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

Explanation:

Step1: Calculate sample proportion $\hat{p}$

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

Step2: Calculate $q$

Since $q = 1-\hat{p}$, then $q=1 - 0.2881=0.7119$.

Step3: Find $z$-value for 95% confidence interval

For a 95% confidence interval, the $z$-value $z_{\alpha/2}=1.96$.

Step4: Calculate margin of error $E$

The formula for margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}q}{n}}$.
Substitute the values: $E=1.96\sqrt{\frac{0.2881\times0.7119}{118}}$.
First, calculate $\frac{0.2881\times0.7119}{118}\approx\frac{0.2052}{118}\approx0.00174$.
Then, $\sqrt{0.00174}\approx0.0417$.
Multiply by $1.96$: $E = 1.96\times0.0417\approx0.0817$.

Step5: Calculate confidence interval

The confidence interval for proportion $p$ is $\hat{p}-ESubstitute $\hat{p}=0.2881$ and $E = 0.0817$.
Lower limit: $0.2881- 0.0817=0.2064$ (or $20.64\%$).
Upper limit: $0.2881+0.0817 = 0.3698$ (or $37.0\%$).

Answer:

C. $20.6\%