Chapter 8, Problem 93E

To determine

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

Answer to Problem 93E

Solution: The test statistic is obtained as $\underset{\_}{z=2.25}$ and P-value is obtained as $\underset{\_}{P-\text{value}=0.025}$.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of $n_1\text{and }{n}_2$ and hence $n_1=60,n_2=60$. The hypotheses are provided as:

$\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1\ne p_2\end{array}$

The proportions are provided as ${\widehat{p}}_1=0.65,{\widehat{p}}_2=0.45$, hence X₁ and X₂ are obtained as:

$\begin{array}{c}{X}_1=0.65\times 60\\ =39\end{array}$

$\begin{array}{c}{X}_2=0.45\times 60\\ =27\end{array}$

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 $P-\text{value}=0.025$.

To determine

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

Answer to Problem 93E

Solution: The test statistic is obtained as $\underset{\_}{z=2.43}$ and P-value is obtained as $\underset{\_}{P-\text{value}=0.015}$.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of $n_1\text{and }{n}_2$ and hence $n_1=70,n_2=70$. The hypotheses are provided as:

$\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1\ne p_2\end{array}$

The proportions are provided as ${\widehat{p}}_1=0.65,{\widehat{p}}_2=0.45$, hence X₁ and X₂ are obtained as:

$\begin{array}{c}{X}_1=0.65\times 70\\ =45.5\\ \approx 46\end{array}$

$\begin{array}{c}{X}_2=0.45\times 70\\ =31.5\\ \approx 32\end{array}$

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 $P-\text{value}=0.015$.

To determine

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

Answer to Problem 93E

Solution: The test statistic is obtained as $\underset{\_}{z=2.60}$ and P-value is obtained as $\underset{\_}{P-\text{value}=0.009}$.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of $n_1\text{and }{n}_2$ and hence $n_1=80,n_2=80$. The hypotheses are provided as:

$\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1\ne p_2\end{array}$

The proportions are provided as ${\widehat{p}}_1=0.65,{\widehat{p}}_2=0.45$, hence X₁ and X₂ are obtained as:

$\begin{array}{c}{X}_1=0.65\times 80\\ =52\end{array}$

$\begin{array}{c}{X}_2=0.45\times 80\\ =36\end{array}$

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 $P-\text{value}=0.009$.

To determine

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

Answer to Problem 93E

Solution: The test statistic is obtained as $\underset{\_}{z=2.90}$ and P-value is obtained as $\underset{\_}{P-\text{value}=0.004}$.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of $n_1\text{and }{n}_2$ and hence $n_1=100,n_2=100$. The hypotheses are provided as:

$\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1\ne p_2\end{array}$

The proportions are provided as ${\widehat{p}}_1=0.65,{\widehat{p}}_2=0.45$, hence X₁ and X₂ are obtained as:

$\begin{array}{c}{X}_1=0.65\times 100\\ =65\end{array}$

$\begin{array}{c}{X}_2=0.45\times 100\\ =45\end{array}$

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 $P-\text{value}=0.004$.

To determine

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

Answer to Problem 93E

Solution: The test statistic is obtained as $\underset{\_}{z=5.80}$ and P-value is obtained as $\underset{\_}{P-\text{value}=0.000}$.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of $n_1\text{and }{n}_2$ and hence $n_1=400,n_2=400$. The hypotheses are provided as:

$\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1\ne p_2\end{array}$

The proportions are provided as ${\widehat{p}}_1=0.65,{\widehat{p}}_2=0.45$, hence X₁ and X₂ are obtained as:

$\begin{array}{c}{X}_1=0.65\times 400\\ =260\end{array}$

$\begin{array}{c}{X}_2=0.45\times 400\\ =180\end{array}$

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 $P-\text{value}=0.000$.

To determine

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

Answer to Problem 93E

Solution: The test statistic is obtained as $\underset{\_}{z=6.49}$ and P-value is obtained as $\underset{\_}{P-\text{value}=0.000}$.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of $n_1\text{and }{n}_2$ and hence $n_1=500,n_2=500$. The hypotheses are provided as:

$\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1\ne p_2\end{array}$

The proportions are provided as ${\widehat{p}}_1=0.65,{\widehat{p}}_2=0.45$, hence X₁ and X₂ are obtained as:

$\begin{array}{c}{X}_1=0.65\times 500\\ =325\end{array}$

$\begin{array}{c}{X}_2=0.45\times 500\\ =225\end{array}$

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 $P-\text{value}=0.000$.

To determine

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

Answer to Problem 93E

Solution: The test statistic is obtained as $\underset{\_}{z=9.18}$ and P-value is obtained as $\underset{\_}{P-\text{value}=0.000}$.

Explanation of Solution

Calculation:

The assumption made that the sample size n is the common value of $n_1\text{and }{n}_2$ and hence $n_1=1000,n_2=1000$. The hypotheses are provided as:

$\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1\ne p_2\end{array}$

The proportions are provided as ${\widehat{p}}_1=0.65,{\widehat{p}}_2=0.45$, hence X₁ and X₂ are obtained as:

$\begin{array}{c}{X}_1=0.65\times 1000\\ =650\end{array}$

$\begin{array}{c}{X}_2=0.45\times 1000\\ =450\end{array}$

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 $P-\text{value}=0.000$.

To determine

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

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.

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.