ANOVA statistics project

ANOVA STATISTICS PROJECT

ANOVAstatistics project

Student`sname

STEP1

sex

Frequency

Percent

Valid Percent

Cumulative Percent

Valid

Female

64

61.0

61.0

61.0

Male

41

39.0

39.0

100.0

Total

105

100.0

100.0

Thedataset is composed of both males and females that are the predictorvariables and their different scores in quiz3 that is the outcome ofinterest. The unit of measurement of sex is nominal while for quiz3is scale with a sample size of 105.

STEP2

ASSUMPTIONSOF ANOVA

  • Observations must be independently distributed normally with mean(μ σ2)

  • Variances are equal

Itis negatively skewed with a mean of 7.98 and a standard deviation of2.308.

Descriptive Statistics

N

Mean

Skewness

Kurtosis

Statistic

Statistic

Statistic

Std. Error

Statistic

Std. Error

quiz3

105

7.98

-1.134

.236

.750

.467

Valid N (listwise)

105

Thequiz3 has a kurtosis value less than 3.This shows that thedistribution is flatter than normal with a wider peak. Values arewidely spread around the mean.

Thequiz3 is skewed to the left. This shows a greater number of largervalues concentrated on the right of the mean, with extreme values tothe left.

Tests of Normality

sex

Kolmogorov-Smirnova

Shapiro-Wilk

Statistic

df

Sig.

Statistic

df

Sig.

quiz3

Female

.211

64

.000

.817

64

.000

Male

.212

41

.000

.845

41

.000

a. Lilliefors Significance Correction

TheShapiro-wilk test is significant with p-value of 0.0001&lt0.05.LEVENETEST

Test of Homogeneity of Variance

Levene Statistic

df1

df2

Sig.

quiz3

Based on Mean

3.436

1

103

.067

Based on Median

2.274

1

103

.135

Based on Median and with adjusted df

2.274

1

99.760

.135

Based on trimmed mean

3.346

1

103

.070

Basedon the mean, Levene test is non-significant for quiz3 because the sigvalue is more than 0.05.The value indicates that the variances arenot equal, and hence the assumption of homogeneity of variance is nottenable.

F(1,103)=3.436,P&gt0.05 Conclusion: The assumptions of normality and homogeneityof variance are not met.

STEP3

Researchquestion: Is there a difference in means of males and females?

Nullhypothesis: There is no difference in means of males and females at95% confidence interval

(α=5%)

Alternativehypothesis: There is difference in means of males and females at 95%confidence interval

(α=5%)

MEANPLOTS

Themean plot indicates that females have a higher mean of 8.19 comparedto the males with a mean of 7.66 and a deviation of 0.53.

Descriptive

N

Mean

Std. Deviation

Std. Error

95% Confidence Interval for Mean

Minimum

Maximum

Lower Bound

Upper Bound

Female

64

8.19

2.130

.266

7.66

8.72

2

10

Male

41

7.66

2.555

.399

6.85

8.47

0

10

Total

105

7.98

2.308

.225

7.53

8.43

0

10

Thefemale have the highest mean of 8.19 with the least standarddeviation of 2.13 compared to males with a mean of 7.66 and astandard deviation of 2.555. The males have the least minimum of 0and a maximum of 10 compared to the females with a minimum of 2 and amaximum of 10. Concluding that females had better scores compared tomen.

ANOVA

quiz3

Sum of Squares

df

Mean Square

F

Sig.

Between Groups

6.992

1

6.992

1.317

.254

Within Groups

546.970

103

5.310

Total

553.962

104

Theresults were insignificant with p-value 0.254&gt0.05 and F-statisticof 1.317 and (1,103) degrees of freedom. The calculated effect size(0.53) between groups and within groups is a minimal change .it issignificant. The post-hoc analysis was not valid since the resultswere insignificant.

Step5

Conclusion:We reject the null hypothesis and conclude that there is a differencein means in males and females.

STRENGTHS

Theobservations must be independent

Limitations

Itassumes that the samples have equal variances which are not valid.

References

Anderson,D. R., Burnham, K. P., &amp Thompson, W. L. (2000). Null hypothesistesting: problems, prevalence, and an alternative.&nbspThejournal of wildlife management,912-923.