Chapter 11.2, Problem 37E

a)

To determine

To summarize data in a two-way table.

Explanation of Solution

Given:

Group	Treatment	Subjects	Successes
1	Nicotine patch	244	40
2	Drug	244	74
3	Patch plus drug	245	87
4	Placebo	160	25

Calculation:

For constructing a two-table, we need successes and failures with each treatment. Therefore, we already know about the treatment and successes. So, failure will be calculated by subtracting successes from each treatment. Following table shows the two-way table:

Treatment	Successes	Failures	Total
Nicotine patch	40	204	244
Drug	74	170	244
Patch plus drug	87	158	245
Placebo	25	135	160
Total	25	135	893

b)

To determine

To perform chi square test of independence.

Answer to Problem 37E

There is convincing evidence that the difference in the effectiveness of the four treatments at the 0.05 level of significance.

Explanation of Solution

Given:

Treatment	Successes	Failures	Total
Nicotine patch	40	204	244
Drug	74	170	244
Patch plus drug	87	158	245
Placebo	25	135	160
Total	25	135	893

Calculation:

Null and alternative hypotheses:

  $H_0$ : The treatment and successes are independent.

  $H_a$ : The treatment and successes are not independent.

Using excel formula,

		Successes	Failures	Total
Nicotine patch	Observed	40	204	244
	Expected	61.75	182.25	244.00
	(O - E)² / E	7.66	2.60	10.26
Drug	Observed	74	170	244
	Expected	61.75	182.25	244.00
	(O - E)² / E	2.43	0.82	3.25
Patch plus drug	Observed	87	158	245
	Expected	62.00	183.00	245.00
	(O - E)² / E	10.08	3.41	13.49
Placebo	Observed	25	135	160
	Expected	40.49	119.51	160.00
	(O - E)² / E	5.93	2.01	7.94
Total	Observed	226	667	893
	Expected	226.00	667.00	893.00
	(O - E)² / E	26.10	8.84	34.94
		34.94	chi-square
		3	df
		1.26E-07	p-value

Decision: P-value = 1.26E-07< 0.05, reject H0.

Conclusion: There is convincing evidence that the difference in the effectiveness of the four treatments at the 0.05 level of significance.