Question

A researcher wants to estimate the mean weekly family expenditure on clothing purchases and maintenance. First, she needs to determine the number of families that must be sampled, in order to estimate the mean weekly expenditure on clothing to within $15 at alpha = 0.01. From prior similar studies she is confident that the standard deviation of weekly family expenditures on clothing purchases and maintenance is $125. The required sample size to be taken is___________.

A. 21
B. 22
C. 461
D. 460
E. 460.46

Answer 1

Answer: C. 461

Explanation:

We know that the formula to find the sample size is given by :-

`n=((z^*\cdot \sigma)/(E))^2`

, where
`\sigma` = Population standard deviation from prior study.

E = margin of error.

z* = Critical value.

As per given , we have

`\sigma=125`

E= 15

Significance level :
`\alpha=0.01`

Critical value (Two tailed)=
`z^*=z_(\alpha/2)=z_(0.005)=2.576`

Now , Required sample size =
`n=(((2.576)\cdot 125)/(15))^2`

`n=(21.4666666667)^2\\\\ n=460.817777779\approx461` [Round to next integer]

Hence, the required sample size to be taken is 461.

Correct answer = C. 461