Statistics
posted by Brenna on .
A recent study was designed to compare smoking habits of young women with those of young men. A random sample of 150 women revealed that 45 smoked. A random sample of 100 men indicated that 25 smoked. At the 0.05 significance level does the evidence show that a higher proportion of women smoke? Compute the pvalue.

Try a binomial proportion 2sample ztest using proportions.
Hypotheses:
Ho: pF = pM (F = female; M = male)
Ha: pF > pM >Onetailed test (shows a specific direction)
The formula is:
z = (pF  pM)/√[pq(1/n1 + 1/n2)]
...where 'n' is the sample sizes, 'p' is (x1 + x2)/(n1 + n2), and 'q' is 1p.
I'll get you started:
p = (45 + 25)/(150 + 100) = ? >once you have the fraction, convert to a decimal (decimals are easier to use in the formula)
q = 1  p
pF = 45/150
pM = 25/100
Convert all fractions to decimals. Plug those decimal values into the formula and find z. Once you have this value, you will be able to determine the pvalue or the actual level of this test statistic by using a ztable. Finally, determine whether or not to reject the null. If the null is rejected, then you can conclude that pF > pM.
I hope this will help.