Question

1. Extracts of St. John’s Wort are widely used to treat depression. An article in the Journal of American Medical Association, April 2001, compared the efficacy of a standard extract of St. John’s Wort with a Placebo in 200 outpatients diagnosed with major depression. Patients were randomly assigned to two groups, each with 100 patients; one group received St. John’s Wort, and the other group received Placebo. After 8 weeks, 19 of the Placebo-treated patients showed improvements, whereas 27 of those treated with St. John’s Wort improved. We need to test if St. John’s Wort is effective in treating major depression.

1. a. What is the type of statistical test procedure that should be used to test the hypotheses? Explain.

1. b. Construct a 95% confidence interval. Test the hypotheses using the confidence interval. Interpret the test result clearly.

1. c. Test the hypotheses using the Test Statistic / Critical Value method (you must clearly indicate the test statistic, and the critical value(s) use Take α=5%). Do you get the same answer as part (c)? If your answer is the same as (c), you do not have to interpret again. Otherwise, interpret the test findings.

1. d. What is the P value of the test? Test the hypotheses using the P value. Take α=5%. Do you get the same answer as parts (c) and (d)? If your answer is the same as (c) and (d), you do not have to interpret again. Otherwise, interpret the test findings.

Answer 1

(a)

the type of statistical test procedure that should be used to test the hypotheses : Two independent Samples Proportions Z test

Explanation:

In this experimetation, the 2 samples: Sample 1: 100 patients received St. John’s Wort and Sample 2: 100 patients received Placebo are independent.

(b)

n1 = 100

1 = 27/100 = 0.27

n2 = 100

2 = 19/100 = 0.19

=0.05

From Table, critical values of Z = 1.96

Confidence Interval:

95% confidence interval. :

( - 0.036, 0.196)

Since 0 is included in the Confidence Interval: ( - 0.036, 0.196) , we conclude that :

the difference is not significant. Fail to reject null hypothesis.

Conclusion:
The data do not support the claim that St. John’s Wort is effective in treating major depression.

Interpretation:
The 95% confidence interval ( - 0.036, 0.196) is a range of values we are 95% confident will cotain the true unkown population proportion of difference between St. John’s Wort and Placebo.

(c)

H0:Null Hypothesis:p1 p2 ( St. John’s Wort is not effective in treating major depression. )

HA: Alternative Hypothesis:p1 > p2 ( St. John’s Wort is effective in treating major depression. ) (Claim)

n1 = 100

1 = 27/100 = 0.27

n2 = 100

2 = 19/100 = 0.19

Pooled Proportion is given by:

=0.05

From Tble, critical values of Z = 1.94

Test Statistic is given by:

Since calculated value of Z = 1.344 is less than critical value of Z = 1.96, we conclude that:

the difference is not significant. Fail to reject null hypothesis.

Conclusion:
The data do not support the claim that St. John’s Wort is effective in treating major depression.

We get the same answer as part (b).

(d)

By Technology,

p - value = 0.0894.

Since p - value = 0.0894. is greater than = 0.05, we conclude that:

the difference is not significant. Fail to reject null hypothesis.

Conclusion:
The data do not support the claim that St. John’s Wort is effective in treating major depression.

We get the same answer as part (b).