Question

A student believes that less than 50% of students at his college receive financial aid. A random sample of 120 students was taken. Sixty-five percent of the students in the sample receive financial aid. Test the hypothesis at the 2% level of significance. What are the p-value and conclusion?

a. .50; Reject H0
b. .999; Do not reject H0
c. 3.29; Do not reject H0
d. .02; Reject H0

Answer 1

P-value = 0.999 ; Conclusion = Do not reject
`H_0`

Explanation:

We are given that a student believes that less than 50% of students at his college receive financial aid. A random sample of 120 students was taken. Sixty-five percent of the students in the sample receive financial aid.

Let Null Hypothesis,
`H_0` : p
`\geq` 0.50 {means that more than or equal to 50% of students at his college receive financial aid}

Alternate Hypothesis,
`H_a` : p < 0.50 {means that less than 50% of students at his college receive financial aid}

The test statics that will be used here is One-sample proportions test;

T.S. =
`\frac{\hat p -p}{\sqrt{(\hat p(1-\hat p))/(n) } }` ~ N(0,1)

where,
`\hat p` = % of the students in the sample that receive financial aid = 65%

n = sample of students = 120

So, test statistics =
`\frac{0.65-0.50}{\sqrt{(0.65(1-0.65))/(120) } }`

= 3.45

Now, at 0.02 level of significance the z table gives critical value of -2.054. Since our test statistics is more than the critical value of z so we have insufficient evidence to reject null hypothesis as it will not fall in the rejection region.

P-value is given by = P(Z < 3.45) = 0.99972 or 0.999 {using z table}

So, we conclude that null hypothesis is not rejected and p-value is 0.999.