# T-TEST

T-test Example

Rosenthal and Jacobson (1968) informed classroom teachers that some of their students showed

unusual potential for intellectual gains. Eight months later the students identified to teachers as

having potentional for unusual intellectual gains showed significiantly greater gains performance

on a test said to measure IQ than did children who were not so identified. Below are the data for

the students in the first grade:

Table 1: Scores for First Graders

 Experimental Comparison 35 2 40 27 12 38 15 31 21 1 14 19 46 1 10 34 28 3 48 1 16 2 30 3 32 2 48 1 31 2 22 1 12 3 39 29 19 37 25 2 Mean=27.15 Mean=11.95 SD = 12.51 14.62

From the table derived above, the mean of of experiment is X = 27.15 and the comparison mean X = 11.95, the corresponding standard deviations are 12.51 and 14.62

The procedure of this work is derived from the pdf attached to show how t-test is applied in statistical research.

Formula for T-test for indepentdent groups

t = X1-X2

var1  – var2

n         n                                        Substituting our values t = 27.15 – 11.95

12.52     –  14.62

20            20

t = 3.54 this is the computed t value

the computed t value is 3.54. With  degrees of freedom that equals to   total group size (40)

substract 2, to get 38. Entering a t table with 38 degrees of freedom, for  alpha = .05 the

tabulated value is 2.03 and for alpha = .01, the tabled value is 2.72.

In this case, 1% is the limit of expected error and 2.72 is the tabulated t value.

The computed value is bigger than the tabulated value at alpha = .01, hence we reject the null hypothesis and do not reject the alternative hypothesisthis shows , that the difference in gain scores is likely the result to  the experimental treatment and not as a  result of chance variation.

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