Chapter 7: Q.3 (page 459)

Suppose we select an SRS of size $n = 100$ from a large population having proportion p of successes. Let $\overset{\land}{p}$ 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 $\overset{\land}{p}$ ?
(a) $0.01$
(b) $\frac{1}{11}$
(c) $0.85$
(d) $0.975$
(e) $0.999$

Short Answer

Expert verified

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 $n p$ and $n (1 - p)$ are at least $10 .$

For option (a):

localid="1650262259734" $n p = 100 \times 0.01 = 1 n (1 - p) = 100 \times (1 - 0.01) = 99$

For option (b):

localid="1650262265572" $n p = 100 \times \frac{1}{11} \approx 9.0909 n (1 - p) = 100 \times (1 - \frac{1}{11}) \approx 90.9091$

For option (c):

localid="1650262271533" $n p = 100 \times 0.85 = 85 n (1 - p) = 100 \times (1 - 0.85) = 15$

Explanation of (d) and (e)

For option (d):

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

For option (e):

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

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

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Short Answer

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Given Information

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

Explanation of (d) and (e)

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Suppose we select an SRS of size n=100from a large population having proportion p of successes. Let 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 p∧?(a) 0.01(b) 111(c) 0.85(d) 0.975(e)0.999

Short Answer

Step by step solution

Given Information

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

Explanation of (d) and (e)

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