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

(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.

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\\eq 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{(\hat p_1(1-\hat p_1))/(n_1)+(\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 =
`(197)/(1647)` = 0.12

`\hat p_2` = sample proportion of people using placebo that stopped smoking =
`(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{(0.12(1-0.12))/(1647)+(0.05(1-0.05))/(1620) } }`

= 7.242

The value of z test statistics is 7.242.