TABLE 12-6
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, the hypotheses the dean should use are:
A) H0: ¦Ð1 - ¦Ð2 = 0 versus H1: ¦Ð1 - ¦Ð2 ¡Ù 0
B) H0: ¦Ð1 - ¦Ð2 ¡Ù 0 versus H1: ¦Ð1 - ¦Ð2 = 0
C) H0: ¦Ð1 - ¦Ð2 ¡Ü 0 versus H1: ¦Ð1 - ¦Ð2 > 0
D) H0: ¦Ð1 - ¦Ð2 ¡Ý 0 versus H1: ¦Ð1 - ¦Ð2 < 0
(b) Referring to Table 12-6, the value of the test statistic is ________.
A) ¦Ö2 = 3.4806
B) ¦Ö2 = 34.806
C) ¦Ö2 = 0.05
D) ¦Ö2 = 348.06
I am not familiar with some of the symbols you are using, and I do not have access to Table 12-6. However:
Ho: Mean 1 = Mean 2
Ha: Mean 1 ≠ Mean 2
I hope this helps.
To determine the appropriate hypothesis test for this scenario, we need to consider the research question and the null and alternative hypotheses. In this case, the dean wants to investigate if there is a difference in the proportion of graduates with job offers between accounting and economics majors.
The appropriate hypotheses for this test are:
H0: μ1 - μ2 = 0
H1: μ1 - μ2 ≠ 0
The null hypothesis (H0) states that there is no difference in the proportion of graduates with job offers between accounting and economics majors. The alternative hypothesis (H1) states that there is a significant difference.
Now, let's move on to determining the value of the test statistic from Table 12-6. Unfortunately, you haven't provided Table 12-6, so I'm unable to calculate the value for you. However, once you have the data from Table 12-6, you can use the formula for the test statistic for comparing two proportions:
test statistic = (p̂1 - p̂2) / √[(p̂(1-p̂) / n1) + (p̂(1-p̂) / n2)]
Here, p̂1 and p̂2 are the sample proportions, n1 and n2 are the sample sizes, and p̂ is the overall pooled proportion.
Using the provided information from the question, you would substitute the values into the formula to calculate the test statistic. Once you have calculated it, you can compare it to the critical value(s) from the appropriate distribution to determine if you reject or fail to reject the null hypothesis.
Unfortunately, without the values from Table 12-6, I cannot calculate the test statistic for you. Please refer to the table and use the formula provided to find the correct value.
(a) The correct answer for the hypotheses the dean should use is:
A) H0: ¦Ð1 - ¦Ð2 = 0 versus H1: ¦Ð1 - ¦Ð2 ¡Ù 0.
(b) To perform the appropriate hypothesis test, we need to calculate the test statistic. In this case, we will be using the chi-square test statistic. The formula for the chi-square test statistic for independent proportions is:
χ2 = ( (O1 - E1)^2 / E1 ) + ( (O2 - E2)^2 / E2 )
where O1 and O2 are the observed frequencies (number of accounting and economics majors with job offers) and E1 and E2 are the expected frequencies (assuming the null hypothesis is true).
In this case, the observed frequencies are 65 accounting majors with job offers (Group 1) and 52 economics majors with job offers (Group 2) out of samples sizes of 100 for each group.
The expected frequencies can be calculated assuming the null hypothesis is true, and assuming that the proportions of accounting and economics majors with job offers are the same. Since the null hypothesis states that ¦Ð1 - ¦Ð2 = 0, we can assume that the proportions ¦Ð1 and ¦Ð2 are equal.
The expected frequencies can be calculated as:
E1 = (65 + 52) * (100 / 200) = 58.5
E2 = (65 + 52) * (100 / 200) = 58.5
Now we can substitute the observed and expected frequencies into the chi-square formula to calculate the test statistic.
χ2 = ( (65 - 58.5)^2 / 58.5 ) + ( (52 - 58.5)^2 / 58.5 )
Calculating the test statistic, we get:
χ2 = 1.966
Therefore, the value of the test statistic is approximately 1.966. None of the given options A), B), C), or D) match the correct value.