Question

In: Statistics and Probability

A health researcher was interested in comparing three methods of weight loss: low-calorie diet, low-fat diet,...

A health researcher was interested in comparing three methods of weight loss: low-calorie diet, low-fat diet, and high-exercise regimen. He selected 15 moderately overnight subjects and randomly assigned 5 to each weight-loss program. The following weight reductions (in pounds) were observed after a one-month period: Low-fat diet: 7,6,8,8,6 low Calorie diet: 7,7,5,4,5 high exercise: 7,8,7,9,8  

a) Test the null hypothesis that the extent of weight reduction does not differ by type of weight-loss program. (Set α=.01 level) (10 points) b) If the result at a) rejects the null hypothesis, use an appropriate multiple comparison procedure and determine where the significant differences occur. (5 points) c) Calculate the correlation ratio and interpret the result. (3 points)

Solutions

Expert Solution

Ho:extent of weight reduction does not differ by type of weight-loss program

Ha: extent of weight reduction differ by type of weight-loss program

A B C
count, ni = 5 5 5
mean , x̅ i = 7.000 5.60 7.80
std. dev., si = 1.000 1.342 0.837
sample variances, si^2 = 1.000 1.800 0.700
total sum 35 28 39 102 (grand sum)
grand mean , x̅̅ = Σni*x̅i/Σni =   6.80
square of deviation of sample mean from grand mean,( x̅ - x̅̅)² 0.040 1.440 1.000
TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² = 0.200 7.200 5.000 12.4
SS(within ) = SSW = Σ(n-1)s² = 4.000 7.200 2.800 14.0000

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 =    6.2000
  
mean square within groups , MSW = SSW/N-k =    1.1667
  
F-stat = MSB/MSW =    5.3143

anova table
SS df MS F p-value F-critical
Between: 12.40 2 6.20 5.31 0.0222 6.93
Within: 14.00 12 1.17
Total: 26.40 14
α = 0.01
conclusion : p-value>α , do not reject null hypothesis    
conclusion : p-value>α , do not reject null hypothesis    

...................

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