# Solved: In a Gallup poll, 1025 randomly selected adults were surveyed and 29% of them said that they used the internet for shopp

#### ByDr. Raju Chaudhari

Oct 9, 2020

In a Gallup poll, 1025 randomly selected adults were surveyed and 29% of them said that they used the internet for shopping at least a few times a year.

a. Find the point estimate of the percentage of adults who use the internet for shopping.
b. Find a 99% confidence interval estimate of the percentage of adults who use the internet for shopping.

### Solution

a. Given that sample size $n = 1025$.

The sample proportion is $\hat{p}=0.29$.

So the point estimate of the percentage of adults who use the internet for shopping is $\hat{p}=0.29$.

b. For 99% confidence interval estimate of the percentage of adults who use the internet for shopping

#### Step 1 Specify the confidence level $(1-\alpha)$

Confidence level is $1-\alpha = 0.99$. Thus, the level of significance is $\alpha = 0.01$.

#### Step 2 Given information

Given that sample size $n =1025$.

The estimate of the proportion of success is $\hat{p} =0.29$.

#### Step 3 Specify the formula

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

 \begin{aligned} & \hat{p} - E \leq p \leq \hat{p} + E. \end{aligned}

where

 \begin{aligned} E=Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \end{aligned}

and $Z_{\alpha/2}$ is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

#### Step 4 Determine the critical value

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

Thus $Z_{\alpha/2} = Z_{0.005} = 2.576$.

#### Step 5 Compute the margin of error

The margin of error for proportions is
 \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & = 2.576 \sqrt{\frac{0.29*(1-0.29)}{1025}}\\ & =0.037. \end{aligned}

#### Step 6 Determine the confidence interval

$99$% confidence interval estimate for population proportion is

 \begin{aligned} \hat{p} - E & \leq p \leq \hat{p} + E\\ 0.29 - 0.037 & \leq p \leq 0.29 + 0.037\\ 0.2535 & \leq p \leq 0.3265. \end{aligned}

Thus, $99$% confidence interval estimate for population proportion $p$ is $(0.2535,0.3265)$.