Chapter 8.1, Problem 15E

(a)

Section 1:

To determine

To find: The sample proportion on the usage of cell phone while in a store.

Answer to Problem 15E

Solution: The sample proportion of the usage of cell phone while in a store is $\underset{\_}{\widehat{p}=0.461}$.

Explanation of Solution

Given: A Pew Internet poll surveyed 1003 adults. 462 responded that they used their cell phone while in the store within the last 30 days to make call.

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 $X=462\text{ and }n=\text{1003}$ in the above formula, to get the required sample proportion. So,

$\begin{array}{c}\widehat{p}=\frac{X}{n}\\ =\frac{462}{1003}\\ =0.461\end{array}$

Therefore, the sample proportion $\widehat{p}$ is obtained as 0.461.

Section 2:

To determine

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

Answer to Problem 15E

Solution: The standard error $S{E}_{\widehat{p}}$ of sample proportion $\widehat{p}$ is $\underset{\_}{S{E}_{\widehat{p}}=0.0157}$.

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 0.461 in the section 1. Substitute the obtained sample proportion of 0.461 and sample size of 1003 in the standard error formula. So,

$\begin{array}{c}S{E}_{\widehat{p}}=\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\\ =\sqrt{\frac{0.461\left(1-0.461\right)}{1003}}\\ =\sqrt{\frac{0.248479}{1003}}\\ =0.0157\end{array}$

Therefore, the standard error is obtained as 0.0157.

Section 3:

To determine

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

Answer to Problem 15E

Solution: The margin of error for 95% confidence level is $\underset{\_}{m=0.0308}$.

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 $S{E}_{\widehat{p}}=0.0157$ 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:

$\begin{array}{c}m=z^{*}\times S{E}_{\widehat{p}}\\ =1.96\times 0.0157\\ =0.0308\end{array}$

Therefore, the margin of error is obtained as 0.0308.

(b)

To determine

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

Answer to Problem 15E

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 count of successes and the count of failures both should be at least 10.

In the provided problem of cell phone survey, the number of successes is defined as the number of respondents who used their cell phone while in the store within the last 30 days to make call So, the number of successes is 462.

The number of failures is obtained as,

$\begin{array}{c}\text{Number of failures}=1003-462\\ =541\end{array}$

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

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

(c)

To determine

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

Answer to Problem 15E

Solution: The 95% large-sample confidence interval is $\underset{\_}{\left(0.4302,0.4918\right)}$.

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 is the margin of error.

The sample proportion $\widehat{p}$ is obtained as 0.461 and the margin of error is obtained as 0.0308 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:

$\begin{array}{c}\widehat{p}\pm m=0.461\pm 0.0308\\ =\left(0.461-0.0308,0.461+0.0308\right)\\ =\left(0.4302,0.4918\right)\end{array}$

Therefore, the large-sample confidence interval for the population proportion is obtained as $\left(0.4302,0.4918\right)$.

(d)

To determine

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

Answer to Problem 15E

Solution: The obtained confidence interval shows that it is 95% confident that between 43.02% an 49.18% of cell phone owners used their cell phone while in a store within last 30 days to make call to any friend or family member for advice on their purchase.

Explanation of Solution

The 95% confidence interval is obtained as $\left(0.4302,0.4918\right)$ in the previous part. So, the 95% confidence interval in the form of percents is calculated as:

$\left(0.4302,0.4918\right)=\left(43.02\%,49.18\%\right)$

This shows that there is 95% confidence that the percentage of people who used their cell phone while in a store within the last 30 days to make call to any friend or family member for advice on their purchase will lie between 43.02% and 49.18%.