Question

According to a recent​ report, 46​% of college student internships are unpaid. A recent survey of

60college interns at a local university found that 34had unpaid internships.

a. Use the​ five-step p-value approach to hypothesis testing and a

0.05level of significance to determine whether the proportion of college interns that had unpaid internships is different from 0.46

b.Assume that the study found that 37of the 60college interns had unpaid internships and repeat​ (a). Are the conclusions the​ same?

a. Let πbe the population proportion. Determine the null​ hypothesis,H0​,

and the alternative​ hypothesis, H1.

What is the test statistic? ZSTAT=_____(round to two decimal places)

What is the p- value? Round to three decimal places

What is the final conclusion? (Reject/Do not reject) the null hypothesis. There (is/is not) sufficient evidence that the proportion of college interns that had unpaid internships is (less than/different than/greater than) 0.46 because the​ p-valueis (greater than/less than) the level of significance

b.Assume that the study found that 37 of the 60college interns had unpaid internships and repeat​ (a). What is the test​ statistic? ZSTAT= (round to two decimal places)

What is the p value? Round to 3 decimal places

The result is (the same as/different from) part a. (Reject/ Do not reject) the null hypothesis. There (is/is not) sufficient evidence that the proportion of college interns that had unpaid internships is(greater than/less than/different from) 0.46 because the p-value is (less than/greater than) the level of significance.

Answer 1

Here

n=60

x=34

a)

we have to test that

H0: P=0.46 Ha: P 0.46

we have

test statistics is given by

test is two tailed so

P-Value =2*P(Z>1.71)=2*0.044=0.088

since P value is more than level of significance hence we failed to reject H0 Hence

Do not reject the null hypothesis. There is not sufficient evidence that the proportion of college interns that had unpaid internships is different than 0.46 because the​ p-valueis greater than the level of significance

b)

Here

n=60

x=37

so

test statistics is given by

test is two tailed so

P-Value =2*P(Z>2.49)=2*0.006=0.012

since P value is less than level of significance hence we reject H0 Hence

reject the null hypothesis. There is sufficient evidence that the proportion of college interns that had unpaid internships is different than 0.46 because the​ p-valueis less than the level of significance