A new medication has been created to treat osteo-arthritis inflammation and pain. Doctors prescribing the drug claim that the drug is not as effective for women as men since women taking the medication have higher pain levels than men. The pharmaceutical company conducts a clinical trial with 52 women and 47 men having osteo-arthritis in order to determine if women taking the drug have higher pain levels than men. After treatment, the 52 women have a mean pain level of 5.6 with a standard deviation of 1.2 while the 47 men have a mean pain level of 4.8 with a standard deviation of 1.5. Higher pain level measures indicate more inflammation and pain.

Test the claim that women have higher pain levels than men when taking this drug to treat osteo-arthritis. Use a 5% significance level.

Determine if the decision is Reject or Fail to reject Ho and write the conclusion for the result.

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.
B) Fail to 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.
C) Reject Ho. The sample evidence supports the claim that women have higher pain levels than men when taking this drug to treat osteo-arthritis.
D) Fail to reject Ho. The sample evidence supports the claim that women have higher pain levels than men when taking this drug to treat osteo-arthritis.

Ho: women's level > men's

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.

I'll let you do the calculations.

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.