Question

The Dean of Students at UTC has said that the average grade of UTC students is higher than that of the students at GSU. Random samples of grades from the two schools are selected, and the results are shown below.

	UTC 91)	GSU (2)
Sample Size	14	12
Sample Mean	2.85	2.61
Sample Standard Deviation	0.40	0.35
Sample Mode	2.5	3.0

At a 0.1 level of significance, conduct a test of hypothesis to assist the Dean evaluate his hypothesis. If the test statistic value for this problem is 1.61, what statement can be made about the Dean of Student's statement?

Select one:

A. Reject null hypothesis, therefore, reject Dean's statement

B. Fail to reject null hypothesis, therefore, don't reject Dean's statement

C. Fail to reject null hypothesis, therefore, reject Dean's statement

D. Reject null hypothesis, therefore, don't reject Dean's statement

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_UTC< u_GSU
Alternative hypothesis: u_UTC > u_GSU

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 0.1471
DF = 24

t = [ (x₁ - x₂) - d ] / SE

t = 1.61

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is thesize of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

The observed difference in sample means produced a t statistic of 1.61

Therefore, the P-value in this analysis is 0.06.

Interpret results. Since the P-value (0.06) is less than the significance level (0.10), we have to reject the null hypothesis.

D. Reject null hypothesis, therefore, don't reject Dean's statement.