SOLUTION: Marketing companies are interested in knowing the population percent of women who make the majority of household purchasing decisions. Suppose the marketing company did do a survey

Question 1205076: Marketing companies are interested in knowing the population percent of women who make the majority of household purchasing decisions. Suppose the marketing company did do a survey. They randomly surveyed 200 households and found that in 120 of them, the woman made the majority of the purchasing decisions.
Construct a 95% confidence interval for the population proportion of households where the women make the majority of the purchasing decisions .Interpret your answer in part a above?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
smple size is 200
p = proportion of household where the woman makes the majority of the purchasing decisions is 120.
p = 120/200 = .6
q = 1-0 = .4
sample mean proportion is .6
standard error is sqrt(.6 * .4 / 200) = .034641.
z-score formula is z = (x-m)/s
z is the z-score
x is the raw score
m is the mean
s is the standard error.
at 95% confidence interval, the critical z-score is z = plus or minus 1.96.
on the low end, the formula becomes -1.96 = (x-.6)/.034641.
solve for x to get x = -1.96 * .034641 + .6 = .5321
on the high end, the formula becomes 1.96 = (x-.6)/.034641.
solve for x to get x = 1.96 * .034641 + .6 = .6679.
your 95% confidence interval is from .5321 to .6679.
here's what it looks like on a graph.

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