Question

Hi please write what's in the comments as the answers! I want people to know the answers!! Tai Chi is often recommended as a way to improve balance and flexibility in the elderly. Below are
before-and-after flexibility ratings (on a 1 to 10 scale, 10 being most flexible) for a random sample of 8 men in their 80's who took Tai Chi lessons for six months.


Subject A B C D E F G H


Flexibility rating after Tai Chi 2 4 3 3 3 4 5 10


Flexibility rating before Tai Chi 1 2 1 2 1 4 2 10


(a) Explain why these are paired data.


(b) Calculate and interpret the mean difference.


(c) Researchers would like to know if the true mean difference (after-before) in flexibility rating for men in their 80's who take Tai Chi lessons for 6 months is greater than zero. The P-value of this test is 0.004. Interpret this value.


(d) If the result of this study is statistically significant, can you conclude that the difference in the mean flexibility rating was caused by the Tai Chi lessons? Why or why not?

Answer 1

Answer:

a) Because same men were sampled before and after Tai Chi

b) d = 1.375

c) If we assume that the mean difference is not greater than zero, the probability of obtaining the mean difference equal to 1.375 is 0.004

d) Yes.

Step-by-step explanation:

Part a)

We are given the data of 8 men and their flexibility rating before and after practicing Tai Chi Lessons. This data is an example of paired data because the same men were sampled before and after practicing Tai Chi for this study. Whenever the subjects of the study remain same, we call it a paired data and use paired data t-test for carrying out the hypothesis testing.

Part b) Mean Difference

We have to calculate the difference in flexibility rating of each man, and then find the mean of those differences. This mean would be termed as the mean difference.

The difference we will calculate is After - Before, as said in the question.

Subject A: 2 - 1 = 1

Subject B: 4 - 2 = 2

Subject C: 3 - 1 = 2

Subject D: 3 - 2 = 1

Subject E: 3 - 1 = 2

Subject F: 4 - 4 = 0

Subject G: 5 - 2 = 3

Subject H: 10 - 10 = 0

Mean of these differences would be sum of these differences divided by number of sums which is 8.

So, mean difference = 11/8 = 1.375

Interpretation: Since the mean difference is greater than 0, it seems that Tai Chi improved the flexibility rating of these 8 men in general.

Part c)

Researchers want to test is the mean value is greater than 0. So, the null and alternate hypothesis would be:

Null Hypothesis: d ≤ 0

Alternate Hypothesis: d > 0

Here, d stands for the mean difference (after -before)

The p-value of this hypothesis test came out to be 0.004.

p-value is the probability of obtaining the result atleast as extreme as the actual test result, assuming the Null Hypothesis to be true. So, we can say that:

If we assume that the mean difference is not greater than zero, the probability of obtaining the mean difference equal to 1.375 is 0.004, which indeed is a very low value. Such a low p-value does not favor the Null Hypothesis.

Part d)

The p-value of this hypothesis test came out to be 0.004, as mentioned in the question. Since, the p-value is very very small, close to zero, we can reject the Null Hypothesis.

Conclusion: We have enough confidence to support the claim that the true mean difference(after - before) in flexibility rating for men in their 80's who take Tai Chi is greater than Zero. This means, Tai Chi helps in increasing the flexibility rating.

So, Yes we can conclude that the difference in the mean flexibility rating was caused by the Tai Chi lessons