Question

mimi was curious if regular excise really helps weight loss, hence she decided to perform a hypothesis test. a random sampleof 5 umuc students was chosen. the students took a 30-minute exercise every day for 6 months. the weight was recorded for each individual before and after the exercise regimen. does the data below suggest that the regular exercise helps weight loss? assume mimi wants to use a .05 significance level to test the claim.

	Weight (lbs)
Subject	Before	After
1	190	180
2	170	160
3	185	190
4	160	160
5	200	190

a. what is the appropriate hypothesis test to use for this analysis: z-test for two proportions, t-test for two proportions, t-test for two dependent samples or t-test or two independent samples? please identify and explain why it is appropriate.

b.Let u1 = mean weight before the exercise regime. Let u2 = mean weight after the exercise regime. which of the following statements correctly defines the null hypothesis?

i. u1-u2 > 0(ud>0)

ii. u1 - u2 = 0 (ud=0)

iii. u1 - u2 < 0 (ud <0)

c. let u1 = mean weight before the exercise regime. Let u2 = mean weight after the exercise regime. which of the following statements correctly defines the alternative hypothesis?

a. u1 - u2 > 0 (ud>0)

b. u1 - u2 = 0 (ud = 0)

c. u1 - u2 (ud <0)

d) determine the test statistic. round your answer to three decimal places. show all work e) determine the p-value. round your answer to three decimal places. f) compare p-value and significance level. what decision should be made regarding the null hypothesis and why? g) is there sufficient evidence to support the claim that regular exercise helps weight loss?

Answer 1

a)

since, samples are dependent so answer is

t-test for two dependent samples

b)

ii. u1 - u2 = 0 (ud=0)

c)

a. u1 - u2 > 0 (ud>0)

d)

Sample #1	Sample #2	difference , Di =sample1-sample2	(Di - Dbar)²
190	180	10	25.00
170	160	10	25.00
185	190	-5	100.00
160	160	0	25.00
200	190	10	25.00

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	905	880	25	200.000

mean of difference ,    D̅ =ΣDi / n =   5.000
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    7.071

std error , SE = Sd / √n =    7.0711   / √   5   =   3.1623      
                          
t-statistic = (D̅ - µd)/SE = (   5.0000   -   0   ) /    3.1623   =   1.581

e)

Degree of freedom, DF=   n - 1 =    4      
  
p-value =        0.095 [excel function: =t.dist.rt(t-stat,df) ]   

f)

decision: p-value>α , Do not reject null hypothesis  

g)

Conclusion: there is not sufficient evidence to support the claim that regular exercise helps weight loss