A researcher is investigating the effectiveness of acupuncture treatment for chr

Question

A researcher is investigating the effectiveness of acupuncture treatment for chronic back pain. A sample n=4 participants is obtained from a pain clinic. Each individual ranks the current level of pain and then begins a 6 week program of acupuncture treatment. At the end of the program, the pain level is rated again and the researcher records the amount of difference between the two ratings. For this sample, pain level decreased by an average of M=4.5 points with SS=27. Use this to answer a and b.

a.) Are data sufficient to conclude that acupuncture has a significant effect on back pain? Use a two-tailed test with alpha=0.05. (state all 4 steps)

b.) Can you conclude that acupuncture significantly reduces back pain? Use a one-tailed test with alpha=0.05. (state all 4 steps)

Explanation / Answer

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: d = 0

Alternative hypothesis: d 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

s = sqrt [ ((di - d)2 / (n - 1) ]

SS = (di - d)2 = 27

s = 3

SE = s / sqrt(n)

S.E = 1.5

DF = n - 1 = 4 -1

D.F = 3

t = [ (x1 - x2) - D ] / SE

t = 3

where di is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 3 degrees of freedom is more extreme than 3; that is, less than - 3 or greater than 3.

Thus, the P-value = 0.0577

Interpret results. Since the P-value (0.0577) is greater than the significance level (0.05), we have to accept the null hypothesis.

b)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: d< 0

Alternative hypothesis: d > 0

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

s = sqrt [ ((di - d)2 / (n - 1) ]

SS = (di - d)2 = 27

s = 3

SE = s / sqrt(n)

S.E = 1.5

DF = n - 1 = 4 -1

D.F = 3

t = [ (x1 - x2) - D ] / SE

t = 3

where di is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a one-tailed test, the P-value is the probability that a t statistic having 3 degrees of freedom is more than 3

Thus, the P-value = 0.029

Interpret results. Since the P-value (0.029) is less than the significance level (0.05), we have to reject the null hypothesis.

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