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 σ^{2})

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 
KolmogorovSmirnov^{a} 
ShapiroWilk 

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 
TheShapirowilk test is significant with pvalue of 0.0001<0.05.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 nonsignificant 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>0.05 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 pvalue 0.254>0.05 and Fstatisticof 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 posthoc 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., & Thompson, W. L. (2000). Null hypothesistesting: problems, prevalence, and an alternative. Thejournal of wildlife management,912923.