2.2 The weights, in kilograms, of twenty men before and after participation in a
ID: 3206848 • Letter: 2
Question
2.2 The weights, in kilograms, of twenty men before and after participation in a waist loss' program are shown in Table (Egger al., 1999) want to know if, on average, they a weight loss twelve months after the program. Table 2.8 Weights of twenty men before and after participation in a waist loss' program Man Before After Man Before After 1 100.8 97.0 11 105.0 105.0 2 102.0 107.5 12 85.0 82.4 3 105.9 97.0 13 107.2 98.2 108.0 108.0 14 80.0 83.6 92.0 84.0 15 115.1 115.0 16 103.5 103.0 6 116.7 111.5 7 110.2 102.5 17 82.0 80.0 18 101.5 101.5 8 135.0 127.5 19 103.5 102.6 9 123.5 118.5 95.0 94.2 93.0 93.00 Let Yik denote the weight of the kth man at the jth time where 1 before the program and j 2 twelve months later. Assume the Yak's are independent random variables with Y N(wi,o2) for j 1,2 and 20 (a) Use an unpaired t-test to test the hypothesis Ho versus (b) Let D 1k 2k for k 1,..., 20. Formulate models for testing Ho against Hi using the Dk's. Using analogous methods to Exercise 2.1 above, assuming o2 is a known constant, test Ho against H1. (c) The analysis in (b) is a paired t-test which uses the natural relationship between weights of the same person before and after the program. Are the conclusions the same from (a) and (b)?Explanation / Answer
a)
Here we perform unpaired t-test.
Null hypotheisis- H0: Mean of weight before= Mean of weight after
Alternative hypotheisis- Ha: Mean of weight before!= Mean of weight after
H0: M1 = M2
Ha: M1 != M2
X1=98.235
X2=100.6
S1=23.6165
S2=12.4745
n1=n2=20
using unpaired t-test,
Sp2=((n1-1)*S12+(n2-1)*S22)/n1+n2-2
=19*(23.6165)2+19*(12.4745)2/38
Sp2 = 356.68
Sp = 18.18
t = (X1-X2)/(Sp*sqrt((1/n1)+(1/n2)))
=(98.235-100.6)/(18.18* sqrt(0.05+0.05))
t =-0.4
At 95% confidence, confidence interval is -14.582628 9.852628
And p-value= 0.695
Since p value is very high and t is in the confidence interval, we accept the null hypotheisis that mean of weight before is equal to the mean of weight after weight loss program
b)
By taking difference between before and after weights and then taking their mean and standard deviation,
Null hypoteisis = H0: Mean M= 0
Alternate hypotheisis= Ha: Mean M != 0
X= -6.657
S=27.960
t= (X-M)/(s/sqrt(n))
=(-6.657-0)/(27.96/sqrt(20))
=-1.07
At 95% confidence interval, confidence interval is -19.384423 6.070137
p-value is 0.288
Since t is in the confidence interval and p-value is very high, we can accept null hypothesis that mean of differences is equal to 0.
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