution

Answer 31

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=100,n2=100. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×100=65

X2=0.45×100=45

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=2.90 and the P-value is obtained as P−value=0.004.

Answer 34

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=400.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=5.80_ and P-value is obtained as P−value=0.000_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=400,n2=400. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×400=260

X2=0.45×400=180

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=5.80 and the P-value is obtained as P−value=0.000.

Answer 35

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=400.

Answer 36

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=5.80_ and P-value is obtained as P−value=0.000_.

Answer 37

Expert Solution

Answer 38

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=400,n2=400. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×400=260

X2=0.45×400=180

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=5.80 and the P-value is obtained as P−value=0.000.

Answer 41

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=500.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=6.49_ and P-value is obtained as P−value=0.000_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=500,n2=500. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×500=325

X2=0.45×500=225

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=6.49 and the P-value is obtained as P−value=0.000.

Answer 42

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=500.

Answer 43

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=6.49_ and P-value is obtained as P−value=0.000_.

Answer 44

Expert Solution

Answer 45

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=500,n2=500. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×500=325

X2=0.45×500=225

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=6.49 and the P-value is obtained as P−value=0.000.

Answer 48

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=1000.

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=9.18_ and P-value is obtained as P−value=0.000_.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=1000,n2=1000. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×1000=650

X2=0.45×1000=450

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=9.18 and the P-value is obtained as P−value=0.000.

Answer 49

To determine

To find: The test statistic and the corresponding P- value for the sample of size n=1000.

Answer 50

Expert Solution

Answer to Problem 93E

Solution: The test statistic is obtained as z=9.18_ and P-value is obtained as P−value=0.000_.

Answer 51

Expert Solution

Answer 52

Expert Solution

Answer 53

Expert Solution

Answer 54

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of n1and n2 and hence n1=1000,n2=1000. The hypotheses are provided as:

H0:p1=p2Ha:p1≠p2

The proportions are provided as p^1=0.65,p^2=0.45, hence X₁ and X₂ are obtained as:

X1=0.65×1000=650

X2=0.45×1000=450

To compute the statistic and the corresponding P-value, Minitab is used.

Follow the below mentioned steps in Minitab to perform the significance test,

Step 1: Go to ‘Stat’ then select ‘Basic Statistics’ and click on ‘2-Proportions’.

Step 2: In the dialogue box that appears select Summarized data and enter the provided sample size and number of successes in the respective fields.

Step 3: Click on ‘Options’ button and enter the confidence level as 95.

Step 4: Choose the Alternative as “Not equal” and click on OK twice.

The output shows that the test statistic is obtained as z=9.18 and the P-value is obtained as P−value=0.000.

Answer 55

To determine

To explain: The obtained results of test statistic and the corresponding P-value.

Expert Solution

Answer to Problem 93E

Solution: The obtained results of test statistic and the corresponding P- value are summarized as:

n	z	P
60	2.25	0.025
70	2.43	0.015
80	2.60	0.009
100	2.90	0.004
400	5.80	0.000
500	6.49	0.000
1000	9.18	0.000

Explanation of Solution

The statistic and the associated P-value is obtained for different sample sizes and tabulated as below.

n	z	P
60	2.25	0.025
70	2.43	0.015
80	2.60	0.009
100	2.90	0.004
400	5.80	0.000
500	6.49	0.000
1000	9.18	0.000

Answer 56

To determine

To explain: The obtained results of test statistic and the corresponding P-value.

Answer 57

Expert Solution

Answer to Problem 93E

Solution: The obtained results of test statistic and the corresponding P- value are summarized as:

n	z	P
60	2.25	0.025
70	2.43	0.015
80	2.60	0.009
100	2.90	0.004
400	5.80	0.000
500	6.49	0.000
1000	9.18	0.000

Answer 58

Expert Solution

Answer 59

Expert Solution

Answer 60

Expert Solution

Answer 61

Explanation of Solution

The statistic and the associated P-value is obtained for different sample sizes and tabulated as below.

n	z	P
60	2.25	0.025
70	2.43	0.015
80	2.60	0.009
100	2.90	0.004
400	5.80	0.000
500	6.49	0.000
1000	9.18	0.000

Answer 62

To determine

To explain: The effect of the sample size on the statistical significance for the same sample proportions.

Expert Solution

Answer to Problem 93E

Solution: The difference of proportions becomes more significant as the sample size increases.

Explanation of Solution

The different values for the statistic and the associated P-value are obtained for different sample sizes.

The obtained values show that for small sample sizes, the results are not significant but for larger sample sizes, the result are significant and show that the sample proportions are significantly different.

Therefore, conclude that the difference becomes more significant as the sample size increases.

Answer 63

To determine

To explain: The effect of the sample size on the statistical significance for the same sample proportions.

Answer 64

Expert Solution

Answer to Problem 93E

Solution: The difference of proportions becomes more significant as the sample size increases.

Answer 65

Expert Solution

Answer 66

Expert Solution

Answer 67

Expert Solution

Answer 68

Explanation of Solution

The different values for the statistic and the associated P-value are obtained for different sample sizes.

The obtained values show that for small sample sizes, the results are not significant but for larger sample sizes, the result are significant and show that the sample proportions are significantly different.

Therefore, conclude that the difference becomes more significant as the sample size increases.

Answer 69

Ch. 8.1 - Prob. 1UYK Ch. 8.1 - Prob. 2UYK Ch. 8.1 - Prob. 3UYK Ch. 8.1 - Prob. 4UYK Ch. 8.1 - Prob. 5UYK Ch. 8.1 - Prob. 6UYK Ch. 8.1 - Prob. 7UYK Ch. 8.1 - Prob. 8UYK Ch. 8.1 - Prob. 9UYK Ch. 8.1 - Prob. 10UYK

The test statistic and the corresponding P- value for the sample of size n = 60 .

Concept explainers

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To find: The test statistic and the corresponding P- value for the sample of size n=60.

Answer to Problem 93E

Explanation of Solution

To find: The test statistic and the corresponding P- value for the sample of size n=70.

Answer to Problem 93E

Explanation of Solution

To find: The test statistic and the corresponding P- value for the sample of size n=80.

Answer to Problem 93E

Explanation of Solution

To find: The test statistic and the corresponding P- value for the sample of size n=100.

Answer to Problem 93E

Explanation of Solution

To find: The test statistic and the corresponding P- value for the sample of size n=400.

Answer to Problem 93E

Explanation of Solution

To find: The test statistic and the corresponding P- value for the sample of size n=500.

Answer to Problem 93E

Explanation of Solution

To find: The test statistic and the corresponding P- value for the sample of size n=1000.

Answer to Problem 93E

Explanation of Solution

To explain: The obtained results of test statistic and the corresponding P-value.

Answer to Problem 93E

Explanation of Solution

To explain: The effect of the sample size on the statistical significance for the same sample proportions.

Answer to Problem 93E

Explanation of Solution

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