In: Statistics and Probability
A researcher hypothesizes that students’ performance in a test is better in situations where the surrounding noise is kept lowest. She collects data on her students under three different levels of noise. A higher score indicates better performance. Her data follow. SHOW ALL YOUR WORK.
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 High Noise  | 
 Mod Noise  | 
 Low Noise  | 
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 3  | 
 4  | 
 6  | 
 G =  | 
|
| 
 5  | 
 6  | 
 8  | 
 ΣX2 =  | 
 393  | 
| 
 4  | 
 4  | 
 6  | 
 N =  | 
|
| 
 3  | 
 2  | 
 7  | 
 k =  | 
|
| 
 2  | 
 3  | 
 8  | 
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| 
 T = 17  | 
 T = 19  | 
 T =  | 
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 M = 3.4  | 
 M = 3.8  | 
 M =  | 
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| 
 SS = 5.2  | 
 SS = 8.8  | 
 SS =  | 
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| 
 n = 5  | 
 n = 5  | 
 n =  | 
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 Source  | 
 SS  | 
 df  | 
 MS  | 
 F  | 
| 
 Between  | 
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| 
 Within  | 
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| 
 Total  | 
| treatment | A1 | A2 | A3 | |||
| count, ni = | 5 | 5 | 5 | |||
| mean , x̅ i = | 3.400 | 3.80 | 7.000 | |||
| std. dev., si = | 1.140 | 1.483 | 1.000 | |||
| sample variances, si^2 = | 1.300 | 2.200 | 1.000 | |||
| total sum | 17 | 19 | 35 | 71 | (grand sum) | |
| grand mean , x̅̅ = | Σni*x̅i/Σni = | 4.73 | ||||
| square of deviation of sample mean from grand mean,( x̅ - x̅̅)² | 1.778 | 0.871 | 5.138 | |||
| TOTAL | ||||||
| SS(between)= SSB = Σn( x̅ - x̅̅)² = | 8.889 | 4.356 | 25.689 | 38.93333333 | ||
| SS(within ) = SSW = Σ(n-1)s² = | 5.200 | 8.800 | 4.000 | 18.0000 | 
no. of treatment , k =   3
df between = k-1 =    2
N = Σn =   15
df within = N-k =   12
  
mean square between groups , MSB = SSB/k-1 =   
19.4667
  
mean square within groups , MSW = SSW/N-k =   
1.5000
  
F-stat = MSB/MSW =    12.9778
| anova table | ||||||
| SS | df | MS | F | p-value | F-critical | |
| Between: | 38.9333 | 2 | 19.467 | 12.978 | 0.0010 | 3.885 | 
| Within: | 18.0000 | 12 | 1.500 | |||
| Total: | 56.9333 | 14 | ||||
| α = | 0.05 | 
critical F = 3.885
reject the null hypothesis
conclusion : there is enough evidence of significant mean difference among three treatments
==========
tukey's test
| Level of significance | 0.05 | 
| no. of treatments,k | 3 | 
| DF error =N-k= | 12 | 
| MSE = | 1.5000 | 
| q-statistic value(α,k,N-k) | 3.7728 | 
critical value = q*√(MSE/2*(1/ni+1/nj)) = 2.07
| population mean difference | critical value | result | |||
| µ1-µ2 | 0.40 | 2.07 | means are not different | ||
| µ1-µ3 | 3.600 | 2.07 | means are different | ||
| µ2-µ3 | 3.20 | 2.07 | means are different |