Chapter 11.2, Problem 38E

a)

To determine

To summarize data in a two-way table.

Explanation of Solution

Given:

Group	Treatment	Number of patients	Having stroke
1	Placebo	1649	250
2	Aspirin	1649	206
3	Dipyridamole	1654	211
4	Both	1650	157

Calculation:

For constructing a two-table, we need successes and failures with each treatment. Therefore, we already know about the treatment and number of patients had a stroke. So, number of patients had a no stroke will be calculated by subtracting number of patients had a stroke from number of patients. Following table shows the two-way table:

Treatment	Having stroke	Having no stroke	Total
Placebo	250	1399	1649
Aspirin	206	1443	1649
Dipyridamole	211	1443	1654
Both	157	1493	1650
Total	824	5778	6602

b)

To determine

To perform chi square test of independence.

Answer to Problem 38E

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	Having stroke	Having no stroke	Total
Placebo	250	1399	1649
Aspirin	206	1443	1649
Dipyridamole	211	1443	1654
Both	157	1493	1650
Total	824	5778	6602

Calculation:

Null and alternative hypotheses:

  $H_0$ : The treatment and stroke are independent.

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

Using excel formula,

		Having strock	Having no stroke	Total
Placebo	Observed	250	1399	1649
	Expected	205.81	1443.19	1649.00
	(O - E)² / E	9.49	1.35	10.84
Aspirin	Observed	206	1443	1649
	Expected	205.81	1443.19	1649.00
	(O - E)² / E	0.00	0.00	0.00
Dipyridamole	Observed	211	1443	1654
	Expected	206.44	1447.56	1654.00
	(O - E)² / E	0.10	0.01	0.12
Both	Observed	157	1493	1650
	Expected	205.94	1444.06	1650.00
	(O - E)² / E	11.63	1.66	13.29
Total	Observed	824	5778	6602
	Expected	824.00	5778.00	6602.00
	(O - E)² / E	21.22	3.03	24.24
		24.24	chi-square
		3	df
		2.22E-05	p-value

Decision: P-value = 2.22E-05< 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.