Question

The data (3-2 WeightLoss (Links to an external site.)) shows the results of a weight-loss contest sponsored by a local newspaper. Participants were encouraged to compete over a one-month period. Was there a significant weight loss? That is, TEST at the 1% level of significance if the mean weights after the contest are lower than the weights before the contest? Use “After” as Population 1 and “Before” as Population 2.

Name	After	Before
Michael M.	202.5	217
Tracy S.	178	188
Gregg G.	210	225
Boydea P.	157	168
Donna I.	169	178
Elizabeth C.	173.5	182
Carole K.	163.5	174.5
Candace G.	153	161.5
Jo Anne M.	170.5	177.5
Willis B.	336	358.5
Marilyn S.	174	181
Tim B.	197.5	210

Answer 1

Solution:
Given in the question
the mean weights after the contest are lower than the weights before the contest so null and alternative hypothesis can be calculated s
Null hypothesis H0: d = 0
Alternative hypothesis Ha: d <0
Differences of samples can be calculated as

After(Population1)	Before(Population2)	After-Before
202.5	217	-14.5
178	188	-10
210	225	-15
157	168	-11
169	178	-9
173.5	182	-8.5
163.5	174.5	-11
153	161.5	-8.5
170.5	177.5	-7
336	358.5	-22.5
174	181	-7
197.5	210	-12.5

Mean of differences can be calculated as
Dbar = (-14.5-10-15-11-9-8.5-11-8.5-7-22.5-7-12.5)/12 = -11.375
Standard deviation of differences can be calculated as
Standard deviation (Sd)= sqrt((Di-mean)^2)/(n-1))

After(Population1)	Before(Population2)	After-Before(D)	Di-mean	(Di-mean)^2
202.5	217	-14.5	-3.125	9.765625
178	188	-10	1.375	1.890625
210	225	-15	-3.625	13.140625
157	168	-11	0.375	0.140625
169	178	-9	2.375	5.640625
173.5	182	-8.5	2.875	8.265625
163.5	174.5	-11	0.375	0.140625
153	161.5	-8.5	2.875	8.265625
170.5	177.5	-7	4.375	19.140625
336	358.5	-22.5	-11.125	123.765625
174	181	-7	4.375	19.140625
197.5	210	-12.5	-1.125	1.265625
			Sum(Di-mean)^2	210.5625

Standard deviation (Sd)= sqrt(210.5625/11) = 4.375
Here we will use paired sample t test because both samples are same
Test statistic can be calculated as
Test stat = (Dbar -d)/sd/sqrt(n) = (-11.375-0)/4.375/sqrt(12) = -9.01
At degree of freedom = n-1 =12-1 =11 and this is left tailed and one tailed test, p-value from t table is 0.00001
At alpha = 0.01, we can reject the null hypothesis as p-value is less than alpha value. so we can conclude that the mean weights after the contest are lower than the weights before the contest.