Question

Five students take statistics one semester and college algebra the next semester. Their overall course grades (%) are listed in the table.

Student	Statistics	College Algebra
1	80.0%	85.5%
2	72.6%	71.0%
3	99.0%	93.2%
4	91.3%	93.0%
5	68.9%	74.8%

a. Which statistical procedure, listed in the Assessment, would be most appropriate to test the claim "student overall course grades are the same in both courses"? Options are:
t-Test: Paired Two Sample for Means

t-Test: Two-Sample Assuming Equal Variances

t-Test: Two-Sample Assuming Unequal Variances

z-Test: Two Sample for Means

b. Is there sufficient evidence at the 95% significance level to reject the null hypothesis? Explain

Answer 1

a. Which statistical procedure, listed in the Assessment, would be most appropriate to test the claim "student overall course grades are the same in both courses"?  
t-Test: Paired Two Sample for Means

As each observation is paired

The following table is obtained:

	Sample 1	Sample 2	Difference = Sample 1 - Sample 2
	80.0	85.5	-5.5
	72.6	71.0	1.6
	99.0	93.2	5.8
	91.3	93	-1.7
	68.9	74.8	-5.9
Average	82.36	83.5	-1.14
St. Dev.	12.637	10.25	4.941
n	5	5	5

From the sample data, it is found that the corresponding sample means are:

Also, the provided sample standard deviations are:

and the sample size is n = 5. For the score differences we have

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μD​ = 0

Ha: μD​ ≠ 0

This corresponds to a two-tailed test, for which a t-test for two paired samples be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df = 4

Hence, it is found that the critical value for this two-tailed test is t_c = 2.776 , for \ α=0.05 and df = 4

(3) Test Statistics

The t-statistic is computed as shown in the following formula:

(4) Decision about the null hypothesis

Since it is observed that |t| = 0.516 < t_c = 2.776, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.6331, and since p = 0.6331 > 0.05 , it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is different than μ2​, at the 0.05 significance level.