Chapter 6, Problem 6.11P

A random sample of 1496 respondents of a major metropolitan area was questioned about a number of issues. Construct estimates to the population at the 95% level for each of the results reported next. Express the final confidence interval in percentages (e.g., “between 40% and 45% agreed that premarital sex was always wrong”).

a. When asked to agree or disagree with the statement "Internet pornography leads to rape and other sex crimes,” 823 agreed.

b. When asked to agree or disagree with the statement "Hand guns should be outlawed,” 650 agreed.

c. 375 of the sample agreed that marijuana should be legalized.

d. 1023 of the sample said that they had attended a church, synagogue, mosque, or other place of worship at least once within the past month.

e. 800 agreed that public elementary schools should have sex education programs starting in the fifth grade.

To determine

(a)

To find:

The 95% confidence interval for the population mean.

Answer to Problem 6.11P

Solution:

The 95% confidence interval suggests that, between 52% and 58% of the people agree with the statement that internet pornography leads to rape and other sex crimes.

Explanation of Solution

Given:

The given information is,

A random sample of 1496 respondents of a major metropolitan area was questioned about a number of issues.

When asked to agree or disagree with the statement "Internet pornography leads to rape and other sex crimes,” 823 agreed.

Approach:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For large samples and given sample proportion, the confidence interval (c.i.) is given by,

$c.i.=P_s\pm Z\left(\sqrt{\frac{P_u\left(1-P_u\right)}{N}}\right)$

Where, $P_s$ is the sample proportion,

$P_u$ is the population proportion,

Z is the Z score determined by the alpha level and,

N is the sample size.

Calculation:

From the given information, the sample size is 1496,

Sample proportion is,

$\frac{823}{1496}=0.55$

And the population proportion is $0.5$.

For 95%, the Z value is given as,

$Z_{\left(95\%\right)}=1.96$

For large samples and given sample proportion, the confidence interval (c.i.) is given by,

$c.i.=P_s\pm Z\left(\sqrt{\frac{P_u\left(1-P_u\right)}{N}}\right)$……$\left(1\right)$

Substitute 0.55 for $P_s$, 0.5 for $P_u$, 1.96 for Z and 1496 for $N$ in equation(1) to obtain the confidence interval.

$\begin{array}{rll}c.i.& =& 0.55\pm 1.96\left(\sqrt{\frac{0.5\times 0.5}{1496}}\right)\\ c.i.& =& 0.55\pm 0.03\end{array}$

Thus, the 95% confidence interval is,

$c.i.=\left(0.52,0.58\right)$

Conclusion:

The 95% confidence interval suggests that, between 52% and 58% of the people agree with the statement that internet pornography leads to rape and other sex crimes.

To determine

(b)

To find:

The 95% confidence interval for the population mean.

Answer to Problem 6.11P

Solution:

The 95% confidence interval suggests that, between 40% and 46% of the people agree with the statement that hand guns should be outlawed.

Explanation of Solution

Given:

The given information is,

A random sample of 1496 respondents of a major metropolitan area was questioned about a number of issues.

When asked to agree or disagree with the statement "Hand guns should be outlawed,” 650 agreed.

Approach:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For large samples and given sample proportion, the confidence interval (c.i.) is given by,

$c.i.=P_s\pm Z\left(\sqrt{\frac{P_u\left(1-P_u\right)}{N}}\right)$

Where, $P_s$ is the sample proportion,

$P_u$ is the population proportion,

Z is the Z score determined by the alpha level and,

N is the sample size.

Calculation:

From the given information, the sample size is 1496,

Sample proportion is,

$\frac{650}{1496}=0.43$

And the population proportion is $0.5$.

For 95%, the Z value is given as,

$Z_{\left(95\%\right)}=1.96$

For large samples and given sample proportion, the confidence interval (c.i.) is given by,

$c.i.=P_s\pm Z\left(\sqrt{\frac{P_u\left(1-P_u\right)}{N}}\right)$……$\left(1\right)$

Substitute 0.43 for $P_s$, 0.5 for $P_u$, 1.96 for Z and 1496 for $N$ in equation(1) to obtain the confidence interval.

$\begin{array}{rll}c.i.& =& 0.43\pm 1.96\left(\sqrt{\frac{0.5\times 0.5}{1496}}\right)\\ c.i.& =& 0.43\pm 0.03\end{array}$

Thus, the 95% confidence interval is,

$c.i.=\left(0.40,0.46\right)$

Conclusion:

The 95% confidence interval suggests that, between 40% and 46% of the people agree with the statement that hand guns should be outlawed.

To determine

(c)

To find:

The 95% confidence interval for the population mean.

Answer to Problem 6.11P

Solution:

The 95% confidence interval suggests that, between 22% and 28% of the people agree with the statement that marijuana should be legalized.

Explanation of Solution

Given:

The given information is,

A random sample of 1496 respondents of a major metropolitan area was questioned about a number of issues.

375 of the sample agreed that marijuana should be legalized.

