In: Statistics and Probability
With all the fad diets advertised on TV, a researcher wanted to know if there was a significant difference in weight loss depending on which new fad diet you were on. Over the course of 2 weeks, a total of 12 male volunteers who weighed the same and had similar lifestyles were broken up into 3 groups: Diet 1, Diet 2, and Diet 3. Conduct an ANOVA on the following table with alpha = 0.5 to determine if at least one of the diets has significantly different weight loss compared to the other two. Diet 1 -1 5 3 4 Diet 2-13 12 8 3 Diet 3-6 2 1 2 1. What are the null and alternative hypotheses? What is the critical value? (Round to 4 decimal places) a. 3.8853 b. 3.9823 c. 4.2565 d. 3.5874
diet 1 | diet 2 | diet 3 | ||||
count, ni = | 4 | 4 | 4 | |||
mean , x̅ i = | 3.25 | 9.00 | 2.75 | |||
std. dev., si = | 1.71 | 4.55 | 2.22 | |||
sample variances, si^2 = | 2.917 | 20.667 | 4.917 | |||
total sum | 13 | 36 | 11 | 60 | (grand sum) | |
grand mean , x̅̅ = | Σni*x̅i/Σni = | 5.00 | ||||
square of deviation of sample mean from grand mean,( x̅ - x̅̅)² | 3.0625 | 16.00 | 5.0625 | |||
TOTAL | ||||||
SS(between)= SSB = Σn( x̅ - x̅̅)² = | 12.25 | 64.00 | 20.25 | 96.5 | ||
SS(within ) = SSW = Σ(n-1)s² = | 8.75 | 62.00 | 14.75 | 85.5 |
no. of treatment , k = 3
df between = k-1 = 2
N = Σn = 12
df within = N-k = 9
mean square between groups , MSB = SSB/k-1 =
48.250
mean square within groups , MSW = SSW/N-k =
9.500
F-stat = MSB/MSW = 5.079
P value = 0.033
anova table | ||||||
SS | df | MS | F | p-value | F-critical | |
Between: | 96.5 | 2 | 48.3 | 5.08 | 0.033 | 4.2565 |
Within: | 85.5 | 9 | 9.5 | |||
Total: | 182.0 | 11 |
F-criitical = 4.2565(answer)