The Dean of Students at ACME University wants to estimate the proportion of stud

Question

The Dean of Students at ACME University wants to estimate the proportion of students who cheated on a major exam.

How many students does the Dean need to select in the sample in order to be 95% confident that the sample proportion is within two percentage points of the actual proportion who cheated if:

a. previous studies at other universities indicate this proportion is less than 20%?

b. the Dean thinks ACME students have more integrity than students at other universities?

c. the Dean really has no idea what the actual proportion of cheaters might be?

Explanation / Answer

a) n = (z (alpha/2) )^ 2 * p(1-p) / MOE^2

z0.025 = 1.96

p = 0.2

1-p = 0.8

MOE = 0.02

n = (1.96*1.96*0.2*0.8 ) /(0.02*0.02) = 1537

the dean requires 1537 students.

b)

when dean thinks ACME students have more integrity then his estimate of proportion is much lesser than 20%

we can give the cases for 15% 10% and 5%

c)

when the dean has no idea of the actual proportion, he can be conservative and can assume p to be 0.5

in that case

n = (1.96*1.96*0.5*0.5 )/(0.02*0.02) = 2401

Proportion Sample 0.15 =+(1.96*1.96*0.15*0.85 )/(0.02*0.02) 1225 0.1 =+(1.96*1.96*0.1*0.9 )/(0.02*0.02) 865 0.05 =+(1.96*1.96*0.05*0.95 )/(0.02*0.02) 457

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