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Table of Contents

 


 

One-Way Analysis of Variance (ANOVA)

 

Used to evaluate means from two or more subgroups.  A statistically significant ANOVA indicates there is more variation between subgroups than would be expected by chance.  It does not identify which subgroup pairs are significantly different from each other.  Used to evaluate multiple means of one independent variable to avoid conducting multiple t-tests.

 

Problem: You obtained the number of years of education from one random sample of 38 police officers from City A, the number of years of education from a second random sample of 30 police officers from City B, and the number of years of education from a third random sample of 45 police officers from City C. The average years of education for the sample from City A is 15 years with a standard deviation of 2 years. The average years of education for the sample from City B is 14 years with a standard deviation of 2.5 years. The average years of education for the sample from City C is 16 years with a standard deviation of 1.2 years. Is there a statistically significant difference between the education levels of police officers in City A, City B, and City C?

 

 

City A

City B

City C

Mean (years)

15

14

16

S (Standard Deviation)

2

2.5

1.2

S2 (Variance)

4

6.25

1.44

N (number of cases)

38

30

45

Sum of Squares

152

187.5

64.8

Sum of Scores (mean*n)

570

420

720

 

Assumptions

Independent Random sampling

Interval/ratio level data

Population variances equal

Groups are normally distributed

State the Hypothesis

Ho: There is no statistically significant difference among the three cities in the mean years of education for police officers.

Ha: There is a statistically significant difference among the three cities in the mean years of education for police officers.

 

Set the Rejection Criteria

Determine the degrees of freedom for the F Distribution

Numerator Degrees of Freedom

df=k-1 where k=3 (number of independent samples/groups)    df=2

Denominator Degrees of Freedom

df=n-k where n=113 (sum of all independent samples)   df=110

Establish Critical Value

Determine the level of confidence -- alpha

At alpha.05, df=(2,110)

Consult f-distribution, Fcv = 3.072

 

Compute the Test Statistic

  

 

            where

= each group mean

= grand mean for all the groups = ( sum of all scores)/N

 = number in each group

 

Note:  F=         Mean Squares between groups

  Mean Squares within groups

 

 

Estimate Grand Mean

      

 

Estimate F Statistic

          

 

Decide Results of Null Hypothesis

Compare the F statistic to the F critical value. If the F statistic equals or exceeds the Fcv, the null hypothesis is rejected. This suggests that the population means of the groups sampled are not equal -- there is a difference between the group means.

Since the F-statistic (9.931) exceeds the F critical value (3.072), we reject the null hypothesis and conclude there is a statistically significant difference between the three cities in the mean years of education for police officers.

Software Output Example


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