Question

The dean of students affairs at a college wants to test the claim that 40 % of all undergraduate students reside in the college dormitories. If 57 out of 69 randomly selected undergraduate students reside in the dormitories, does this support dean's claim with α = 0.05 ?

Test statistic:

Critical Value:

Do we accept or reject Dean's claim?

A. Reject Dean's claim that 40 % of undergraduatute students live in dormitories.

B. There is not sufficient evidence to reject Dean's claim.

Answer 1

Answer)

Null hypothesis Ho : P = 0.4

Alternate hypothesis Ha : P not equal to 0.4

N = 69.

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 27.6

N*(1-p) = 41.4

Both the conditions are met so we can use standard normal z table to estimate the P-Value

Test statistics z = (oberved p - claimed p)/standard error

Standard error = √{claimed p*(1-claimed p)/√n

Observed P = 57/69

Claimed P = 0.4

N = 69.

After substitution,

Z = 7.22

As the test is two tailed,

We will divide given significance into two parts = 0.05/2 = 0.025.

From z table, P(z<-1.96) = P(z>1.96) = 0.025.

Critical values are -1.96 and 1.96.

Rejection region is, reject Ho if test statistics is greater than 1.96 or less than -1.96.,

As 7.22 > 1.96, we reject Ho.

So answer is,

Reject Dean's claim that 40 % of undergraduatute students live in dormitories.