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To Interpret the SPSS table of Independent sample T-Test, Paired sample T-Test and One way Anova

Assignment III

BYRANJANI B

15AB30MBA (I YEAR)

PSGIM

1. A. Null Hypothesis:There is no significant variance between two batchesAlternate hypothesis:There is a significant variance between two batches

Inter

Levene's Test for Equality of Variances

t-test for Equality of Means

F Sig. t df Sig. (2-tailed)

Mean Difference

Std. Error Difference

95% Confidence Interval of the

DifferenceLower Upper

MARKS Equal variances assumed

1.052 .319 -1.454

18 .163 -14.00 9.629 -34.229 6.229

Equal variances not assumed

-1.454

15.320 .166 -14.00 9.629 -34.486 6.486

Interpretation: Using the Levene’s test for equality of variance, the f-test significant value is 0.319 >

0.05. So t-test for equal variance assumed should be considered. Here, the value of t-test is 0.163 > 0.05. So accept null hypothesis.

It is concluded that there is no significant variance between two batches. The variance are equal. But the mean values are different. Batch B has the high average

then Batch A.Therefore, Batch B is the best.

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Class A

class B

0 10 20 30 40 50 60 70 80

16.421

25.64

21.614

33.748

11.228

17.532

Standard deviation and Std. error

Std. deviation positive deviation negative deviation

Mean0

10203040506070

53.967.9

Mean

class AClass B

marks of batch

N Mean Std. Deviation

Std. Error Mean

MARKS A 10 53.90 16.421 5.193B 10 67.90 25.640 8.108

1. B Null hypothesis: The means are of all the batches are equal. That is µ1=µ2=µ3=µ4.

Alternate hypothesis: At least one batch mean value will be different. That is µ1≠µ2≠µ3≠µ4.

N Mean Std. Deviation Std. Error 95% Confidence Interval for Mean Minimum Maximum

Lower Bound Upper Bound

A 10 53.90 16.421 5.193 42.15 65.65 25 78B 10 67.90 25.640 8.108 49.56 86.24 12 100C 12 43.75 19.786 5.712 31.18 56.32 11 78D 12 42.33 26.756 7.724 25.33 59.33 0 90Total 44 51.16 24.110 3.635 43.83 58.49 0 100

Interpretation: The significant Value is 0.046 < 0.05. So reject null hypothesis and accept alternate

hypothesis. Therefore the means are not equal. To find the highest average, descriptive statistical analysis is done. Batch B has the

highest mean among four batches.

2. Null hypothesis:

2

Mean

Batch ABatch C

53.9

67.9

43.7542.33

meanBatch A Batch B Batch C Batch D

Sum of Squares

df Mean Square F Sig.

Between Groups 4471.170 3 1490.390 2.905 .046Within Groups 20524.717 40 513.118Total 24995.886 43

o The means are of all the sales men are equal. That is µ1=µ2=µ3=µ4.Alternate hypothesis:

o At least one sales man mean will be different. That is µ1≠µ2≠µ3≠µ4.

N Mean Std. Deviation Std. Error 95% Confidence Interval for Mean

Minimum Maximum

Lower Bound Upper Bound

A 4 25.00 4.243 2.121 18.25 31.75 20 29B 4 27.00 4.967 2.483 19.10 34.90 21 32C 4 26.00 7.257 3.629 14.45 37.55 18 35D 4 20.00 4.163 2.082 13.38 26.62 15 25Total 16 24.50 5.502 1.375 21.57 27.43 15 35

- Sum of Squares

df Mean Square F Sig.

Between Groups 116.000 3 38.667 1.373 .298Within Groups 338.000 12 28.167Total 454.000 15

Interpretation: The significant Value is 0.298 > 0.05. So the accept null hypothesis and reject alternate

hypothesis. Therefore the means are equal.

3. Null hypothesis:

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A B C D0

5

10

15

20

25

30 2527 26

20

Mean

o The means are of all the bags are equal. That is µ1=µ2=µ3=µ4.Alternate hypothesis:

o At least one bags mean will be different. That is µ1≠µ2≠µ3≠µ4.N Mean Std.

DeviationStd. Error

95% Confidence Interval for Mean

Minimum Maximum

Lower Bound Upper Bound

Leather bag

5 664.00 61.074 27.313 588.17 739.83 * *

plastic 5 670.00 110.454 49.396 532.85 807.15 * *canvas 5 680.00 115.109 51.478 537.07 822.93 * *Total 15 671.33 91.563 23.642 620.63 722.04 * *

Sum of Squares

df Mean Square F Sig.

Between Groups 653.333 2 326.667 .034 .967Within Groups 116720.00

012 9726.667

Total 117373.333

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Interpretation: The significant Value is 0.967 > 0.05. So the accept null hypothesis and reject alternate

hypothesis. Therefore the means are equal.

4. Null hypothesis:

The perception of the product before and after trying will be equal. That is µ1=µ2

4

Mean655660665670675680

664670

680

Mean

Leather bag plastic canvas

Alternate hypothesis:

The perception of the product before and after trying will not be equal. That is µ1≠µ2

Mean N Std. Deviation Std. Error Mean

Pair 1 Perception before

2.20 25 1.323 .265

Perception after

3.76 25 1.508 .302

Since N = 25, we don't have any missing values on the test variables.

As expected, the mean reaction about the perception of the product after trying

(3.76) is higher than before trying it (2.20).

Paired Differences t df Sig. (2-tailed)

Mean Std. Deviation

Std. Error Mean

95% Confidence Interval of the Difference

Lower Upper

Pair 1

Perception before - Perception after

-1.56 2.551 .510 -2.61 -.51 -3.058

24 .005

On average, respondents slow down some 1.56.

The p-value denoted by “Sig. (2-tailed)” is 0.005. So if the population means are equal,

there's a 0% chance of finding this result. We therefore reject the null hypothesis.

Therefore the perception of the product get differ after trying it.

5. Null hypothesis:

The performance of the employee before and after the training will be equal. That is

µ1=µ2

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Alternate hypothesis:

The performance of the employee before and after the training will not be equal. That is

µ1≠µ2

Mean N Std. Deviation

Std. Error Mean

Pair 1 before training

61.80 10 5.073 1.604

after training

61.90 10 6.420 2.030

Since N = 10, we don't have any missing values on the test variables.

As expected, the mean reaction time before a training (61.80 minutes) is slightly lower

than after the training (61.90 minutes).

Sig. (2-tailed

Mean Std. Deviation

Std. Error Mean

95% Confidence Interval of the Difference

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Lower UpperPair 1

before training - after training -.10 5.666 1.792 -4.15 3.95 -.05

6 9 .957

On average, respondents slow down by 0.10.

The p-value denoted by “Sig. (2-tailed)” is 0.957 > 0.05. We therefore accept the null

hypothesis. Therefore the performance before and after the training are almost equal.

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