Question 1181427: Drug A was prescribed for a random sample of 12 patients complaining of insomnia. An independent random sample of 16 patients with the same complaint received drug B.The number of hours of sleep experienced during the second night after treatment began were as follows:
A: 3.5, 5.7, 3.4, 6.9, 17.8, 3.8, 3.0, 6.4, 6.8, 3.6, 6.9, 5.7
B: 4.5, 11.7, 10.8, 4.5, 6.3, 3.8, 6.2, 6.6, 7.1, 6.4, 4.5, 5.1, 3.2, 4.7, 4.5, 3.0
Construct a 95 percent confidence interval for the difference between the population means. Assume that the population variances are equal.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to construct a 95% confidence interval for the difference between the population means, assuming equal variances:
**1. Calculate the Sample Means:**
* **Drug A:** (3.5 + 5.7 + 3.4 + 6.9 + 17.8 + 3.8 + 3.0 + 6.4 + 6.8 + 3.6 + 6.9 + 5.7) / 12 = 6.275
* **Drug B:** (4.5 + 11.7 + 10.8 + 4.5 + 6.3 + 3.8 + 6.2 + 6.6 + 7.1 + 6.4 + 4.5 + 5.1 + 3.2 + 4.7 + 4.5 + 3.0) / 16 = 5.8
**2. Calculate the Sample Standard Deviations:**
* **Drug A:** Use your calculator or statistical software. The sample standard deviation is approximately 4.22.
* **Drug B:** Use your calculator or statistical software. The sample standard deviation is approximately 2.34.
**3. Calculate the Pooled Standard Deviation:**
Since we assume equal variances, we calculate a pooled standard deviation to get a better estimate of the common population standard deviation.
* s_p = sqrt[((n_A - 1) * s_A^2 + (n_B - 1) * s_B^2) / (n_A + n_B - 2)]
* s_p = sqrt[((12 - 1) * 4.22^2 + (16 - 1) * 2.34^2) / (12 + 16 - 2)]
* s_p ≈ sqrt[(11 * 17.8084 + 15 * 5.4756) / 26]
* s_p ≈ sqrt(277.6224 / 26)
* s_p ≈ 3.27
**4. Determine the Critical t-Value:**
* Degrees of freedom (df) = n_A + n_B - 2 = 12 + 16 - 2 = 26
* For a 95% confidence interval and a two-tailed test, the alpha level is 1 - 0.95 = 0.05. We then divide alpha by 2 since it is a two-tailed test, giving us 0.025.
* Using a t-table or calculator, find the t-value associated with df = 26 and α/2 = 0.025. The critical t-value is approximately 2.056.
**5. Calculate the Margin of Error:**
* Margin of Error = t * s_p * sqrt(1/n_A + 1/n_B)
* Margin of Error = 2.056 * 3.27 * sqrt(1/12 + 1/16)
* Margin of Error ≈ 2.056 * 3.27 * 0.354
* Margin of Error ≈ 2.37
**6. Construct the Confidence Interval:**
* (x̄_A - x̄_B) ± Margin of Error
* (6.275 - 5.8) ± 2.37
* 0.475 ± 2.37
**7. Final Answer:**
The 95% confidence interval for the difference between the population means is approximately (-1.90, 2.85).
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