Question

A medical researcher is comparing the effectiveness of two treatments. Assume that the data is normally distributed. Sample data is as follows: Treatment A: Time to recover = 8 days s = 2.5 days n = 20 Treatment B: Time to recover = 10 days s = 3.5 days n = 16 Perform a hypothesis test at 95% confidence to determine whether there is a significant difference in the time to recover between the two treatments. To get full credit, do the following: - State your null and alternate hypothesis. - The value of your test statistic. - Provide either one of the following: your p-value or the boundary (or boundaries) of your critical region. - Do you reject the null hypothesis? YES or NO - Could the time to recover be the same for the two groups? YES or NO

Answer 1

The two-sample t-test comparing recovery times for Treatment A M = 8, SD = 2.5, n = 20 and Treatment B M = 11, SD = 3.5, n = 16 revealed a significant difference t = -2.885, df = 34, p < 0.05. Consequently, we reject the null hypothesis and conclude that the recovery times for the two treatments are significantly different.

Let's go through the entire process.

1. Null Hypothesis (H₀): There is no significant difference in the time to recover between Treatment A and Treatment B. (μ₁ - μ₂ = 0)

Alternative Hypothesis (H₁): There is a significant difference in the time to recover between Treatment A and Treatment B. (μ₁ - μ₂ ≠ 0)

2. Calculation of the Test Statistic (t-value):

Given data:

Treatment A:
`\(\overline{x}_1 = 8\) days, \(s_1 = 2.5\) days, \(n_1 = 20\)`

Treatment B:
`\(\overline{x}_2 = 11\) days, \(s_2 = 3.5\) days, \(n_2 = 16\)`

`\[ t = \frac{(8 - 11)}{\sqrt{(2.5^2)/(20) + (3.5^2)/(16)}} \]\[ t \approx -2.885 \]`

3. Comparison with Critical Value or P-value:

Degrees of freedom
`(\(df\)) = \(n_1 + n_2 - 2 = 20 + 16 - 2 = 34\)`

Using a t-distribution table or statistical software, compare the calculated t-value to the critical t-value at a 95% confidence level (two-tailed) with (df = 34). Alternatively, find the p-value associated with the t-value.

- Suppose the critical t-value is
`\(t_{\text{critical}} \approx \pm 2.032\) for \(df = 34\)` at a 95% confidence level. Since (-2.885 < -2.032), we reject the null hypothesis.

- Alternatively, suppose the p-value is (p ≈ 0.007). Since (p < 0.05), we reject the null hypothesis.

4. Conclusion:

- If the p-value is less than 0.05 or if the confidence interval for the difference in means does not include zero, you reject the null hypothesis.

- Since either the t-value is beyond the critical value or the p-value is less than 0.05, we reject the null hypothesis.

5. Final Statement:

- Based on the statistical analysis, we reject the null hypothesis. There is a significant difference in the time to recover between Treatment A and Treatment B.

- Answer the question: Are the recovery times for the two treatments different? YES.

This completes the comparison part, providing a clear understanding of how the test statistic is compared to the critical value or p-value to make a decision regarding the null hypothesis.