Chapter 8, Problem 96E

To determine

To find: The 95% margin of error for the difference of two proportions for sample size $n=60$.

Answer to Problem 96E

Solution: The margin of error is 0.1607.

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 $n_1=60\text{and}{n}_2=60$. The margin of error is defined as:

$m=z^{*}\times {\text{SE}}_D$

Where, z^* is the critical value for the standard normal variable and ${\text{SE}}_D$ is the standard error of the difference. Also, the standard error of D is defined as:

${\text{SE}}_D=\sqrt{\frac{{\widehat{p}}_1\left(1-{\widehat{p}}_1\right)}{n_1}+\frac{{\widehat{p}}_2\left(1-{\widehat{p}}_2\right)}{n_2}}$

Where,

$\begin{array}{l}{\widehat{p}}_1=\text{Sample proportion from population 1}\\ {\widehat{p}}_2=\text{Sample proportion from population 2}\\ n_1=\text{Sample size of group 1}\\ n_2=\text{Sample size of group 2}\end{array}$

Substitute the provided values ${\widehat{p}}_1=0.8,{\widehat{p}}_2=0.6$ and $n_1=60,n_2=60$ in the above defined formula to calculate the standard error of D.

$\begin{array}{c}{\text{SE}}_D=\sqrt{\frac{0.8\times \left(1-0.8\right)}{60}+\frac{0.6\times \left(1-0.6\right)}{60}}\\ =\sqrt{0.0027+0.004}\\ =0.082\end{array}$

The value of $z^{*}$ for 95% confidence level is $z^{*}=1.96$ from the standard normal table. So, the margin of error is obtained as:

$\begin{array}{c}m=z^{*}\times {\text{SE}}_{\widehat{D}}\\ =1.96\times 0.082\\ =0.1607\end{array}$

Therefore, the margin of error is obtained as 0.1607.

To determine

To find: The 95% margin of error for the difference of two proportions for sample size $n=70$.

Answer to Problem 96E

Solution: The margin of error is 0.1482.

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 margin of error is defined as:

$m=z^{*}\times {\text{SE}}_D$

Where z^* is the critical value for the standard normal variable and ${\text{SE}}_D$ is the standard error of the difference. Also, the standard error of D is defined as:

${\text{SE}}_D=\sqrt{\frac{{\widehat{p}}_1\left(1-{\widehat{p}}_1\right)}{n_1}+\frac{{\widehat{p}}_2\left(1-{\widehat{p}}_2\right)}{n_2}}$

Where

$\begin{array}{l}{\widehat{p}}_1=\text{Sample proportion from population 1}\\ {\widehat{p}}_2=\text{Sample proportion from population 2}\\ n_1=\text{Sample size of group 1}\\ n_2=\text{Sample size of group 2}\end{array}$

Substitute the provided values ${\widehat{p}}_1=0.8,{\widehat{p}}_2=0.6$ and $n_1=70,n_2=70$ in the above defined formula to calculate the standard error of D.

$\begin{array}{c}{\text{SE}}_D=\sqrt{\frac{0.8\times \left(1-0.8\right)}{70}+\frac{0.6\times \left(1-0.6\right)}{70}}\\ =\sqrt{0.0023+0.0034}\\ =0.0756\end{array}$

The value of $z^{*}$ for 95% confidence level is $z^{*}=1.96$ from the standard normal table. So, the margin of error is obtained as:

$\begin{array}{c}m=z^{*}\times {\text{SE}}_{\widehat{D}}\\ =1.96\times 0.0756\\ =0.1482\end{array}$

Therefore, the margin of error is obtained as 0.1482.

To determine

To find: The 95% margin of error for the difference of two proportions for sample size $n=80$.

Answer to Problem 96E

Solution: The margin of error is 0.1386.

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 $n_1=80\text{and}{n}_2=80$. The margin of error is defined as:

$m=z^{*}\times {\text{SE}}_D$

Where z^* is the critical value for the standard normal variable and ${\text{SE}}_D$ is the standard error of the difference. Also, the standard error of D is defined as:

${\text{SE}}_D=\sqrt{\frac{{\widehat{p}}_1\left(1-{\widehat{p}}_1\right)}{n_1}+\frac{{\widehat{p}}_2\left(1-{\widehat{p}}_2\right)}{n_2}}$

Where

$\begin{array}{l}{\widehat{p}}_1=\text{Sample proportion from population 1}\\ {\widehat{p}}_2=\text{Sample proportion from population 2}\\ n_1=\text{Sample size of group 1}\\ n_2=\text{Sample size of group 2}\end{array}$

Substitute the provided values ${\widehat{p}}_1=0.8,{\widehat{p}}_2=0.6$ and $n_1=80,n_2=80$ in the above defined formula to calculate the standard error of D.

