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Statistics

Name of Student

Name of Institution

Scenario 1

Result section

Since the standard deviation is not given for the population and the sample size is = 30, we will use t test for this analysis.

The formula will be

t =(X^ - pop. Mean)/s/n^0.5

The sample test used is one sample t test

Mean for males: 39.15

Mean for females: 42.176

There is one outlier value i.e. 2.

The value was changed to 42 because it was closer to the overall data

The assumption of normality was not met as the significance of both the tests below is less than 0.05. this means that the null hypothesis of normality is rejected.

Tests of Normality

Kolmogorov-Smirnova

Shapiro-Wilk

Statistic

df

Sig.

Statistic

df

Sig.

Score

.263

30

.000

.598

30

.000

The degrees of freedom were 29. Calculated as n-1

One-Sample Test

Test Value = 25

t

df

Sig. (2-tailed)

Mean Difference

95% Confidence Interval of the Difference

Lower

Upper

Score

10.654

29

.000

15.867

12.82

18.91

The value of t statistic is 10.654

P<0.001

Cohenâ€™s d = 10.654/30^0.5 = 1.945

This value is interpreted in a sense that the sample mean is 1.945 standard deviations higher than the population mean

Task 2

Answer 4 because the test is carried out just to see the differences between the sample and population means. Nothing like lower or higher.

Task 3

The data was not normal

Scenario 2

The scenario will require the use of dependent samples t test and the following formula will be used.

T = d^/s/n^0.5

d^ is the mean of differences between values before and after the tests.

2. The paired samples t test was applied.

3.Males = 12.841

Females = 12.171

4. There were no outliers in the series weight before but there are two outliers in the series weight after. These were 127 and 146.

5. The numbers were changed to 56 and 72.

6. End weight

Tests of Normality

Kolmogorov-Smirnova

Shapiro-Wilk

Statistic

df

Sig.

Statistic

df

Sig.

EndWeight

.268

34

.000

.646

34

.000

Start Weight

Tests of Normality

Kolmogorov-Smirnova

Shapiro-Wilk

Statistic

df

Sig.

Statistic

df

Sig.

StartWeight

.103

35

.200*

.970

35

.434

The above tables show that the normality assumption is fulfilled in case of start weight only.

7. The degrees of freedom are 33.

8. The value of t statistic was -2.868

9. The probability value is 0.007.

10. The effect size is not applicable

11. Not applicable

Task 2

Option 4

Task 3

Experimenter effect

Scenario 3

The scenario is the case for independent samples t test with sample 1 is for the group 1 and sample 2 is for group 2.

T = (X^1- X^2)/ (S1/n1+S2/n2)^0.5

The independent t test was used

There are 4 levels of IV. M1, M2, F1 and F2.

Male avg = 42.5

Female avg = 38.28

The assumption of normality is met as the significance scores of the test are greater than 0.05.

Tests of Normalitya,b,c,d,e,g,h,i,j,k,l,m,n,o,p

Age

Kolmogorov-Smirnovf

Shapiro-Wilk

Statistic

df

Sig.

Statistic

df

Sig.

Output

29

.260

2

.

31

.358

3

.

.812

3

.144

37

.260

2

.

43

.260

2

.

56

.260

2

.

64

.161

4

.

.990

4

.959

b. Output is constant when Age = 20. It has been omitted.

c. Output is constant when Age = 22. It has been omitted.

d. Output is constant when Age = 25. It has been omitted.

e. Output is constant when Age = 28. It has been omitted.

f. Lilliefors Significance Correction

g. Output is constant when Age = 30. It has been omitted.

h. Output is constant when Age = 32. It has been omitted.

i. Output is constant when Age = 34. It has been omitted.

j. Output is constant when Age = 36. It has been omitted.

k. Output is constant when Age = 38. It has been omitted.

l. Output is constant when Age = 39. It has been omitted.

m. Output is constant when Age = 42. It has been omitted.

n. Output is constant when Age = 54. It has been omitted.

o. Output is constant when Age = 55. It has been omitted.

p. Output is constant when Age = 63. It has been omitted.

Since the values are different, the variances will not be homogenous for the two groups.

There will be 36 degrees of freedom

The t statistic is -4.448

Independent Samples Test

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

Output

Equal variances assumed

3.588

.066

-3.073

36

.004

-172.167

56.032

Equal variances not assumed

-4.448

23.484

.000

-172.167

38.704

P< 0.001

This is not applicable in this scenario

Not applicable

Task 2

Option 3 as the significance value is less than 0.05.

Task 3

Experimenter effects

Scenario 4

The difference of mean in case of independent samples is applied here

T = (X^1- X^2)/ (S1/n1+S2/n2)^0.5

The independent samples t test is applied here in this scenario.

Males = 9.0746

Females = 12.362

There is only one outlier 5 in c2.

5.

Tests of Normality

Kolmogorov-Smirnova

Shapiro-Wilk

Statistic

df

Sig.

Statistic

df

Sig.

C1

.089

43

.200*

.983

43

.751

For C1 the value of significance is greater than 0.05 thus the normality assumption holds

Tests of Normality

Kolmogorov-Smirnova

Shapiro-Wilk

Statistic

df

Sig.

Statistic

df

Sig.

C2

.132

43

.059

.789

43

.000

The significance value is less than 0.05, thus , the normality assumption does not hold.

The homogeneity of variance assumption does not exist.

The data provided for this part is insufficient to solve it.

Scenario 5

The single sample t test is applied in this scenario

T = (X^ - pop. Mean) / s/n^0.5

This is the case of single sample t test

Males = 24.55

Females = 24.5

Standard deviation

Males = 3.64

Females =3.64

No outlier was found

No outlier

The normality assumption was met as the significance value is greater than 0.05

Tests of Normality

Kolmogorov-Smirnova

Shapiro-Wilk

Statistic

df

Sig.

Statistic

df

Sig.

Score

.169

40

.005

.961

40

.185

Yes the variances were the same

T value is 58.96

It was less than 0.05

0.05

Task 2

Option 3

Task 3

Lack of reliability

Scenario 6

The difference of means for independent samples will be calculated here

T = (X^1- X^2)/ (S1/n1+S2/n2)^0.5

This is a case of difference of means for independent samples

Mean age females = 44.41

Mean age Males = 43.69

There was 1 outlier in group 1 with score 2 and one in group 2 with score 3

The following table is for group 1 which shows that the normality assumption has not been met.

Tests of Normality

Kolmogorov-Smirnova

Shapiro-Wilk

Statistic

df

Sig.

Statistic

df

Sig.

Score

.205

25

.008

.883

25

.008

The following table shows the normality aspects for group 2, the normality assumption has been fulfilled.

Tests of Normality

Kolmogorov-Smirnova

Shapiro-Wilk

Statistic

df

Sig.

Statistic

df

Sig.

Score

.174

25

.050

.949

25

.240

This concept is not applicable

The degrees of freedom used were 48

The value of t statistic is 0.819

The probability value is 0.417

Independent Samples Test

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

Score

Equal variances assumed

.024

.879

.819

48

.417

.3200

.3907

Equal variances not assumed

.819

47.996

.417

.3200

.3907

This is not applicable

This is not applicable

Task 2

Option 3

Task 3

Lack of validity

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