Question

QUESTION 5

In a large midwestern university (the class of entering freshmen is 6000 or more students), an SRS of 100 entering freshmen in 1999 found that 25 finished in the bottom third of their high school class. Admission standards at the university were tightened in 2000. In 2001, an SRS of 100 entering freshmen found that 15 finished in the bottom third of their high school class.

Let p1 and p2 be the proportion of all entering freshmen in 1999 and 2001, respectively, who graduated in the bottom third of their high school class.  

Is there evidence that the proportion of freshmen who graduated in the bottom third of their high school class in 2001 has been reduced, as a result of the tougher admission standards adopted in 2000, compared to the proportion in 1999?

Use this information to make the best match below as it relates to an appropriate significance test.

-       A.       B.       C.       D.       E.       F.       G.       H.       I.       J.    The null and alternative hypotheses are       -       A.       B.       C.       D.       E.       F.       G.       H.       I.       J.    p?, the pooled estimate, is about       -       A.       B.       C.       D.       E.       F.       G.       H.       I.       J.    The D =p?1 - p?2 is about       -       A.       B.       C.       D.       E.       F.       G.       H.       I.       J.    The test statistic is about       -       A.       B.       C.       D.       E.       F.       G.       H.       I.       J.    The P-value for the test is about       -       A.       B.       C.       D.       E.       F.       G.       H.       I.       J.    The decision, using ? = 0.01, is to

A. reject the null hypothesis. B. 1.96 C. 0.9616 D. 0.20. E. H0: p1 = p2. , Ha: p1 < p2. F. not reject the null hypothesis. G. H0: p1 = p2. , Ha: p1 > p2. H.   0.1. I. 1.77. J. 0.0384

Answer 1

H₀: P1 = P2

H₁: P1 > P2

p1 = 25/100 = 0.25

p2 = 15/100 = 0.15

The pooled proportion(P) = (p1 * n1 + p2 * n2)/(n1 + n2)

                                          = (0.25 * 100 + 0.15 * 100)/(100 + 100)

                                           = 0.20

p1 - p2 = 0.25 - 0.15 = 0.10

The test statistic z = (p1 - p2)/sqrt(P(1 - P)(1/n1 + 1/n2))

                              = 0.1/sqrt(0.2 * (1 - 0.2) * (1/100 + 1/100))

                               = 1.77

P-value = P(Z > 1.77)

              = 1 - P(Z < 1.77)

              = 1 - 0.9616

               = 0.0384

At = 0.01, as the P-value > , we should not reject the null hypothesis.

G - The null and alternative hypothesis

D - The pooled estimate

H - p1 - p2

I - The test statistic

J - P-value

F - The decision