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

Answer:

The large  sample n = 174.24

Step-by-step 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  = $\frac{2S.D}{\sqrt{n} }$

Given standard deviation = 33% = 0.33

$M .E = \frac{2S.D}{\sqrt{n} }$

$0.05 = \frac{2X0.33}{\sqrt{n} }$

cross multiplication , we get

$\sqrt{n} = \frac{2X0.33}{0.05}$

√n = 13.2

squaring on both sides, we get

n = 174.24

Conclusion:-

The large  sample n = 174.24