# 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.