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,
[tex]\begin{aligned}\hat{p}&=\frac{x}{n}\\\hat{p}&=\frac{867}{1050}\\ \hat{p}&=0.826 \end[/tex]
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 [tex]Z_{\frac{\alpha}{2}}[/tex] by dividing alpha with 2,
[tex]\begin{aligned}Z_{\frac{\alpha}{2}}&=Z_{\frac{0.03}{2}}\\ Z_{\frac{\alpha}{2}}&=Z_{0.015}\end[/tex]
From this table which is shown in the figure it says that the critical value is
[tex]Z_{crit}=\pm 2.17[/tex]
Use this formula to find the confidence interval [tex]P=\hat{p}\pm Z_{Crit}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}[/tex]
Substitute the values in the formula,
[tex]\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}[/tex]
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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