Chapter 8.1, Problem 18E

(a)

Section 1:

To determine

To find: The sample proportion of the students who takes breakfast regularly.

Answer to Problem 18E

Solution: The sample proportion of the students who takes breakfast regularly is $\underset{\_}{\widehat{p}=0.3633}$.

Explanation of Solution

Given: The randomly selected samples of 300 students are asked on their regular eating habits of breakfast. The survey showed that 109 students eat their breakfast regularly.

Explanation:

Calculation: The formula for sample proportion is defined as:

$\widehat{p}=\frac{X}{n}$

Here,

$\begin{array}{c}X=\text{number of succeses in the sample}\\ n=\text{sample size}\end{array}$

Substitute $X=109\text{ and }n=300$ in the above defined formula to get the required sample proportion. So,

$\begin{array}{c}\widehat{p}=\frac{X}{n}\\ =\frac{109}{300}\\ =0.3633\end{array}$

Therefore, the sample proportion $\widehat{p}$ is obtained as 0.3633.

Section 2:

To determine

To find: The standard error $S{E}_{\widehat{p}}$ of sample proportion $\widehat{p}$.

Answer to Problem 18E

Solution: The standard error $S{E}_{\widehat{p}}$ of sample proportion $\widehat{p}$ is $\underset{\_}{S{E}_{\widehat{p}}=0.0278}$.

Explanation of Solution

Calculation: The formula for standard error $S{E}_{\widehat{p}}$ of sample proportion $\widehat{p}$ and sample size n is defined as:

$S{E}_{\widehat{p}}=\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}$

The sample proportion $\widehat{p}$ is obtained as 0.3633 in the previous part. Substitute the obtained sample proportion of 0.3633 and sample size of 300 in the standard error formula. So,

$\begin{array}{c}S{E}_{\widehat{p}}=\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\\ =\sqrt{\frac{0.3633\left(1-0.3633\right)}{300}}\\ =\sqrt{\frac{0.231313}{300}}\\ =0.0278\end{array}$

Therefore, the standard error is obtained as 0.0278.

Section 3:

To determine

To find: The margin of error for 95% confidence level.

Answer to Problem 18E

Solution: The margin of error for 95% confidence level is $\underset{\_}{m=0.0545}$.

Explanation of Solution

Calculation: The formula for margin of error m is defined as:

$m=z^{*}\times S{E}_{\widehat{p}}$

Here, $z^{*}$ is the critical value of the standard normal density curve.

The standard error is obtained as $S{E}_{\widehat{p}}=0.0278$ in previous part. The value of $z^{*}$ for 95% confidence level is $z^{*}=1.96$ from the standard normal table provided in the book.

So, the margin of error is obtained as:

$\begin{array}{c}m=z^{*}\times S{E}_{\widehat{p}}\\ =1.96\times 0.0278\\ =0.0545\end{array}$

Therefore, the margin of error is obtained as 0.0545.

(b)

To determine

Whether the guidelines to use the large-sample confidence interval for population proportion are satisfied.

Answer to Problem 18E

Solution: Yes, the guidelines are satisfied to use the large-sample confidence interval for the population proportion.

Explanation of Solution

The guidelines required to be satisfied to use large-sample confidence interval for the population proportion is that the number of successes and the number of failures both should be at least 10.

In the provided problem of eating breakfast, the number of successes is defined as the number of students who eat their breakfast regularly. So, the number of successes is 109.

The number of failures is obtained as,

$\begin{array}{c}\text{Number of failures}=300-109\\ =191\end{array}$

The obtained number of successes and failures shows that they are more than 10.

Therefore, the guidelines to use the large-sample confidence interval for a population proportion are satisfied.

(c)

To determine

To find: The 95% large-sample confidence interval for the population proportion.

Answer to Problem 18E

Solution: The 95% large-sample confidence interval is $\underset{\_}{\left(0.3088,0.4178\right)}$.

Explanation of Solution

Calculation: The formula for large-sample confidence interval for population proportion p is defined as:

$\widehat{p}\pm m$

Here, $\widehat{p}$ is the sample proportion and m is the margin of error.

The sample proportion $\widehat{p}$ is obtained as 0.3633 and the margin of error is obtained as 0.0545 in part (a).

Substitute the values of margin of error and sample proportion in the formula for confidence interval. Therefore, the large-sample confidence interval is obtained as:

$\begin{array}{c}\widehat{p}\pm m=0.3633\pm 0.0545\\ =\left(0.3633-0.0545,0.3633+0.0545\right)\\ =\left(0.3088,0.4178\right)\end{array}$

Therefore, the required confidence interval is obtained as $\left(0.3088,0.4178\right)$.

(d)

To determine

To explain: A short statement on the meaning of the obtained confidence interval.

Answer to Problem 18E

Solution: The obtained confidence interval shows that it is 95% confident that between 30.88% and 41.78% of students responded that they eat their breakfast regularly.

Explanation of Solution

The 95% confidence interval is obtained as $\left(0.3088,0.4178\right)$ in the previous part. So, the 95% confidence interval in the form of percents is calculated as:

$\left(0.3088,0.4178\right)=\left(30.88\%,41.78\%\right)$

This shows that there is 95% confidence that the percentage of the students who responded that they eat their breakfast regularly is lie between 30.88% and 41.78%.