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.

a)

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

b)

The large sample size 'n' = 864

Step-by-step 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

Cross multiplication , we get

                      √n  = 29.4

Squaring on both sides , we get

                     n = 864.36

Conclusion:-

The large sample size 'n' = 864

-13

step-by-step explanation:

f(x)=2x-5 for f(-4)

f(-4)   means let x=-4

f(-4) = 2(-4) -5

f(-4) =-8-5

      = -13