Question

A researcher wishes to prove that less than 40% of students support the changes announced by the Ford
government in January 2019 to tuition, the Ontario Student Assistance Program, and student fees. If 32% of all
students support the changes, what is the chance that a random sample of 250 students provides insufficient proof
to meet a 5% significance level? In other words, what is the probability of a Type II error? Answer with hypotheses in
formal notation, TWO fully-labelled graphs, a quantitative analysis & the requested probability

Answer 1

The null and alternative hypotheses

H0: p 0.40

Ha: p < 0.40

We have to find probability of type II error

Type II error is the probability of accepting the null hypothesis given that the null hypothesis is false, that is probability of getting insignificant result given p < 0.40

Test statistic

At

left tailed critical value of z is , zc = - 1.65

Reject the null hypothesis if z < -1.65

Fail to reject the null hypothesis (accept the null hypothesis) if z > -1.65

zc = -1.65

that is we reject H0 , if and accept if

P (type II error) =P( accept H0 I H0 false )

= P( accept H0 I H0 false )

=P( I p=0.32 )

= P( z > 1.02)

=0.1538

Therefore , probability of type II error =0.1538