Question

please show all work:

68% of students at a university live on campus. A random sample found that 23 of 45 male students and 42 of 55 of female students live on campus. At the 0.05 level of significance, is there sufficient evidence to support the claim that a difference exists between the proportions of male and female students who live on campus?

Evaluate using EITHER a Traditional or P-value Hypothesis Test (your choice.)

* If using Traditional - draw the normal curve, labeling the CV, TV, and shading the critical region. * If using P-value - explain why you chose your decision. Next, find the confidence interval using the same alpha.

Last, explain how your confidence interval supports your hypothesis test.

Answer 1

Ans:

pooled proportion=(23+42)/(45+55)=0.65

Test statistic:

z=(0.5111-0.7636)/sqrt(0.65*(1-0.65)*((1/45)+(1/55)))

z=-2.63

p-value(2 tailed)=2*P(z<-2.63)=0.0085

critical z values=+/-1.96

As,p-value<0.05,Reject the null hypothesis.

There is sufficient evidence to conclude that a difference exists between the proportions of male and female students who live on campus.

95% confidence interval for difference in proportions

=(0.5111-0.7636)+/-1.96*sqrt((0.5111*(1-0.5111)/45)+(0.7636*(1-0.7636)/55))

=-0.2525+/-0.1842

=(-0.437, -0.068)