# Solved: A marketing research firm wishes to estimate the proportion of adults who are planning to buy a new car in the next 6 months

#### ByDr. Raju Chaudhari

Feb 23, 2021

A marketing research firm wishes to estimate the proportion of adults who are planning to buy a new car in the next 6 months. A simple random sample of 100 adults led to 22 who were planning to buy a new car in the next 6 months.

a. Compute the 95% confidence interval for the proportion of adults who are planning to buy a new car in the next 6 months.
b. Interpret this confidence interval.
c. How large a sample size will need to be selected if we wish to have a 95% confidence interval that is accurate to within 1%.

#### Solution

Given that sample size $n = 100$. The sample proportion is
$\hat{p}=\frac{22}{100}=0.22$.

a. Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.

$100(1-\alpha)$% confidence interval for population proportion is

\begin{aligned} \hat{p} - E \leq p \leq \hat{p} + E. \end{aligned}
where $E=Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ and $Z_{\alpha/2}$ is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$.

Thus $Z_{\alpha/2} = Z_{0.025} = 1.96$.

The margin of error for proportions is

\begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & = 1.96 \sqrt{\frac{0.22*(1-0.22)}{100}}\\ & =0.081. \end{aligned}

$95$% confidence interval estimate for population proportion is

\begin{aligned} \hat{p} - E & \leq p \leq \hat{p} + E\\ 0.22 - 0.081 & \leq p \leq 0.22 + 0.081\\ 0.1388 & \leq p \leq 0.3012. \end{aligned}

Thus, $95$% confidence interval estimate for population proportion is $(0.3012,0.1388)$.

b. We are 95% confident that the true proportion of of adults who are planning to buy a new car in the next 6 months lies between $(0.3012,0.1388)$.

c. The formula to estimate the sample size required to estimate the proportion is

$$n =p*(1-p)\bigg(\frac{z}{E}\bigg)^2$$

where $p$ is the proportion of success, $z$ is the $Z_{\alpha/2}$ and $E$ is the margin of error.

Given that margin of error $E =0.01$. The confidence coefficient is $1-\alpha=0.95$. The sample proportion is $p =0.22$.

The critical value of $Z$ is $Z_{\alpha/2} =Z_{0.025}= 1.96$.

The minimum sample size required to estimate the proportion is

\begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.22(1-0.22)\bigg(\frac{1.96}{0.01}\bigg)^2\\ &=6592.186\\ &\approx 6593. \end{aligned}