Question

Suppose we select an SRS of size $n=100$from a large population having proportion p of successes. Let $\stackrel{\wedge }{p}$ be the proportion of successes in the sample. For which value of p would it be safe to use the Normal approximation to the sampling distribution of $\stackrel{\wedge }{p}$?

(a) $0.01$

(b) $\frac{1}{11}$

(c) $0.85$

(d) $0.975$

(e)$0.999$

Short answer

The correct answer is (c) $0.85$

Step-by-step solution

Given Information

Given information:

The sample size for SRS, $n=100$

The proportion of successes in the sample $=\hat{p}$

Explanation of (a) (b) and (c)

Know that the sampling distribution of $\hat{p}$is approximately normal if both $np$and $n(1-p)$are at least $10.$

For option (a):

localid="1650262259734" $$\begin{gathered} np=100\times 0.01=1 \\ n(1-p)=100\times (1-0.01) \\ =99 \end{gathered}$$

For option (b):

localid="1650262265572" $$\begin{gathered} np=100\times \frac{1}{11}\approx 9.0909 \\ n(1-p)=100\times \left(1-\frac{1}{11}\right) \\ \approx 90.9091 \end{gathered}$$

For option (c):

localid="1650262271533" $$\begin{gathered} np=100\times 0.85=85 \\ n(1-p)=100\times (1-0.85) \\ =15 \end{gathered}$$

Explanation of (d) and (e)

For option (d):

$$\begin{gathered} np=100\times 0.975=97.5 \\ n(1-p)=100\times (1-0.975) \\ =2.5 \end{gathered}$$

For option (e):

$$\begin{gathered} np=100\times 0.999=99.9 \\ n(1-p)=100\times (1-0.999) \\ =0.1 \end{gathered}$$

It is clear that in option (c) both $np$and $n(1-p)$are greater than localid="1650262292919" $10$. Hence, the correct answer is (c).