Question

A study used nicotine gum to help people quit smoking. The study was​ placebo-controlled, randomized, and​ double-blind. Each participant was interviewed after 28​ days, and success was defined as being abstinent from cigarettes for 28 days. The results showed that 197 out of 1647 people using the nicotine gum​ succeeded, and 85 out of 1620 using the placebo succeeded. Although the sample was not​ random, the assignment to groups was randomized. a. Find the proportion of people using nicotine gum that stopped smoking and the proportion of people using the placebo that stopped​ smoking, and compare them. Is this what the researchers had​ expected? b. Find the observed value of the test​ statistic, assuming that the conditions for a​ two-proportion z-test hold.

Answer 1

Answer:

(a) The proportion of people using nicotine gum that stopped smoking is 0.12 and the proportion of people using the placebo that stopped​ smoking is 0.05.

(b) The value of z test statistics is 7.242.

Step-by-step explanation:

We are given that a study used nicotine gum to help people quit smoking. The study was​ placebo-controlled, randomized, and​ double-blind.

The results showed that 197 out of 1647 people using the nicotine gum​ succeeded, and 85 out of 1620 using the placebo succeeded.

Let $p_1$ = population proportion of people who quit smoking using nicotine gum.

$p_2$ = population proportion of people who quit smoking using placebo.

SO, Null Hypothesis, $H_0$ : $p_1=p_2$    

Alternate Hypothesis, $H_A$ : $p_1\neq p_2$  

The test statistics that would be used here Two-sample z test for proportions;

                        T.S. =  $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$  ~ N(0,1)

where, $\hat p_1$ = sample proportion of people using nicotine gum that stopped smoking = $\frac{197}{1647}$ = 0.12

$\hat p_2$ = sample proportion of people using placebo that stopped smoking = $\frac{85}{1620}$ = 0.05

$n_1$ = sample of people using the nicotine gum = 1647

$n_2$ = sample of people using the placebo = 1620

So, the test statistics  =  $\frac{(0.12-0.05)-(0)}{\sqrt{\frac{0.12(1-0.12)}{1647}+\frac{0.05(1-0.05)}{1620} } }$

                                     =  7.242

The value of z test statistics is 7.242.