Which statement is incorrect? Explain.
A)If p = .50 and n = 100, the standard error of the sample proportion is .05.
B)In a sample size calculation for estimating π, it is conservative to assume π = .50.
C)If n = 250 and p = .06, we cannot assume normality in a confidence interval for π.

Question

Quizplus · Accepted Answer

The standard error of the sample proportion is calculated using the formula $\sqrt{\frac{p(1-p)}{n}}$. For A) using p = .50 and n = 100, the standard error is indeed $\sqrt{\frac{.50(1-.50)}{100}} = .05$, making A correct. For B), assuming π = .50 is conservative because it maximizes the product p(1-p), which in turn maximizes the standard error, making the sample size calculation conservative. This is because .50 is the point where the variability of the sample proportion is highest. For C), the statement is incorrect because even with n = 250 and p = .06, we can assume normality in a confidence interval for π under the central limit theorem, especially since np and n(1-p) are both greater than 5 (15 and 235, respectively), which is a common rule of thumb for assuming normality of the sampling distribution.

Which Statement Is Incorrect? Explain