Question

4. In a simple random sample of 120 women and 150 men from
a population of 150,000, 25% of the women and 37% of the men
are smokers. Suppose you are interested in comparing the
proportion of smokers by sex, compute the Z statistic.

Answer 1

The z-test compares the two sample to determine the existence of a

difference in the two proportion.

• The Z-statistic or test of the difference in proportion is approximately -2.106.

Reasons:

The given parameter are;

Number of women, n₁ = 120 women

Number of men, n₂ = 150 men

The population size = 150,000

Percentage of women that are smokers in population, $\hat p_1$ = 25%

Percentage of men that are smokers in population, $\hat p_2$ = 37%

The Z statistic is given as follows;

$\displaystyle Z= \mathbf{\frac{\hat{p}_1-\hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})\left (\frac{1}{n_{1}}+\frac{1}{n_{2}} \right )}}}$

Number of men = 0.37 × 150 = 55.5

Number of women = 0.25 × 120 = 30

$\displaystyle \hat p = \frac{55.5 + 30}{150 + 120} = \frac{19}{60}$

Which gives;

$\displaystyle Z=\frac{0.25-0.37}{\sqrt{\frac{19}{60} (1-\frac{19}{60} )\left (\frac{1}{120}+\frac{1}{150} \right )}} \approx -2.106$

The Z-statistic is approximately -2.106

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