In: Statistics and Probability
A sports researcher gave a standard questionnaire of eating habits to 12 randomly selected male professional athletes, four each from baseball, football, and basketball. The results were:
Baseball: 34, 18, 21, 65
Football: 27, 28, 67, 42
Basketball: 35, 44, 47, 61
Is there a significant difference in eating habits among professionals in the three sports? (Use the .05 sig level.)
H0: μ1 = μ2 = μ3 |
H1: There is at least one mean difference among the populations. |
Applying one way ANOVA:
Baseball | Football | basketball | |
34 | 27 | 35 | |
18 | 28 | 44 | |
21 | 67 | 47 | |
65 | 42 | 61 | |
count ni= | 4 | 4 | 4 |
Average==ΣXi/ni= | 34.5000 | 41.0000 | 46.7500 |
Total=ΣX2i = | 6146.000 | 7766.0000 | 9091.0000 |
SS=ΣX2i -(Σxi)2/ni= | 1385.000 | 1042.0000 | 348.7500 |
Grand average Xgrand =∑xi/N = | 489/8= | 40.7500 |
i | ni | x̅i | ni*(Xi-Xgrand)2 | SS=(ni-1)*s2 |
treatment 1 | 4 | 34.5000 | 156.250 | 1385 |
treatment 2 | 4 | 41.0000 | 0.250 | 1042 |
treatment 3 | 4 | 46.7500 | 144.000 | 348.75 |
Total | 12 | 300.500 | 2775.7500 | |
SSTr | SSE | |||
df treatments = | number of treatments-1= | 2 | ||
df error = | N-k= | 12-3= | 9 | |
df total= | N-1= | 12-1= | 11 | |
MS(treatment) =SSTr/df(Tr)=300.5/2= | 150.25000 | |||
MS(error) =SSE/df(error)=2775.75/9= | 308.41667 | |||
test statistic F =MSTr/MSE =150.25/308.42= | 0.487 |
for (2,11) degree of freedom and 0.05 level critical value =4.26
since test statistic < critical value, we fail to reject null hypothesis
we do not have significant evidence to conclude that there is a significant difference in eating habits among professionals in the three sports