A researcher wants to obtain a confidence interval for the difference between th

Question

A researcher wants to obtain a confidence interval for the difference between the proportions of women and men who approve of a proposed smoking ban for all city parks. You wish to obtain a common sample size that will ensure a margin of error of at most 0.05 for a 95% confidence interval, if you can reasonably assume that at most 47% of men (pg1) and between 55% and 60% of women (pg2) approve of the proposed smoking ban.

            What proportions (guesses) should you use for pg1 and for pg2 in the formula to determine the required sample sizes for the CI? You do not need to calculate the required sample size.

Explanation / Answer

Here for women (p^ g2) i will guess = (0.55 + 0.60)/2 = 0.575

Now we know that margin of error = 0.05

mArgin of error = critical test statistic * standard error of difference between probportion.

0.05 = 1.96 * se0

se0 = 0.05/1.96 = 0.0255 = sqrt [p1^q1/N + p2^q2/N]

so we can assume that both will contribute similarly in the standard error of the difference of the proportion.

Here for p^g1 it is given that at most 47% of men are proposing smoking ban. Here margin of error is around 0.05 for difference

so,margin of error of sample proportion of men = 0.05 / sqrt(2) = 0.0354

so p^ g2 = 0.47 - 0.0354 = 0.4346 or 0.435 around

so i will guess proportion for men = 0.435 = 43.5% and for women = 0.575 = 57.5 %

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