In: Statistics and Probability
Researchers were curious on how accurate people are at judging distances in different body orientations. Participants viewed a target shown in front of them while "standing" (upright), "supine" (lying down on their back), or "prone" (lying on their stomach). They had to judge how far away the target was from them (in centimetres), then the researchers recorded the errorsthey made.
Using the data below, conduct a repeated measure ANOVA to determine whether different body orientations make any difference in people's judgment of distance. Use α = .05. Report η², and if the ANOVA is significant conduct a post-hoc test using Tukey’s HSD. Finally, please state all the steps and the formulas you used and report your conclusion clearly in words.
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Subject Standing Supine Prone
1 -1 -4 -2
2 -3 -2 -3
3 2 2 -1
4 0 -2 -3
5 2 -1 0
6 -3 -5 -4
7 4 1 0
8 -2 -3 -2
9 0 -1 -3
10 -1 -4 -3
∑X2 48 81 61
SS 47.6 44.9 16.9
column data , i (Factor B) | standing | supine | prone | |
count, ni = | 10 | 10 | 10 | |
mean , x̅ i =Σxi / ni | -0.200 | -1.900 | -2.100 | |
std. dev., si = | 2.30 | 2.23 | 1.37 | |
sample variances, si^2 = | 5.289 | 4.989 | 1.878 | |
SS ,Σ(x-Xbar)² | 47.6 | 44.9 | 16.9 | |
grand mean , x̅̅ = | ΣXij/Σn = | |||
square of deviation of sample mean from grand mean,( x̅ - x̅̅)² | 1.4400 | 0.2500 | 0.4900 | |
SS(between)= SSB = Σni( x̅ i - x̅̅)² = | 14.4000 | 2.5000 | 4.9000 |
Factor A | |||||
count, n j | mean , x̅ j =Σxj/nj | std. dev., sj | sample variances, sj^2 | ( x̅ j - x̅̅)² | nj( x̅j - x̅̅)² |
3 | -2.33 | 1.5275 | 2.333 | 0.871 | 2.613 |
3 | -2.67 | 0.5774 | 0.333 | 1.604 | 4.813 |
3 | 1.00 | 1.7321 | 3.000 | 5.760 | 17.280 |
3 | -1.67 | 1.5275 | 2.333 | 0.071 | 0.213 |
3 | 0.3 | 1.5275 | 2.333 | 3.004 | 9.013 |
3 | -4.0 | 1.0000 | 1.000 | 6.760 | 20.280 |
3 | 1.6666667 | 2.0817 | 4.333 | 9.404 | 28.213 |
3 | -2.333333 | 0.5774 | 0.333 | 0.871 | 2.613 |
3 | -1.333333 | 1.5275 | 2.333 | 0.004 | 0.013 |
3 | -2.666667 | 1.5275 | 2.333 | 1.604 | 4.813 |
SSA = total= | 89.867 | |
SSA=Σnj( x̅j - x̅̅)² = | 89.867 | |
SST = | Σ(Xij - X̅̅ ) = | 131.2 |
SSE = | SST-SSA-SSB = | 19.533 |
ΣXi² = | 190 | |
SST=Σxi² - (Σxi)²/N= | 131.2 |
so, SSA= | 89.867 |
SSB= | 21.8 |
SSE= | 19.533 |
df factor A(row) = r-1 = | 9 |
df factor B(column) = c-1 = | 2 |
here,total observations, N = | 30 |
df error = (r-1)(c-1) = | 18 |
mean square factor A , MSA = SSA/df= | 9.99 |
mean square(factor B) =MSB= SSB/df = | 10.90 |
mean square error = MSE =SSE/df = | 1.09 |
ANOVA | ||||||||
Source of Variation | SS | df | MS | F-stat | p-value | F-critical | Result | |
Rows | 89.8667 | 9 | 9.9852 | 9.201 | 0.0000 | 2.46 | significant | |
Columns | 21.8000 | 2 | 10.9000 | 10.044 | 0.0012 | 3.555 | significant | |
Error | 19.5333 | 18 | 1.0852 | |||||
Total | 131.2000 | 29 | ||||||
α = | 0.05 |
/.......................
tukey test:
Level of significance | 0.05 |
no of treatments | 3 |
df error | 18 |
MSE | 1.0852 |
q-statistic value | 3.6093 |
confidence interval | ||||||
population mean difference | critical value | lower limit | upper limit | result | ||
µ1-µ2 | 1.7 | 1.19 | 0.51 | 2.89 | means are different | |
µ1-µ3 | 1.9 | 1.19 | 0.71 | 3.09 | means are different | |
µ2-µ3 | 0.2 | 1.19 | -0.99 | 1.39 | means are not different |
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