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=0.05α=0.05 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	36	49	44	22	44	40	48	28	46	27
Sit-ups after	53	59	59	32	47	49	51	41	50	41

Step 1 of 5: State the null and alternative hypotheses for the test.

Step 2 of 5:Find the value of the standard deviation of the paired differences. Round your answer to one decimal place.

Step 3 of 5:Compute the value of the test statistic. Round your answer to three decimal places

Step 4 of 5:Determine the decision rule for rejecting the null hypothesis H 0   . Round the numerical portion of your answer to three decimal places.

Step 5 of 5: Make the decision for the hypothesis test. (accept or reject null)

Answer 1

Step 1

To Test :-

H0 :- µd = 0

H1 :- µd > 0

Step 2

S(d) = √(Σ (di - d̅)2 / n-1)
S(d) = √(233.6 / 10-1) = 5.1

Step 3

Test Statistic :-

t = d̅ / ( S(d) / √(n) )
t = 9.8 / ( 5.0947 / √(10) )
t = 6.083

Step 4

Test Criteria :-
Reject null hypothesis if t > t(α)
Critical value t(α) = t(0.05) = 1.833

t > 1.833

Step 5

t > t(α) = 6.0829 > 1.8331
Result :- Reject null hypothesis