Question

A fitness course claims that it can improve an individual's physical ability. To test the effect of a physical fitness course on one's physical ability, the number of sit-ups that a person could do in one minute, both before and after the course, was recorded. Ten individuals are randomly selected to participate in the course. The results are displayed in the following table. Can it be concluded, from the data, that participation in the physical fitness course resulted in significant improvement?

Let d=(number of sit-ups that can be done after taking the course)−(number of sit-ups that can be done prior to taking the course)d=(number of sit-ups that can be done after taking the course)−(number of sit-ups that can be done prior to taking the course). Use a significance level of α=0.01 for the test. Assume that the numbers of sit-ups are normally distributed for the population both before and after taking the fitness course.

Sit-ups before	52	49	42	50	28	38	43	36	38	34
Sit-ups after	56	57	58	53	43	47	50	40	48	45

Copy Data
Step 1:

H0: μd ≤ 0

Ha: μd > 0

Step 2: standard deviation= 4.5

Step 3: t test statistic= 6.114

Step 4 of 5 :  Determine the decision rule for rejecting the null hypothesis H₀. Round the numerical portion of your answer to three decimal places.

ANSWER: Reject H₀ if t > 2.821

Step 6: Reject Null hypothesis

Answer 1

1) Null and Alternative hypothesis

H0: μd ≤ 0

Ha: μd > 0

Person	Sit-ups before	Sit-ups after	diff(d)=sample1-sample2	(d-dbar)^2
1	52	56	-4	22.09
2	49	57	-8	0.49
3	42	58	-16	53.29
4	50	53	-3	32.49
5	28	43	-15	39.69
6	38	47	-9	0.09
7	43	50	-7	2.89
8	36	40	-4	22.09
9	38	48	-10	1.69
10	34	45	-11	5.29
		SUM	-87	180.1

2)

Mean(d)=dbar=sum(d)/n	-8.7
standard deviation(sd)=sqrt(sum(d-dbar)^2/n-1)	4.4734
n	10

Mean=-8.7

standard deviation= 4.5

3) t test statistic= -8.7/(4.5/sqrt(10))=-6.114

4) Use a significance level of α=0.01

t=critical value at 99 % with (n-1) degree of freedom   =2.8214

5) | t test statistic| >t-critical value hence we reject Ho

6) Conclusion:

reject Null hypothesis

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