Question

A prior study found that 43% of Ph.D. students in the program for Mechanical Engineering were satisfied with the program. The University of "I want to Learn Here" wants to conduct its own research of how satisfied Ph.D. students are in the program for Mechanical Engineering. The University surveyed 461 randomly chosen Ph.D. students in the program for Mechanical Engineering and found that 298 were satisfied with the program.

Calculate the margin of error, lower limit, upper limit for a 90% confidence interval for the proportion of satisfied Ph.D. students in the program for Mechanical Engineering. Then interpret the 90% Confidence Interval comprised from lower limit, upper limit, and margin of error. Using the 90% Confidence Interval which comprises from your calculations, is it plausible that 75% of Ph.D. students in the program for Mechanical Engineering are satisfied with the program?

Answer 1

a)

p̂ = X / n = 298/461 = 0.646

Margin of Error = Z(α/2) √ ( (p̂*( 1 - p̂) ) / n) =

= 1.645 * sqrt [ 0.646 ( 1 - 0.646) / 461)

= 0.037

Lower limit = p̂ - E = 0.646 - 0.037 = 0.609

Upper limit = p̂ + E = 0.646 + 0.037 = 0.683

b)

Interpretation = We are 90% confident that the true proportion of satisfied Ph.D. students are in the program for Mechanical Engineering is between 0.609 and 0.683

c)

Since 0.75 does not contain in confidence interval,

It is not plausible that 75% of Ph.D. students in the program for Mechanical Engineering are satisfied with the program