A student claims that 20 of MU students plan to attend our f

A student claims that 20% of MU students plan to attend our football bowl game. Another student believes that the actual percentage is less than that. To prove his point, the second student surveyed 400 MU students and conducted a test at the .05 level of significance. Find the probability of committing a type II error if, in reality, the true proportion is .15. Also, find the power of the test.

Solution

Null Hypothesis : Mean < 0.2

Alternate Hypothesis : Mean >=0.2

n=400

p = 0.20

P = 0.15

= sqrt[ P * ( 1 - P ) / n ]

z = (p - P) /

= (0.2 -0.15) / sqrt( 0.15 * ( 1 - 0.15 ) / 400 )

=2.8

Therefore , P value = 0.9974

Therefore , P > 0.05

Hence , failed to reject null hypothesis.