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.

Hypotheses:
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.

To determine whether there is evidence to show that a higher proportion of women smoke compared to men, we can perform a hypothesis test.

Let's set up the null and alternative hypotheses:

Null hypothesis (H0): The proportion of women who smoke is equal to the proportion of men who smoke.
Alternative hypothesis (H1): The proportion of women who smoke is higher than the proportion of men who smoke.

Next, we need to compute the test statistic. For this, we can use the formula for the test statistic for comparing proportions:

z = (p1 - p2) / √[ (p * (1-p) / n1) + (p* (1-p) / n2) ]

where:
p1 and p2 are the proportions of women and men who smoke, respectively,
p is the pooled proportion,
n1 and n2 are the sample sizes for women and men, respectively.

First, we calculate the pooled proportion:
p = (x1 + x2) / (n1 + n2)
where:
x1 is the number of women who smoke,
x2 is the number of men who smoke,
n1 is the sample size for women, and
n2 is the sample size for men.

In this case:
x1 = 45, x2 = 25
n1 = 150, n2 = 100

p = (45 + 25) / (150 + 100)
p = 70 / 250
p = 0.28

Next, we can calculate the test statistic using the formula mentioned earlier:
z = (p1 - p2) / √[ (p * (1-p) / n1) + (p* (1-p) / n2) ]

For this case:
p1 = 45 / 150
p2 = 25 / 100
n1 = 150, n2 = 100

z = (45/150 - 25/100) / √[ (0.28 * (1-0.28) / 150) + (0.28 * (1-0.28) / 100) ]
z = (0.3 - 0.25) / √[ (0.28*0.72 / 150) + (0.28*0.72 / 100) ]
z = 0.05 / 0.0506
z ≈ 0.99

Now, to find the p-value, we need to compare the test statistic to the critical value from the standard normal distribution. At a significance level of 0.05 (equivalent to a confidence level of 95%), the critical value is approximately 1.645 (obtained from a standard normal distribution table).

Since the calculated test statistic (z = 0.99) is less than the critical value (1.645), we fail to reject the null hypothesis. This means that we do not have sufficient evidence to conclude that a higher proportion of women smoke compared to men.

To find the p-value, we can use a standard normal distribution table or a statistical software. The p-value corresponds to the probability of obtaining a test statistic as extreme as the one observed (or more extreme) under the null hypothesis.

Using software, the p-value is approximately 0.163.

Therefore, the final conclusion is that at the 0.05 significance level, there is not enough evidence to show that a higher proportion of women smoke compared to men. The p-value is approximately 0.163.