Question

In the past decades there have been intensive antismoking campaigns sponsored by both federal and private agencies. In one study of national smoking trends, two random samples of U.S. adults were selected in different years: The first sample, taken in 1995, involved 4276 adults, of which 1642 were smokers. The second sample, taken in 2010, involved 3908 adults, of which 1415 were smokers. The samples are to be compared to determine whether the proportion of U.S. adults that smoke declined during the 15-year period between the samples.Let p1 be the proportion of all U.S. adults that smoked in 1995. Let p2 denote the proportion of all U.S. adults that smoked in 2010.The value of the z statistic for testing equality of the proportion of smokers in 1995 and 2010 isA. z = 1.39.B. z = 1.66.C. z = 2.05.D. z = 4.23.

Answer 1

Answer:

C. z = 2.05

Step-by-step explanation:

We have to calculate the test statistic for a test for the diference between proportions.

The sample 1 (year 1995), of size n1=4276 has a proportion of p1=0.384.

$p_1=X_1/n_1=1642/4276=0.384$

The sample 2 (year 2010), of size n2=3908 has a proportion of p2=0.3621.

$p_2=X_2/n_2=1415/3908=0.3621$

The difference between proportions is (p1-p2)=0.0219.

$p_d=p_1-p_2=0.384-0.3621=0.0219$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{1642+1415}{4276+3908}=\dfrac{3057}{8184}=0.3735$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.3735*0.6265}{4276}+\dfrac{0.3735*0.6265}{3908}}\\\\\\s_{p1-p2}=\sqrt{0.00005+0.00006}=\sqrt{0.00011}=0.0107$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.0219-0}{0.0107}=\dfrac{0.0219}{0.0107}=2.05$

z=2.05