Using this TABLE 12-6 for the following questions:

The dean of a college is interested in the proportion of graduates from his college who have a job offer on graduation day. He is particularly interested in seeing if there is a difference in this proportion for accounting and economics majors. In a random sample of 100 of each type of major at graduation, he found that 65 accounting majors and 52 economics majors had job offers. If the accounting majors are designated as "Group 1" and the economics majors are designated as "Group 2," perform the appropriate hypothesis test using a level of significance of 0.05.

(a) Referring to Table 12-6 (above), the hypotheses the dean should use are:
A) H0: pi one -pi2 = 0 versus H1: pi one -pi2 ¡Ù 0
B) H0: pi one - pi2 ¡Ù 0 versus H1: pi one - pi2 = 0
C) H0: pi one - pi2 ¡Ü 0 versus H1: pi one - pi2 > 0
D) H0: pi one - pi2 ¡Ý 0 versus H1: pi one - pi2 < 0

(b) Referring to Table 12-6 (above), the value of the test statistic is ________.
A) ¦Ö2 = 3.4806
B) ¦Ö2 = 34.806
C) ¦Ö2 = 0.05
D) ¦Ö2 = 348.06

To perform the appropriate hypothesis test, we need to compare the proportions of accounting majors and economics majors who have job offers on graduation day.

(a) The correct hypothesis the dean should use is:
A) H0: pi1 - pi2 = 0 versus H1: pi1 - pi2 ≠ 0

Here, pi1 represents the proportion of accounting majors with job offers, and pi2 represents the proportion of economics majors with job offers. The null hypothesis (H0) assumes that there is no difference in the proportions, while the alternative hypothesis (H1) assumes that there is a difference.

(b) To calculate the test statistic, we can use the formula for the chi-square test statistic:

χ2 = Σ((O - E)^2 / E)

where O is the observed frequency and E is the expected frequency. In this case, the observed frequencies are 65 accounting majors with job offers and 52 economics majors with job offers.

To calculate the expected frequencies, we can use the formula:

E = (n1 * n2) / N

where n1 and n2 are the sample sizes for each group (100 in this case), and N is the total sample size (200).

E = (100 * 100) / 200 = 50

Then, we can calculate the test statistic:

χ2 = ((65 - 50)^2 / 50) + ((52 - 50)^2 / 50)
= (15^2 / 50) + (2^2 / 50)
= 225/50 + 4/50
= 4.5 + 0.08
= 4.58

The value of the test statistic is 4.58.

Therefore, the answer is:
A) χ2 = 3.4806