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By using excel sheet we can get the answers for the difference t test.

The two kinds of formula are used in the excel . One from the data analysis the t test assuming equal variances 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

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.

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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