Question

Test the claim that the mean GPA of Orange Coast students is significantly different than the mean GPA of Coastline students at the 0.05 significance level.
The null and alternative hypothesis would be:

The test is: two-tailed, right-tailed, left-tailed?

The sample consisted of 60 Orange Coast students, with a sample mean GPA of 3.38 and a standard deviation of 0.04, and 60 Coastline students, with a sample mean GPA of 3.41 and a standard deviation of 0.06.
The test statistic is:  (to 2 decimals)
The p-value is:  (to 2 decimals)
Based on this we: Fail to reject the null hypothesis or Reject the null hypothesis?

Answer 1

Null hypothesis: H:44 = 12 | Alternative hypothesis: H:14 #lly The test is two tailed test Level of significance: a = 0.05 Use the Minitab software to find the test statistic for the given data with the help of following steps: 1. Enter the data in to the Minitab work sheet, 2. Select stat from the Minitab menu bar. 3. Choose stat-> basic statistics->2 sample t test. 4. In the 2 sample t test dialog box, click on summarized data, and enter the values of sample size, sample mean, sample standard deviations. 5. Choose options ->Confidence level (95.0->alternative hypothesis (<). 6. Click ok. The obtained output as follows:

Two-Sample T-Test and CI NU Sample 1 2 N Mean St Dev SE Mean 60 3.3800 0.0400 0.0052 60 3.4100 0.0600 0.0077 Difference = u(1) - (2) Estimate for difference: -0.03000 95% CI for difference: (-0.04847, -0.01153) T-Test of difference = 0 (vs #): I-Value = -3.22 P-value The value of test statistic is, t = -3.22 The p-value is 0.00 (Round to 2 decimal places) 0.002 DF = 102 Decision rule: Reject Ho, if p sa otherwise do not reject H,. Since the p-value less than the level of significance, so we reject the null hypothesis at specified significance level. Conclusion: There is sufficient evidence to support the claim that the mean GPA of Orange Coast students is significantly different than the mean GPA of Coastline students