Question

of 118 randomly selected adults, 34 were found to have high blood pressure. construct a 95% confidence interval for the true percentage of all adults that have high blood pressure.

○ a. 19.1% < p < 38.5%
○ b. 18.1% < p < 39.6%
○ c. 21.9% < p < 35.7%
○ d. 20.6% < p < 37.0%

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 34$ (number of successes) and $n=118$ (sample size).
$\hat{p}=\frac{34}{118}\approx0.2881$

Step2: Find $z -$ value

For a 95% confidence interval, the $z -$ value $z_{\alpha/2}=1.96$ (from standard normal distribution table).

Step3: Calculate margin of error

The margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$
Substitute $\hat{p}=0.2881$, $n = 118$, and $z_{\alpha/2}=1.96$

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Step4: Calculate confidence interval

The confidence interval is $\hat{p}-ESubstitute $\hat{p}=0.2881$ and $E = 0.0817$
$0.2881- 0.0817$0.2064Convert to percentage: $20.64\%

Answer:

D. $20.6\%