$\begin{array}{c}{\text{SE}}_D=\sqrt{\frac{0.8\times \left(1-0.8\right)}{80}+\frac{0.6\times \left(1-0.6\right)}{80}}\\ =0.0707\end{array}$

The value of $z^{*}$ for 95% confidence level is $z^{*}=1.96$ from the standard normal table. So, the margin of error is obtained as:

$\begin{array}{c}m=z^{*}\times {\text{SE}}_{\widehat{D}}\\ =1.96\times 0.0707\\ =0.1386\end{array}$

Therefore, the margin of error is obtained as 0.1386.

To determine

To find: The 95% margin of error for the difference of two proportions for sample size $n=100$.

Answer to Problem 96E

Solution: The margin of error is 0.1239.

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 $n_1=100\text{and}{n}_2=100$. The margin of error for confidence level C is defined as:

$m=z^{*}\times {\text{SE}}_D$

Where z^* is the value for the standard normal density curve and ${\text{SE}}_D$ is the standard error of the difference. Also, the standard error of D is defined as:

${\text{SE}}_D=\sqrt{\frac{{\widehat{p}}_1\left(1-{\widehat{p}}_1\right)}{n_1}+\frac{{\widehat{p}}_2\left(1-{\widehat{p}}_2\right)}{n_2}}$

Where

$\begin{array}{l}{\widehat{p}}_1=\text{Sample proportion from population 1}\\ {\widehat{p}}_2=\text{Sample proportion from population 2}\\ n_1=\text{Sample size of group 1}\\ n_2=\text{Sample size of group 2}\end{array}$

Substitute the provided values ${\widehat{p}}_1=0.8,{\widehat{p}}_2=0.6$ and $n_1=100,n_2=100$ in the above defined formula to calculate the standard error of D.

$\begin{array}{c}{\text{SE}}_D=\sqrt{\frac{0.8\times \left(1-0.8\right)}{100}+\frac{0.6\times \left(1-0.6\right)}{100}}\\ =0.0632\end{array}$

The value of $z^{*}$ for 95% confidence level is $z^{*}=1.96$ from the standard normal table. So, the margin of error is obtained as:

$\begin{array}{c}m=z^{*}\times {\text{SE}}_{\widehat{D}}\\ =1.96\times 0.0632\\ =0.1239\end{array}$

Therefore, the margin of error is obtained as 0.1239.

To determine

To find: The 95% margin of error for the difference of two proportions for sample size $n=400$.

Answer to Problem 96E

Solution: The margin of error is 0.0619.

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 $n_1=400\text{and}{n}_2=400$. The margin of error is calculated as,

$m=z^{*}\times {\text{SE}}_D$

Where z^* is the critical value for the standard normal variable and ${\text{SE}}_D$ is the standard error of the difference. The standard error of D is defined as:

${\text{SE}}_D=\sqrt{\frac{{\widehat{p}}_1\left(1-{\widehat{p}}_1\right)}{n_1}+\frac{{\widehat{p}}_2\left(1-{\widehat{p}}_2\right)}{n_2}}$

Where

$\begin{array}{l}{\widehat{p}}_1=\text{Sample proportion from population 1}\\ {\widehat{p}}_2=\text{Sample proportion from population 2}\\ n_1=\text{Sample size of group 1}\\ n_2=\text{Sample size of group 2}\end{array}$

Substitute the provided values ${\widehat{p}}_1=0.8,{\widehat{p}}_2=0.6$ and $n_1=400,n_2=400$ in the above defined formula to calculate the standard error of D.

$\begin{array}{c}{\text{SE}}_D=\sqrt{\frac{0.8\times \left(1-0.8\right)}{80}+\frac{0.6\times \left(1-0.6\right)}{80}}\\ =0.0316\end{array}$

The value of $z^{*}$ for 95% confidence level is $z^{*}=1.96$ from the standard normal table. So, the margin of error is obtained as:

$\begin{array}{c}m=z^{*}\times {\text{SE}}_{\widehat{D}}\\ =1.96\times 0.0316\\ =0.0619\end{array}$

Therefore, the margin of error is obtained as 0.0619.

To determine

To find: The 95% margin of error for the difference of two proportions for sample size $n=500$.

Answer to Problem 96E

Solution: The margin of error is 0.0555.

