Question

Students at a certain school were​ surveyed, and it was estimated that 10​% of college students abstain from drinking alcohol. To estimate this proportion in your​ school, how large a random sample would you need to estimate it to within 0.02 with probability 0.95​, if before conducting the study​ (a) you are unwilling to predict the proportion value at your school and​ (b) you use the results from the surveyed school as a guideline.

Answer 1

a)

you are unwilling to predict the proportion value at your school = 0.90

b)

The large sample size 'n' = 864

Explanation:

Given estimated proportion 'p' = 10% = 0.10

Given Margin of error M.E = 0.02

Level of significance α = 0.05

a)

you are unwilling to predict the proportion value at your school

q = 1- p = 1- 0.10 =0.90

b)

The Margin of error is determined by

`M.E = \frac{Z_{(\alpha )/(2) } √(p(1-p)) }{√(n) }`

`0.02 = (1.96 X √(0.10 (0.90)) )/(√(n) )`

Cross multiplication , we get

`√(n) = (1.96 X √(0.10 (0.90)) )/(0.02 )`

√n = 29.4

Squaring on both sides , we get

n = 864.36

Conclusion:-

The large sample size 'n' = 864