Question

hilip Morris wishes to determine if there is a difference between the proportion of women and proportion of men who smoke cigarettes. Random samples of 125 women and 140 men reveal that 13 women and 5 men smoke cigarettes. Does the data indicate that the proportion of women who smoke cigarettes is higher than the proportion of men who do at α=.01? What is the 95% confidence interval for pWomen - pMen?

Answer 1

Answer:

We conclude that the proportion of women who smoke cigarettes is smaller than or equal to the proportion of men at 0.01 significance level.

95% confidence interval for the difference in population proportions of women and men who smoke cigarettes is [0.0062 , 0.1298].

Step-by-step explanation:

We are given that random samples of 125 women and 140 men reveal that 13 women and 5 men smoke cigarettes.

Let $p_1$ = population proportion of women who smoke cigarettes

$p_2$ = population proportion of men who smoke cigarettes

So, Null Hypothesis, $H_0$ : $p_1\leq p_2$      {means that the proportion of women who smoke cigarettes is smaller than or equal to the proportion of men}

Alternate Hypothesis, $H_A$ : $p_1> p_2$      {means that the proportion of women who smoke cigarettes is higher than the proportion of men}

The test statistics that will be used here is Two-sample z proportion test statistics;

                               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 women who smoke cigarettes= $\frac{13}{125}$ =0.104

$\hat p_2$ = sample proportion of men who smoke cigarettes = $\frac{5}{140}$ = 0.036

$n_1$ = sample of women = 125

$n_2$ = sample of men = 140

So, the test statistics  =  $\frac{(0.104-0.036)-(0)}{\sqrt{\frac{0.104(1-0.104)}{125}+ \frac{0.036(1-0.036)}{140}} }$

                                     =  2.158

Now, at 0.01 significance level, the z table gives critical value of 2.3263 for right tailed test. Since our test statistics is less than the critical value of z as 2.158 < 2.3263, so we have insufficient evidence to reject our null hypothesis due to which we fail to reject our null hypothesis.

Therefore, we conclude that the proportion of women who smoke cigarettes is smaller than or equal to the proportion of men.

Now, coming to 95% confidence interval;

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportions is given by;

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

where, $\hat p_1$ = sample proportion of women who smoke cigarettes= $\frac{13}{125}$ =0.104

$\hat p_2$ = sample proportion of men who smoke cigarettes = $\frac{5}{140}$ = 0.036

$n_1$ = sample of women = 125

$n_2$ = sample of men = 140

Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.

So, 95% confidence interval for the difference between population proportions, $(p_1-p_2)}$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                    of significance are -1.96 & 1.96}  

P(-1.96 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < $(p_1-p_2)}$ < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

95% confidence interval for $(p_1-p_2)}$ =

[$(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$,$(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$]

= [ $(0.104-0.036)-1.96 \times {\sqrt{\frac{0.104(1-0.104)}{125}+ \frac{0.036(1-0.036)}{140}} }$ , $(0.104-0.036)+1.96 \times {\sqrt{\frac{0.104(1-0.104)}{125}+ \frac{0.036(1-0.036)}{140}} }$ ]

 = [0.0062 , 0.1298]

Therefore, 95% confidence interval for the difference in population proportions of women and men who smoke cigarettes is [0.0062 , 0.1298].