The two kinds of formula are used in the excel . One from the data analysis the <u><em>t test </em></u><u><em>assuming equal variances </em></u>as nothing was mentioned. But if the t test for the means is used there might be slight variation in the values.
The other formula was of descriptive statistics.
The data was
No of feedback (sec) Feedback (sec) Difference
45 54 9
33 50 17
59 58 -1
32 38 6
30 42 12
27 35 8
29 38 9
59 66 7
64 65 1
38 55 17
The null hypothesis is accepted because t stat= -1.322 is less than t critical which is t critical= ±2.119 and lies in the acceptance region.
The p- value is 2.04 which is greater than alpha= 0.05
The t stat for 0.01= 4.561578674 and t Critical two-tail 3.249835541
The t stat is greater than t critical at alpha 0.01 so it is rejected.
The difference t test gives
Count=N 9 9
Mean 41.2 49.67
Mean Difference 8.4
Standard Deviation 14.967 11.97
Variance 223.944 143.25
Standard Error 2.0824
t = -1.322
df = 16
t Critical two-tail 2.119905285
Confidence Level(95.0%) 4.802039614
Lower CI (95.0%) = 3.6
Upper CI(95.0%)= 13.2
t stat for 0.01= 4.561578674
t Critical two-tail 3.249835541
Confidence Level(99.0%) 6.055710988
Lower CI (99.0%) = 2.345
Upper CI(99.0%)= 14.457
For values between 3.6 and 13.2 we are 95% confident that there is no difference between the means of two given data values.
The null hypothesis is accepted because t stat= -1.322 is less than t critical which is t critical= ±2.119 and lies in the acceptance region.
The p- value is 2.04 which is greater than alpha= 0.05
The t stat for 0.01= 4.561578674 and t Critical two-tail 3.249835541
The t stat is greater than t critical at alpha 0.01 so it is rejected.
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