Question

Critical Values: z0.005 = 2.575, z0.01 = 2.325, z0.025 = 1.96, z0.05 = 1.645

When d.f.=7: t0.005 = 3.499, t0.01 = 2.998, t0.025 = 2.365, t0.05 = 1.895  

When d.f.=69: t0.005 = 2.648, t0.01 = 2.381, t0.025 = 1.994, t0.05 = 1.667

Among 198 smokers who underwent a “sustained care” program, 51 were no longer smoking after 6 months. Among 199 who underwent a “standard care” program, 30 were no longer smoking after 6 months. Use a 0.01 significance level to test the claim that the rate of success for stopping smoking is greater with sustained care.

Answer 1

The hypothesis being tested is:

H0: p1 = p2

Ha: p1 > p2

p1	p2	pc
0.2576	0.1508	0.204	p (as decimal)
51/198	30/199	81/397	p (as fraction)
51.	30.	81.	X
198	199	397	n
	0.1068	difference
	0.	hypothesized difference
	0.0405	std. error
	2.64	z
	.0041	p-value (one-tailed, upper)

The p-value is 0.0041.

Since the p-value (0.0041) is less than the significance level (0.01), we can reject the null hypothesis.

Therefore, we can conclude that the rate of success for stopping smoking is greater with sustained care.