Question

Based on a poll, 867 of 1050 randomly selected adult said that they always wear seat belts. Construct and interpret a 97% confidence interval for the proportion of adults who always wear seat belts.

Answer 1

The 97% confidence interval for the proportion of adults is 0.801<P<0.851.

Given that 867 adults are randomly selected from 1050 who always wear seat belts.

A confidence interval refers to the probability that a population parameter will fall between a set of values for a certain proportion of times.

Let's assume n=1050 and x=867

Find the sample proportion by dividing the given proportion with sample size as,

$\begin{aligned}\hat{p}&=\frac{x}{n}\\\hat{p}&=\frac{867}{1050}\\ \hat{p}&=0.826 \end$

Find the α by subtracting the given confidence interval that is 97% from 1 as

α=1-(97÷100)

α=1-0.97

α=0.03

To find the critical value that is $Z_{\frac{\alpha}{2}}$ by dividing alpha with 2,

$\begin{aligned}Z_{\frac{\alpha}{2}}&=Z_{\frac{0.03}{2}}\\ Z_{\frac{\alpha}{2}}&=Z_{0.015}\end$

From this table which is shown in the figure it says that the critical value is

$Z_{crit}=\pm 2.17$

Use this formula to find the confidence interval $P=\hat{p}\pm Z_{Crit}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Substitute the values in the formula,

$\begin{aligned}P&=0.826\pm 2.17\times \sqrt{\frac{0.826(1-0.826)}{1050}}\\ P&=0.826\pm 2.17\times \sqrt{\frac{0.826\times 0.174}{1050}}\\ P&=0.826\pm 2.17\times 0.0116995\\ P&=0.826\pm 0.025\end{aligned}$

Hence, a 97% confidence interval for the proportion of adults who always wear seat belts is 0.801<P<0.851.

Learn more about confidence interval from here brainly.com/question/13929861

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