Question

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 estimated standard
error.

Answer 1

From the information given, and applying the formula, it is found that the standard error of the difference in proportion of smokers by sex is 0.0558.

The proportion and the standard error for each sample are given by:

$p_W = 0.25, s_W = \sqrt{\frac{0.25(0.75)}{120}} = 0.0395$

$p_M = 0.37, s_M = \sqrt{\frac{0.37(0.63)}{150}} = 0.0394$

The standard error of the difference in proportions is the square root of the sum of the standard error for each sample squared, hence:

$s = \sqrt{s_W^2 + s_M^2} = \sqrt{0.0395^2 + 0.0394^2} = 0.0558$

Hence, the standard error is of 0.0558.

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