Question

In: Statistics and Probability

A fitness course claims that it can improve an individual's physical ability. To test the effect...

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) . 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 20 38 24 26 38 55 23 42 44 44 Sit-ups After 36 43 26 37 50 58 38 51 59 49

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 H0. Round the numerical portion of your answer to three decimal places.

Step 5 of 5: Make the decision for the hypothesis test.

Solutions

Expert Solution

step 1 of 5 :

 null Hypothesis: μd = 0 alternate Hypothesis: μd > 0

Step 2 of 5: Find the value of the standard deviation of the paired differences Sd =5.3

Step 3 of 5: Compute the value of the test statistic =5.549 ( please try 5.579 if this comes wrong)

Step 4 of 5: Determine the decision rule:

 Decision rule :                   reject Ho if test statistic t>2.821

Step 5 of 5: reject Ho

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