Question

To test the effectiveness of a business school preparation course, 8 students took a general business test before and after the course. The results are given below.

Student Last Name	Score Before Course	Score After Course
Garcia	670	730
Lam	710	720
Rodriguez	870	860
Torriero	770	800

1. At the 0.05 level of significance, the conclusion for this hypothesis test is:

a		Do not reject H0: there is sufficient evidence that the business school preparation course has an impact on score.
b		Reject H0: there is sufficient evidence that the business school preparation course has an impact on score.
c		Do not reject H0: there is insufficient evidence that the business school preparation course has an impact on score.
d		Reject H0:  there is insufficient evidence that the business school preparation course has an impact on score.

Answer 1

The answer is:

c

Do not reject H0: there is insufficient evidence that the business school preparation course has an impact on score.

Student	Score Before Course	Score After Course	Difference
1	670	730	-60
2	710	720	-10
3	870	860	10
4	770	800	-30

Sample Size (n)	4
Sample Mean(Xˉ)	-22.500
Sample St. Deviation (s)	29.861

Paired Sample t test Output

For the score differences we have, mean is Dˉ=-22.5, the sample standard deviation is sD=29.8608, and the sample size is n=4. (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. (2a) Critical Value Based on the information provided, the significance level is α=0.05, and the degree of freedom is n-1=4-1=3. Therefore the critical value for this Two-tailed test is tc​=3.1824. This can be found by either using excel or the t distribution table. (2b) Rejection Region The rejection region for this Two-tailed test is |t|>3.1824 i.e. t>3.1824 or t<-3.1824 (3)Test Statistics The t-statistic is computed as follows: (4) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is 0.2289 (5) The Decision about the null hypothesis (a) Using traditional method Since it is observed that |t|=1.507 < tc​=3.1824, it is then concluded that the null hypothesis is Not rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0.2289, and since p=0.2289>0.05, it is concluded that the null hypothesis is Not rejected. (6) 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.