To test the claim that women have higher pain levels than men when taking this drug to treat osteo-arthritis, we can use a two-sample t-test.
First, let's state the null hypothesis (Ho) and the alternative hypothesis (Ha).
Ho: μ1 = μ2 (The mean pain levels of women and men are equal)
Ha: μ1 > μ2 (The mean pain levels of women are higher than men)
We will use a 5% significance level, which means the critical value for this one-tailed test is 1.645 (assuming a one-tailed test as we are only interested in whether women have higher pain levels than men).
Now, we can calculate the test statistic.
t = (x1 - x2) / sqrt((s1^2/n1) + (s2^2/n2))
where x1 and x2 are the sample means for women and men respectively, s1 and s2 are the sample standard deviations for women and men respectively, n1 and n2 are the sample sizes for women and men respectively.
Given the data:
For women, x1 = 5.6, s1 = 1.2, and n1 = 52.
For men, x2 = 4.8, s2 = 1.5, and n2 = 47.
Calculating the test statistic:
t = (5.6 - 4.8) / sqrt((1.2^2/52) + (1.5^2/47))
= 0.8 / sqrt(0.0276923 + 0.0468298)
= 0.8 / sqrt(0.0745221)
≈ 0.8 / 0.272812
≈ 2.93
Since the test statistic (2.93) is greater than the critical value (1.645), we can reject the null hypothesis.
The decision is: Reject Ho.
Therefore, the correct answer is:
A) Reject Ho. The sample evidence does not support the claim that women have higher pain levels than men when taking this drug to treat osteo-arthritis.