Question

In a recent health survey, 333 adult respondents reported a history of diabetes out of 3573 respondents. What is the critical value for a 90% confidence interval of the proportion of respondents who reported a history of diabetes

Answer 1

Answer:

The critical value for the 90% confidence interval is $Z_c = 1.645$.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

The critical value for the 90% confidence interval is $Z_c = 1.645$.