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Out of 167 randomly selected adults in the United States who were surveyed, 70 exercise on a regular basis. Construct a 90% confidence interval for the proportion of all adults in the United States who exercise on a regular basis. Round to three decimal places

Sagot :

ANSWER:

(0.356, 0.482)

STEP-BY-STEP EXPLANATION:

The first thing is to calculate the proportion with the data of the statement:

[tex]\begin{gathered} p=\frac{x}{n}=\frac{70}{167} \\ \\ p=0.4192 \end{gathered}[/tex]

For a 90% confidence interval, we have that the value of Z is the following:

[tex]\begin{gathered} \alpha=1-90\% \\ \\ \alpha=1-0.9=0.1 \\ \\ \alpha\text{/2}=\frac{0.1}{2}=0.05 \\ \text{ } \\ \text{The corresponding value of Z would be:} \\ \\ Z_{\alpha\text{/2}}=1.645 \end{gathered}[/tex]

We calculate the interval as follows:

[tex]\begin{gathered} \text{ Upper limit }=p+Z_{\alpha\text{/2}}\cdot\sqrt{\frac{p\cdot(1-p)}{n}}=0.4192+1.645\cdot\sqrt{\frac{0.4192\cdot\left(1-0.4192\right)}{167}}\:=0.482 \\ \\ \text{ Lower limit}=p-Z_{\alpha\text{/2}}\cdot\sqrt{\frac{p\cdot(1-p)}{n}}=0.4192-1.645\cdot\sqrt{\frac{0.4192\cdot\left(1-0.4192\right)}{167}}\:=0.356 \end{gathered}[/tex]

The 90% confidence interval for the proportion of all adults in the United States is (0.356, 0.482)