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

Answer:

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

Step-by-step 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. = $\frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ 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 = $\frac{0.65-0.50}{\sqrt{\frac{0.65(1-0.65)}{120} } }$

                             = 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.