Question

Suppose we select an SRS of size $\mathrm{n}=100n=100$from a large population having proportion p of successes. Let ${\mathrm{p}}^{\wedge \hat{p}}$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 $\mathrm{p}\wedge \hat{p}$?

a. 0.01

b. 0.09

c. 0.85

d. 0.975

e. 0.999

Short answer

Option (c)$0.85$is the right answer.

Step-by-step solution

Given Information

Given,

The value of $n$ is $100$

Explanation for correct option

If both $np$ and $n(1-p)$ are more than 10, the sampling distribution of $\hat{p}$ is close to normal.

(a) $np=100\times 0.01=1n(1-p)=100\times (1-0.01)=99$

(b) $np=100\times \frac{1}{11}=9.0909n(1-p)=100\times \left(1-\frac{1}{11}\right)=90.9091$

(c) $np=100\times 0.85=85n(1-p)=100\times (1-0.85)=15$

(d) $np=100\times 0.975=97.5n(1-p)=100\times (1-0.975)=2.5$

(e) $np=100\times 0.999=99.9n(1-p)=100\times (1-0.999)=0.1$

Hence, the correct option is (c)

Explanation for incorrect option

Option (A) $0.01$is the right answer

Option (B) $0.09$is the right answer

Option (D) $0.975$is the right answer

Option (E) $0.999$is the right answer