Automatically calculate the two sample independent t test. [Two site hints: http://www.usablestats.com/calcs/2samplet ; and http://www.statisticslectures.com/calculators/ttest2/ ]. Using the following two independent samples of scores of five scores each, from Group One the scores are: 4, 5, 4, 4, and 3. From Group Two the five scores are 0, 0, 2, 1, and 2. Using a two-tailed test and alpha equal to .05, calculate the independent t statistic. Assume equal variances. You will find that calculated t is about 5.48.
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The t- statistic can be calculated for the given data sets in the following way (you can check the results from the calculations done in the given websites):
Step 1: Stating the hypotheses
Null hypothesis: The scores in the two Groups are not significantly different.
Alternate hypothesis: The scores in Group A is significantly different from Group B.
Step 2: Calculation of S.D.
(Group 1) `n_1 = 5 `
(Group 2) `n_2 = 5 `
`(S_(n_2-1))^2 =1 `
Step 3: Calculation of S.E. of the mean difference
S.E.mean diff. = `((S_(n_1-1))^2+(S_(n_2-1))^2)/n `
Step 4: Calculation of t
`t =( stackrel(-)(x_1) -stackrel(-)(x_2))/ (S.E.)`
Step 5: Confidence Interval at alpha= 0.05(two tailed test, df=8)
For this, `S_p` , the pooled S.D. has to be calculated first (assuming equal variance).
`S_p=sqrt((n_1-1) (S_(n_1-1))^2+(n_2-1)(S_(n_2-1))^2)/(n_1+n_2-2)``=sqrt((4*0.5+4*1)/(5+5-2)) `
C.I.= Difference in mean `+-(t_(hypo(at alpha=0.05, df=8))*S_p)/sqrt(n)`
Lower limit= 3-0.8931= 2.1069
Upper limit = 3+0.8931 = 3.8931
Interpretation of results:
1. As `t_(calc.)` is greater than `t_(hypo)` , the null hypothesis is rejected. In other words, Marks in Group A differ significantly from the marks in Group B.
2. It can be said with 95% confidence that the difference in mean of these two groups will be in the interval (2.1069, 3.8931).
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