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 $n_1=500\text{and}{n}_2=500$. The margin of error for confidence level C is defined as:

$m=z^{*}\times {\text{SE}}_D$

Where z^* is the critical value for the standard normal variable and ${\text{SE}}_D$ is the standard error of the difference. The standard error of D is defined as:

${\text{SE}}_D=\sqrt{\frac{{\widehat{p}}_1\left(1-{\widehat{p}}_1\right)}{n_1}+\frac{{\widehat{p}}_2\left(1-{\widehat{p}}_2\right)}{n_2}}$

Where

$\begin{array}{l}{\widehat{p}}_1=\text{Sample proportion from population 1}\\ {\widehat{p}}_2=\text{Sample proportion from population 2}\\ n_1=\text{Sample size of group 1}\\ n_2=\text{Sample size of group 2}\end{array}$

Substitute the provided values ${\widehat{p}}_1=0.8,{\widehat{p}}_2=0.6$ and $n_1=500,n_2=500$ in the above defined formula to calculate the standard error of D.

$\begin{array}{c}{\text{SE}}_D=\sqrt{\frac{0.8\times \left(1-0.8\right)}{80}+\frac{0.6\times \left(1-0.6\right)}{80}}\\ =0.0283\end{array}$

The value of $z^{*}$ for 95% confidence level is $z^{*}=1.96$ from the standard normal table. So, the margin of error is obtained as:

$\begin{array}{c}m=z^{*}\times {\text{SE}}_{\widehat{D}}\\ =1.96\times 0.0283\\ =0.0555\end{array}$

Therefore, the margin of error is obtained as 0.0555.

To determine

To find: The 95% margin of error for the difference of two proportions for sample size $n=1000$.

Answer to Problem 96E

Solution: The margin of error is 0.0392.

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 $n_1=1000\text{and}{n}_2=1000$. The margin of error is calculated as:

$m=z^{*}\times {\text{SE}}_D$

Where z^* is the critical value for the standard normal variable and ${\text{SE}}_D$ is the standard error of the difference. The standard error of D is defined as:

${\text{SE}}_D=\sqrt{\frac{{\widehat{p}}_1\left(1-{\widehat{p}}_1\right)}{n_1}+\frac{{\widehat{p}}_2\left(1-{\widehat{p}}_2\right)}{n_2}}$

Where

$\begin{array}{l}{\widehat{p}}_1=\text{Sample proportion from population 1}\\ {\widehat{p}}_2=\text{Sample proportion from population 2}\\ n_1=\text{Sample size of group 1}\\ n_2=\text{Sample size of group 2}\end{array}$

Substitute the provided values ${\widehat{p}}_1=0.8,{\widehat{p}}_2=0.6$ and $n_1=1000,n_2=1000$ in the above defined formula to calculate the standard error of D.

$\begin{array}{c}{\text{SE}}_D=\sqrt{\frac{0.8\times \left(1-0.8\right)}{80}+\frac{0.6\times \left(1-0.6\right)}{80}}\\ =0.02\end{array}$

The value of $z^{*}$ for 95% confidence level is $z^{*}=1.96$ from the standard normal table. So, the margin of error is obtained as:

$\begin{array}{c}m=z^{*}\times {\text{SE}}_{\widehat{D}}\\ =1.96\times 0.02\\ =0.0392\end{array}$

Therefore, the margin of error is obtained as 0.0392.

To determine

The obtained results of margin of error for different sample sizes in a table.

Answer to Problem 96E

Solution: The obtained results of 95% margins of error for different sample sizes are tabulated as:

N	M
60	0.1607
70	0.1482
80	0.1386
100	0.1239
400	0.0619
500	0.0555
1000	0.0392

Explanation of Solution

The 95% margins of error for the difference between two proportions for different sample sizes are calculated as in the previous parts and are tabulated as:

N	M
60	0.1607
70	0.1482
80	0.1386
100	0.1239
400	0.0619
500	0.0555
1000	0.0392

So, it is observed that with an increase in the sample size the margin of error reduces. Hence, a larger sample results in more accurate results.

To determine

To graph: The 95% margins of error obtained for different sample sizes.

Explanation of Solution

Graph: To obtain the graph of 95% margins of error for different sample sizes, Minitab is used. The steps followed to construct the graph are:

Step 1: Enter the data of the sample sizes and the associated margins of error values in the worksheet.

Step 2: Select Graph→Scatterplot.

Step 3: Select Simple from the opened dialog and click on OK.

Step 4: Select the Y variables as margin of error column and X variable as sample size column and click on OK.

The graph is obtained as:

To determine

To explain: A short summary of the obtained results.

Answer to Problem 96E

Solution: The obtained results of margins of error and the graph shows that the margin of error decreases as the sample size increases. But the rate of decrease of margin of error is very less for large n.

Explanation of Solution

The margins of error for different sample sizes are obtained and tabulated. The results show that the margin of error decreases as the sample size increases. This is also shown by the graph.

Hence, a larger sample is desirable for a lower error.