Question

A health club annually surveys its members. Last year, 33% of the members said they use the treadmill at least four times a week. How large a sample should be selected this year to estimate the percentage of members who use the treadmill at least four times a week? The estimate is desired to have a margin of error of 5% with a 95% level of confidence.

Answer 1

The large sample n = 174.24

Explanation:

Given 33% of the members said they use the treadmill at least four times a week.

Given the margin of error is 5% = 0.05

we know that the margin of error at 95% 0f level of confidence =
`(2S.D)/(√(n) )`

Given standard deviation = 33% = 0.33

`M .E = (2S.D)/(√(n) )`

`0.05 = (2X0.33)/(√(n) )`

cross multiplication , we get

`√(n) = (2X0.33)/(0.05)`

√n = 13.2

squaring on both sides, we get

n = 174.24

Conclusion:-

The large sample n = 174.24