Approach:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For large samples and given sample proportion, the confidence interval (c.i.) is given by,

$c.i.=P_s\pm Z\left(\sqrt{\frac{P_u\left(1-P_u\right)}{N}}\right)$

Where, $P_s$ is the sample proportion,

$P_u$ is the population proportion,

Z is the Z score determined by the alpha level and,

N is the sample size.

Calculation:

From the given information, the sample size is 1496,

Sample proportion is,

$\frac{375}{1496}=0.25$

And the population proportion is $0.5$.

For 95%, the Z value is given as,

$Z_{\left(95\%\right)}=1.96$

For large samples and given sample proportion, the confidence interval (c.i.) is given by,

$c.i.=P_s\pm Z\left(\sqrt{\frac{P_u\left(1-P_u\right)}{N}}\right)$……$\left(1\right)$

Substitute 0.25 for $P_s$, 0.5 for $P_u$, 1.96 for Z and 1496 for $N$ in equation(1) to obtain the confidence interval.

$\begin{array}{rll}c.i.& =& 0.25\pm 1.96\left(\sqrt{\frac{0.5\times 0.5}{1496}}\right)\\ c.i.& =& 0.25\pm 0.03\end{array}$

Thus, the 95% confidence interval is,

$c.i.=\left(0.22,0.28\right)$

Conclusion:

The 95% confidence interval suggests that, between 22% and 28% of the people agree with the statement that marijuana should be legalized.

To determine

(d)

To find:

The 95% confidence interval for the population mean.

Answer to Problem 6.11P

Solution:

The 95% confidence interval suggests that, between 65% and 71% of the people attended a church, synagogue, mosque, or other place of worship at least once within the past month.

Explanation of Solution

Given:

The given information is,

A random sample of 1496 respondents of a major metropolitan area was questioned about a number of issues.

1023 of the sample said that they had attended a church, synagogue, mosque, or other place of worship at least once within the past month.

Approach:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For large samples and given sample proportion, the confidence interval (c.i.) is given by,

$c.i.=P_s\pm Z\left(\sqrt{\frac{P_u\left(1-P_u\right)}{N}}\right)$

Where, $P_s$ is the sample proportion,

$P_u$ is the population proportion,

Z is the Z score determined by the alpha level and,

N is the sample size.

Calculation:

From the given information, the sample size is 1496,

Sample proportion is,

$\frac{1023}{1496}=0.68$

And the population proportion is $0.5$.

For 95%, the Z value is given as,

$Z_{\left(95\%\right)}=1.96$

For large samples and given sample proportion, the confidence interval (c.i.) is given by,

$c.i.=P_s\pm Z\left(\sqrt{\frac{P_u\left(1-P_u\right)}{N}}\right)$……$\left(1\right)$

Substitute 0.68 for $P_s$, 0.5 for $P_u$, 1.96 for Z and 1496 for $N$ in equation(1) to obtain the confidence interval.

$\begin{array}{rll}c.i.& =& 0.68\pm 1.96\left(\sqrt{\frac{0.5\times 0.5}{1496}}\right)\\ c.i.& =& 0.68\pm 0.03\end{array}$

Thus, the 95% confidence interval is,

$c.i.=\left(0.65,0.71\right)$

Conclusion:

The 95% confidence interval suggests that, between 65% and 71% of the people attended a church, synagogue, mosque, or other place of worship at least once within the past month.

To determine

(e)

To find:

The 95% confidence interval for the population mean.

Answer to Problem 6.11P

Solution:

The 95% confidence interval suggests that, between 50% and 56% of the people agreed that public elementary schools should have sex education programs starting in the fifth grade.

Explanation of Solution

Given:

The given information is,

A random sample of 1496 respondents of a major metropolitan area was questioned about a number of issues.

800 agreed that public elementary schools should have sex education programs starting in the fifth grade.

Approach:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For large samples and given sample proportion, the confidence interval (c.i.) is given by,

$c.i.=P_s\pm Z\left(\sqrt{\frac{P_u\left(1-P_u\right)}{N}}\right)$

Where, $P_s$ is the sample proportion,

$P_u$ is the population proportion,

Z is the Z score determined by the alpha level and,

N is the sample size.

Calculation:

From the given information, the sample size is 1496,

Sample proportion is,

$\frac{800}{1496}=0.53$

And the population proportion is $0.5$.

For 95%, the Z value is given as,

$Z_{\left(95\%\right)}=1.96$

For large samples and given sample proportion, the confidence interval (c.i.) is given by,

$c.i.=P_s\pm Z\left(\sqrt{\frac{P_u\left(1-P_u\right)}{N}}\right)$……$\left(1\right)$

Substitute 0.53 for $P_s$, 0.5 for $P_u$, 1.96 for Z and 1496 for $N$ in equation(1) to obtain the confidence interval.

$\begin{array}{rll}c.i.& =& 0.53\pm 1.96\left(\sqrt{\frac{0.5\times 0.5}{1496}}\right)\\ c.i.& =& 0.53\pm 0.03\end{array}$

Thus, the 95% confidence interval is,

$c.i.=\left(0.50,0.56\right)$

Conclusion:

The 95% confidence interval suggests that, between 50% and 56% of the people agreed that public elementary schools should have sex education programs starting in the fifth grade.