Question

The following table lists the weight of individuals before and after taking a diet prescribed by a weight-loss company for a month:

Weight-loss Data:

Individual: A, B, C, D, E, F

Weight Before (lb): 123.7, 128.7, 135.6, 194.9, 145.5, 162.3

Weight After (lb): 109.4, 109.7, 123.3, 186.5, 126.8, 151.5

Weight loss (lb): 14.3, 19.0, 12.3, 8.4, 18.7, 10.8

You may find this Student's t distribution table useful in answering the following questions. You may assume that the differences in weight are normally distributed.

a)Calculate the sample variance (sd2) of the changes in individual weights. Give your answer to 2 decimal places.

sd2 =

b)A disgruntled customer states: "This weight-loss company is a complete farce. All the people I know who signed up experienced no changes in their weight at all. I seriously doubt this diet has any effect whatsoever. I want my money back!"

You plan to do a hypothesis test on this claim where the hypotheses are:

H0: the customer's claim is true and the program has no effect on weight

HA: the customer's claim is not true and the program does have an effect on weight, whether it increases or decreases

According to the data given, you should accept, reject, or not reject the null hypothesis at a confidence level of 99%.

Answer 1

sample 1 = weight before

sample 2= weight after

Sample #1	Sample #2	difference , Di =sample1-sample2	(Di - Dbar)²
123.7	109.4	14.300	0.15
128.7	109.7	19.000	25.84
135.6	123.3	12.300	2.61
194.9	186.5	8.400	30.43
145.5	126.8	18.700	22.88
162.3	151.5	10.800	9.71

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	890.7	807.2	83.5	91.6

a)

mean of difference ,    D̅ =ΣDi / n =   13.9167

sample variance,SD² = [ (Di-Dbar)²/(n-1) =18.33
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b)

Ho :   µd=   0
Ha :   µd ╪   0
      
Level of Significance ,    α =    0.01
      
sample size ,    n =    6
mean of difference ,    D̅ =ΣDi / n =   13.917          
                  
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    4.2808          
                  
std error , SE = Sd / √n =    4.2808   / √   6   =   1.7476
                  
t-statistic = (D̅ - µd)/SE = (   13.9167   -   0   ) /    1.7476
                  
Degree of freedom, DF=   n - 1 =    5          
  
p-value =        0.00050   [excel function: =t.dist.2t(t-stat,df) ]       
Conclusion:     p-value <α , Reject null hypothesis