I am trying to do the following question from the textbook Beginning Statistics by Douglas S. Shafer and Zhiyi Zhang:
I computed $n=\frac{z_{0.05}^{2}\hat{p}(1-\hat{p})}{E^{2}}=\frac{(1.65)^{2}(0.5)(1-0.5)}{0.02^{2}}\approx 1702$. The textbook answer is $1692$. Is it fine to have such a difference in answers?
Best Answer
It's the difference between 1.65 and 1.645 as the 95th percentile of the standard normal distribution; 1.645 is shown in some printed tables and is more accurate. The computation using R statistical software is shown below. So technically, 1692 is the more accurate answer. In practice, someone might round this to 1700.
However, this computation uses the 'worst case' scenario by taking $p = 1/2.$ That approach is commonly taken when one has no idea about the correct value of $p.$ Then it is the safe approach because $p = 1/2$ gives the largest projected sample size.
As a reality check, I would hope the college's graduation rate is greater than 50%. If it's around 75%, then the required $n$ to find the exact value within $\pm 0.02$ would be $1269 < 1692.$
Whether you'd get credit for 1702 instead of 1692 depends on the fussiness of the person doing the grading. Because the value 1.65 is shown in some tables, I'd give credit for it.
Note: The figure below shows $n$ required for various prospective values of $p.$ [The normal approximation probably isn't valid near $p = 0$ or $1,$ so the graph doesn't include extreme values of $p.]$