Chapter 21, Problem 18SE

a.

To determine

Obtain a 95% confidence interval for the population proportion.

Answer to Problem 18SE

The 95% confidence interval for the population proportion is $\left(0.1488,0.2312\right)$.

Explanation of Solution

Calculation:

In a nationwide poll, 19% out of 363 adults who have seen the ads M, liked the ads a lot.

Confidence interval:

The approximate 95% confidence interval estimate of the population proportion is $\overline{p}\pm 2s_{\overline{p}}$.

Where $s_{\overline{p}}=\sqrt{\left(\frac{N-n}{N}\right)\left(\frac{\overline{p}\left(1-\overline{p}\right)}{n-1}\right)}$

Substitute $\left(\frac{N-n}{N}\right)$ as 1 and $\overline{p}$ as 0.19 in the above formula.

$\begin{array}{c}{s}_{\overline{p}}=\sqrt{1\left(\frac{0.19\left(1-0.19\right)}{363-1}\right)}\\ =\sqrt{\left(\frac{0.19\left(0.81\right)}{362}\right)}\\ =\sqrt{0.000425}\\ =0.0206\end{array}$

Thus, the standard error of the proportion is 0.0206.

Confidence interval:

$\begin{array}{c}\overline{p}\pm 2s_{\overline{p}}=0.19\pm 2\left(0.0206\right)\\ =0.19\pm 0.0412\\ =\left(0.19-0.0412,0.19+0.0412\right)\\ =\left(0.1488,0.2312\right)\end{array}$

Thus, the approximate 95% confidence interval for the population proportion is $\left(0.1488,0.2312\right)$.

b.

To determine

Obtain a 95% confidence interval for the population proportion.

Answer to Problem 18SE

The 95% confidence interval for the population proportion is $\left(0.2614,0.3586\right)$.

Explanation of Solution

Calculation:

The respondents who disliked the new ads are 31%.

Substitute $\left(\frac{N-n}{N}\right)$ as 1 and $\overline{p}$ as 0.31 in the above formula.

$\begin{array}{c}{s}_{\overline{p}}=\sqrt{1\left(\frac{0.31\left(1-0.31\right)}{363-1}\right)}\\ =\sqrt{\left(\frac{0.31\left(0.69\right)}{362}\right)}\\ =\sqrt{0.000591}\\ =0.0243\end{array}$

Thus, the standard error of the proportion is 0.0243.

Confidence interval:

$\begin{array}{c}\overline{p}\pm 2s_{\overline{p}}=0.31\pm 2\left(0.0243\right)\\ =0.31\pm 0.0486\\ =\left(0.31-0.0486,0.31+0.0486\right)\\ =\left(0.2614,0.3586\right)\end{array}$

Thus, the approximate 95% confidence interval for the population proportion is $\left(0.2614,0.3586\right)$.

c.

To determine

Obtain approximate 95% confidence interval for the proportion of adults who think the ads are very effective.

Answer to Problem 18SE

The approximate 95% confidence interval for the proportion of adults who think the ads are very effective is, $\left(0.1305,0.2095\right)$.

Explanation of Solution

Calculation:

The respondents who felt the ads are very effective is 17%.

Substitute $\left(\frac{N-n}{N}\right)$ as 1 and $\overline{p}$ as 0.17 in the above formula.

$\begin{array}{c}{s}_{\overline{p}}=\sqrt{1\left(\frac{0.17\left(1-0.17\right)}{363-1}\right)}\\ =\sqrt{\left(\frac{0.17\left(0.83\right)}{362}\right)}\\ =\sqrt{0.000390}\\ =0.0197\end{array}$

Thus, the standard error of the proportion is 0.0197.

Confidence interval:

$\begin{array}{c}\overline{p}\pm 2s_{\overline{p}}=0.17\pm 2\left(0.0917\right)\\ =0.17\pm 0.0395\\ =\left(0.17-0.0395,0.17+0.0395\right)\\ =\left(0.1305,0.2095\right)\end{array}$

Thus, the approximate 95% confidence interval for the proportion of adults who think the ads are very effective is $\left(0.1305,0.2095\right)$.

d.

To determine

Explain the statement and find the number they got.

Explanation of Solution

Calculation:

The statement “margin of error is five percentage points” is reported by Person L.

The standard error is largest when $\overline{p}=0.50$.

Now,

$\begin{array}{c}{s}_{\overline{p}}=\sqrt{1\left(\frac{0.5\left(1-0.5\right)}{363-1}\right)}\\ =\sqrt{\left(\frac{0.5\left(0.5\right)}{362}\right)}\\ =\sqrt{0.000691}\\ =0.0263\end{array}$.

Here by multiplying the standard error with 2 the bound is,

$\begin{array}{c}B=2\left(0.0263\right)\\ =0.0526\end{array}$.

Hence, for the sample of 363, the bond would be 5% when $\overline{p}=0.50$ is the worst case.

e.

To determine

Explain whether the non-sampling error biased the results of the survey or not.

Explanation of Solution

When the poll is conducted by calling people at home during day time then the sample results represents for the adults who are not working outside the home. The Person L took precautions against this reason which causes bias.