Question

We are to estimate the confidence interval for a population proportion p, based on two samples A and B of sizes 100 and 200, respectively. Both samples have the same standard deviations. Compare the lengths of the 95% confidence interval based on A to the interval based on B.

A. The length of the confidence interval based on A is the same as the length of the confidence interval based on B since the standard deviations for both samples are the same.
B. The length of the confidence interval based on A is larger than the length of the confidence interval based on B since SE for the sample A is larger than SE for the sample
C. The length of the confidence interval based on A is larger than the length of the confidence interval based on B since SE for the sample A is less than SE for the sample B.
D. The length of the confidence interval based on A is less than the length of the confidence interval based on B since SE of the sample A is larger than SE of the sample B.
E. The length of the confidence interval based on A is less than the length of the confidence interval based on B since SE of the sample A is less than the size of the sample B.

Answer 1

The option B is True

Explanation:

Solution

nA < nB

Thus

1/√nA > 1/√nB

SD/√nA > SD/√nB

So,

SE (A) > SE (B).......(1)

Zα/2 SE (A) > Zα/2 SE (B)

Now

E (A) > E(B) this is the margin error

Since SE fro sample A is larger than SE for sample B

We have that,

The length = ZE

Therefore, option B is correct because here, we know that as the sample size increases or goes higher the length of the confidence interval decreases

In this case, the sample size is large for B then A.

Thus, the length of confidence interval based on A is larger than the length of confidence interval Based on B Since SE is large For A than B