posted by Brenna .
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 p-value.
Try a binomial proportion 2-sample z-test using proportions.
Ho: pF = pM (F = female; M = male)
Ha: pF > pM -->One-tailed 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 1-p.
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 p-value or the actual level of this test statistic by using a z-table. 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.