Question 28 A local non-profit group would like to estimate the proportion of residents of Indian River county who do not have health insurance. To help with this estimate, the non-profit surveyed a random sample of Indian River county residents and found that $72 \%$ did not have heath insurance. Determine the sample size required to limit the margin of error to within 0.079 of the population proportion for a $97 \%$ confidence interval. Round the solution up to the nearest whole number. \[ n= \] Question Help: $\square$ Video Submit Question

Step 1 ：Given values are the margin of error \(E = 0.079\), the proportion of residents without health insurance \(p = 0.72\), and the Z-score for a 97% confidence interval \(Z = 2.17\).

Step 2 ：Use these values in the formula for sample size in a proportion estimation: \(n = \frac{Z^2 \cdot p \cdot (1-p)}{E^2}\).

Step 3 ：Substitute the given values into the formula: \(n = \frac{(2.17)^2 \cdot 0.72 \cdot (1-0.72)}{(0.079)^2}\).

Step 4 ：Calculate the value of \(n\).

Step 5 ：Round up \(n\) to the nearest whole number to get the required sample size.

Step 6 ：The sample size required to limit the margin of error to within 0.079 of the population proportion for a 97% confidence interval is \(\boxed{153}\).