Question

Use the following information for the next 3 questions:

Suppose you want to test for whether there is a difference in starting salaries for those with a finance degree compared to those with an accounting degree. You sample 35 graduates with a finance degree and find the average starting salary to be 58,000 with a sample standard deviation of 5,000. You also take a sample of 40 graduates with an accounting degree and find the average starting salary 56,000 to be and a sample variance of 3000. Suppose you want to test the hypothesis that the starting salary for accounting degree graduates is less than that of finance degree graduates. (D=μaccounting−μfinance) Which of the following is the hypothesis test?

H0:D≥0;HA:D<0H0:D≥0;HA:D<0

H0:D≤0;HA:D>0H0:D≤0;HA:D>0

H0:D<0;HA:D<0

H0:D=0;HA:D≠0

Using the previous information, which of the following is the test statistic for the hypothesis test? Recall: D=μaccounting−μfinance

135.5

-2.13

-0.002

-2.06

Assume the test statistic found was 2.5, and the degrees of freedom found was 50. What is the critical value and outcome of the hypothesis test at the 1% significance level, if the hypothesis was found to be: H0:D=0;HA:D≠0

$LaTeX: \pm$±±2.403; Fail to Reject the Null Hypothesis

$LaTeX: \pm$±±2.678; Reject the Null Hypothesis

$LaTeX: \pm$±±2.403; Reject the Null Hypothesis

$LaTeX: \pm$±±2.678; Fail to Reject the Null Hypothesis

|(D = Haccounting finance Ho: D 0; HA: D0 Ho: D

Answer 1

1. Since we want to test the hypothesis that the starting salary for accounting is greater than that of finance,

$\small H_0:D\leq0, H_A:D>0$

2. To calculate the test statistic, we need to compute the standard deviation:

Standard deviation of the joint sample:

3000 14.76 5000 35 + 40

Test statistic can be calculated as follows:

56000-58000-0-135.5 14.76

3. For a two tailed test, we find the t value corresponding to 50 degrees of freedom and 0.01 significance level. This value is 2.678. Since the calculated test statistic is less than the critical value, we fail to reject the null hypothesis.