Suppose we want to construct a confidence interval for $ p $ and $ n = 50 $ and $\hat{p} = 0.9$. Is the large counts condition met?

A. Yes, $ n\hat{p} $ is at least 10.
B. Yes, $ n(1 - \hat{p}) $ is at least 10.
C. Yes, both $ n\hat{p} $ and $ n(1 - \hat{p}) $ are at least 10.
D. No, $ n\hat{p} $ and $ n(1 - \hat{p}) $ are not both at least 10.

In constructing a confidence interval for proportions, n=50 and p' = 0.9 do not meet the large counts condition because while np' is 45 and satisfies the condition, n(1-p') is only 5, which does not. Therefore, the condition that both np' and n(1-p') must be at least 10 is not met.

Explanation:

Understanding the Condition for a Confidence Interval for Proportions

To check the large counts condition for constructing a confidence interval for p, we need to ensure that both np' (the number of successes) and n(1-p') (the number of failures) are at least 10. In this case, with n = 50 and p' = 0.9, we calculate:

np' = 50 * 0.9

= 45

n(1-p') = 50 * (1 - 0.9)

= 50 * 0.1

= 5

The condition np' ≥ 10 is met because np' = 45. However, n(1-p') ≥ 10 is not met because n(1-p') = 5, which does not satisfy the large counts condition. Thus, the correct answer is: No, np' and n(1-p') are not both at least 10.