Question

A researcher interested in the effect of study habits on academic performance wanted to compare two types of study habits: cramming the night before vs. steady studying. He selected a sample of 5 students to use the cramming method for their midterm and the steady method for their final and compared their grades with the following result. Assume a normal population.

Student	Cramming	Steady
A	75	79
B	87	85
C	88	92
D	79	83
E	65	70

1. Specify the null and alternative hypotheses.

2. Test the hypothesis that studying style affects academic performance with alpha = .10. Report the df, the critical value, the SE, the test statistic and your decision.

3. Without calculating anything, if this test was run as an independent measures design, would it be more or less likely to reject the null?

Answer 1

a) Ho:mean difference (say d) between of cramming group & steady group is zero.

H₁: mean difference(_d) between of cramming group & steady group is non-zero.

i.e Ho:_d=0

H₁: _d 0

b) As two groups results are from same students there is a inherent relationship between each observation of two group.So this is a case of pairwise t-test (as sample size small t test should use)

sample size=5

degrees of freedom=5-1=4

Test Statistic=Xbar/(S_d/) under Ho, where Xbar & S_d is the sample mean and SD of difference between two groups

First Find difference of those 5 observation=(4,-2,4,4,5)

Xbar= (4-2+4+4+5)/5= 3

Similarly S_d= 2.828427

Test Statistic=3/ (2.828427/)=2.371708

Standard error=S_d/=2.828427/ =1.264

Critical value ( t_{0.05, 4} ) =2.776 (from table)

As Test Statistic fall between (-2.776 , 2.776) we failed to reject Ho.That is based on the sample data we don't have enough evidence to say there is any statistical difference between two kind of study.

c) without calculating anything, if this test was run as an independent measures design, it would be less likely to reject the null as test statistic value will be very small.