In: Statistics and Probability
The data below represents the weight loss (in pounds) for people on three different exercise programs. At the 0.01 significance level, does it appear that a difference exists in the true mean weight loss produced by the three exercise programs.
Exercise A |
Exercise B |
Exercise C |
7.1 |
3.1 |
6.7 |
8.0 |
2.2 |
8.1 |
7.5 |
5.0 |
5.0 |
6.8 |
4.4 |
9.0 |
a) State the null and alternative hypotheses.
b) Find the P-value. STATE THE CALCULATOR COMMAND YOU USE.
c) State whether you should reject or fail to reject the null hypothesis. Justify your answer.
d) State your conclusion in non-technical terms. Which exercise program appears to be the least effective based on the sample data?
Mean | n | Std. Dev | |
Exercise A | 7.35 | 4 | 0.520 |
Exercise B | 3.68 | 4 | 1.263 |
Exercise C | 7.20 | 4 | 1.745 |
Total | 6.08 | 12 | 2.118 |
Part a)
To Test :-
H0 :- µ1 = µ2 = µ3 = 0 i.e There is no difference in the treatment
means
H0 :- µ1 = µ2 = µ3 ≠ 0 i.e Some means are different
Part b)
ANOVA table | |||||
Source | SS | df | MS | F | p-value |
Treatment | 34.605 | 2 | 17.3025 | 10.57 | 0.0043 |
Error | 14.738 | 9 | 1.6375 | ||
Total | 49.343 | 11 |
Excel formula
FDIST(10.5664,2,9) = 0.004349242
Part c)
Decision based on P value
Reject null hypothesis if P value < α = 0.01
Since P value = 0.0043 <
0.01, hence we reject the null hypothesis
Conclusion :- Treatment means differs
Part d)
There is sufficient evidence to conclude that a difference exists in the true mean weight loss produced by the three exercise programs.
Lookinat at the averages we can say that Exercise B is lease effective.