Question

Of 136 randomly selected adults, 33 were found to have high blood pressure. Construct a 95% confidence interval for the true of all adults that have high blood pressure.

Answer 1

Answer:

The 95% confidence interval of the proportion of all adults that have high blood pressure is 0.17059 < $\hat{p}$ < 0.314695

Step-by-step explanation:

The confidence interval for a proportion is given by the following formula;

$CI=\hat{p}\pm z\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Where:

x = 33

n = 136

$\hat{p}$ = x/n = 33/136 = 0.243

z value for 95% confidence is 1.96

Plugging in the values, we have;

$CI=0.243\pm 1.96\times \sqrt{\frac{0.243(1-0.243)}{136}}$

Which gives;

0.17059 < $\hat{p}$ < 0.314695

Hence the 95% confidence interval of the proportion of all adults that have high blood pressure = 0.17059 < $\hat{p}$ < 0.314695

From the above we have;

23.2 < x < 42.798

Since we are dealing with people, we round down as follows;

23 < x < 42.