Question

The director of student services at Oxnard College is interested in whether women are less likely to attend orientation than men before they begin their coursework. A random sample of freshmen at Oxnard College were asked what their gender is and whether they attended orientation. The results of the survey are shown below:

Data for Gender vs. Orientation Attendance

	Women	Men
Yes	456	493
No	204	197

What can be concluded at the αα = 0.05 level of significance?

• The test statistic ? z t Correct =  (please show your answer to 3 decimal places.)

• The p-value =  (Please show your answer to 4 decimal places.)

Answer 1

Answer:

Given,

sample proportion p1 = x1/n1 = 456/660 = 0.6909

p2 = x2/n2 = 493/690 = 0.7145

Pooled proportion p = (x1 + x2)/(n1 + n2)

substitute values

= (456+493)/(456+493+204+197)

= 949 / 1350

= 0.7029

q = 1 - p

= 1 - 0.7029

= 0.2971

Standard error = sqrt(pq(1/n1 + 1/n2))

substitute values

= sqrt(0.7029*0.2971(1/660 + 1/690))

= 0.0248

test statistic z = (p1 - p2)/se

substitute values

= (0.6909 - 0.7145)/0.0248

= - 0.9516

P value = P(z < - 0.9516)

= 0.1706499 [since from z table]

= 0.1706

Here we observe that p value > alpha, so we fail to reject Ho.

So there is no sufficient evidence to support the claim.