Question

A researcher wants to see if the dress code used in schools has some impact on the students’ fear of crime [How safe do you feel when at school?]. Fear of crime is measured on a scale from 1 (very unsafe] to 5 [very safe]. She randomly selects 5 students from three schools that have different dress codes:   [Note: you should use Excel to answer the questions; if you wish, you may calculate F by hand using the formula included in your book]                                                                     [15 points]

• School A: no formal dress code
• School B: dress code required
• School C: uniforms required

School A	School B	School C
3	2	4
3	2	4
3	2	3
4	1	4
4	3	3

1. Identify the dependent and the independent variables
2. Formulate the null and the alternative hypotheses  
3. Test the null hypothesis and reach a statistical conclusion based on the calculated F
4. Interpret your results.
5. For extra-credit [2 points]. Conduct a multiple comparison of means using the Bonferroni test and describe your findings
6. For extra-credit [2 points] Calculate the effect size [i.e., what percentage of the variation in fear of crime is explained by differences in dress code?]
7. For extra-credit [2 points] Graph your findings using the samples’ means [line graph]

Answer 1

a) dependent: fear of scrime

independent: difference in dress code

b)

Ho: µ1=µ2=µ3
H1: not all means are equal

c)

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance	std dev
School A	5	17	3.4000	0.3000	0.5477
School B	5	10	2.0000	0.5000	0.7071
School C	5	18	3.6000	0.3000	0.5477
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	7.6000	2	3.8000	10.3636	0.0024	3.89
Within Groups	4.4000	12	0.3667
Total	12.0000	14

value of test stat=   10.36
p value is    0.0024
   Decision:   p-value<α , reject null hypothesis
      
conclusion :    there is enough evidence of significant mean difference among three treatments  

d)

Level of significance= 0.05/3 = 0.0167
no. of treatments,k=   3
DF error =N-k=   12
MSE=   0.3667
t-critical value,t(α/2,df)=   2.7795

critical value=tα/2,df √(MSE(1/ni+1/nj))

if absolute difference of means > critical value,means are significnantly different ,otherwise not

population mean difference	critical value			result
µ1-µ2	1.40	1.06			means are different
µ1-µ3	0.20	1.06			means are not different
µ2-µ3	1.60	1.06			means are different

f)

eta square ,effect size = SSbet/SST=   0.6333 (large)

63.33 percentage of the variation in fear of crime is explained by differences in dress code

g)

Groups	Count	Sum	Average
School A	5	17	3.4000
School B	5	10	2.0000
School C	5	18	3.6000