Question

Q3. A clinical trial is run to investigate the effectiveness of an experimental drug in reducing pre-term delivery to a drug considered standard care and to placebo. Pregnant women are enrolled and randomly assigned to receive either the experimental drug, the standard drug, or placebo. Women are followed through delivery and classified as delivering pre-term (< 37 weeks) or not. The data are shown below.

Pre-term Delivery	Experimental Drug	Standard Drug	Placebo
Yes	20	32	36
No	80	68	64

a) Is there a statistically significant difference in the proportions of women delivering pre-term among the three treatment groups? Run the test at a 5% level of significance. (Hint: χ2 test) (10 points)

(1 point) Ho: VS H1:

Compute: χ2 (7 points):

P-value (1 point)

Conclusion (1 point)

Accept H0 Reject H0

Answer 1

The following cross tablulation have been provided. The row and column total have been calculated and they are shown below:

	Experimental Drug	Standard Drug	Placebo	Total
Yes	20	32	36	88
No	80	68	64	212
Total	100	100	100	300

The expected values are computed in terms of row and column totals. In fact, the formula is , where Ri​ corresponds to the total sum of elements in row i, Cj​ corresponds to the total sum of elements in column j, and T is the grand total. The table below shows the calculations to obtain the table with expected values:

Expected Values	Experimental Drug	Standard Drug	Placebo	Total
Yes				88
No				212
Total	100	100	100	300

Based on the observed and expected values, the squared distances can be computed according to the following formula: (E - O)^2/E . The table with squared distances is shown below:

Squared Distances	Experimental Drug	Standard Drug	Placebo
Yes
No

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

H0​: The two variables are independent

Ha​: The two variables are dependent

This corresponds to a Chi-Square test of independence.

Rejection Region

Based on the information provided, the significance level is α=0.05 , the number of degrees of freedom is df=(2−1)×(3−1)=2, so then the rejection region for this test is

Test Statistics

The Chi-Squared statistic is computed as follows:

Decision about the null hypothesis

Since it is observed that , it is then concluded that the null hypothesis is rejected.

Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the two variables are dependent, at the 0.05 significance level.

The corresponding p-value for the test is