# 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...

**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.

Given Data:

**(Group 1)** `n_1 = 5 `

`stackrel(-)(x_1)=4`

`S_(n_1-1)=0.707107`

`(S_(n_1-1))^2 =0.5`

**(Group 2)** `n_2 = 5 `

`stackrel(-)(x_2)=1`

`S_(n_2-1)=1 `

`(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 `

=`sqrt((0.5+1)/5) `

=`sqrt(0.3)`

=`0.547723 `

**Step 4**: Calculation of t

`t =( stackrel(-)(x_1) -stackrel(-)(x_2))/ (S.E.)`

=(4-1) /0.547723

=5.477226

=5.48 (approx.)

**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)) `

`=sqrt0.75 `

`=0.866025 `

C.I.= Difference in mean `+-(t_(hypo(at alpha=0.05, df=8))*S_p)/sqrt(n)`

`=(4-1)+-(2.306*0.866025)/sqrt5 `

`=3+-0.8931`

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).