Question

To find out whether Medicine X is more effective at managing insomnia than Medicine Y, 11 patients are selected. 6 receive Medicine X, and 5 receive Medicine Y. The time taken to fall asleep, in minutes, is as follows: Medicine X: 22, 24, 18, 28, 16, 20. Medicine Y: 18, 20, 14, 20, 14 Assume the two populations are normally distributed with equal variances. (a) Construct a 99% confidence interval for the difference in average time to fall asleep (in minutes) between medicine X and Y?

Answer 1

The 99% confidence interval for the difference in average time to fall asleep between Medicine X and Medicine Y is approximately (-3.08, 11.34) minutes.

How to construct a 99% confidence interval

To construct a 99% confidence interval for the difference in average time to fall asleep between Medicine X and Medicine Y, use a two-sample t-test.

For Medicine X:

Sample size (n₁) = 6

Sample mean (
`\bar{x_1}`) = (22 + 24 + 18 + 28 + 16 + 20) / 6 = 128 / 6 ≈ 21.33

Sample standard deviation (s₁) =
`\sqrt`(((22-21.33)² + (24-21.33)² + (18-21.33)² + (28-21.33)² + (16-21.33)² + (20-21.33)²) / (6-1)) ≈ 4.16

For Medicine Y:

Sample size (n₂) = 5

Sample mean (
`\bar{x_2}`) = (18 + 20 + 14 + 20 + 14) / 5 = 86 / 5 = 17.2

Sample standard deviation (s₂) =
`\sqrt`(((18-17.2)² + (20-17.2)² + (14-17.2)² + (20-17.2)² + (14-17.2)²) / (5-1)) ≈ 2.64

Now, calculate the pooled standard deviation (sp) using the formula:

`\sqrt`(((n₁ - 1) * s1² + (n₂ - 1) * s₂²) / (n₁ + n₂ - 2))

=
`\sqrt`(((6 - 1) * 4.16² + (5 - 1) * 2.64²) / (6 + 5 - 2))

≈
`\sqrt`((5 * 17.3056 + 4 * 6.9696) / 9)

≈
`\sqrt`((86.528 + 27.8784) / 9)

≈
`\sqrt`(114.4064 / 9)

≈
`\sqrt`(12.7118)

≈ 3.56

The standard error of the difference in means (SE) can be calculated as:

SE = sp *
`\sqrt`(1/n₁ + 1/n₂)

= 3.56 *
`\sqrt`(1/6 + 1/5)

= 3.56 *
`\sqrt`(0.1667 + 0.2)

= 3.56 *
`\sqrt`(0.3667)

≈ 3.56 * 0.6051

≈ 2.157

To construct a 99% confidence interval, we need to calculate the margin of error (ME):

ME = t * SE

Since the sample sizes are small (n₁ = 6, n₂ = 5), use a t-distribution instead of a z-distribution. With a 99% confidence level, the degrees of freedom (df) is (n₁ + n₂ - 2) - 1 = 8. We can consult the t-distribution table or use a statistical software to find the critical t-value.

For a 99% confidence level with 8 df, the critical t-value is approximately 3.355.

ME = 3.355 * 2.157 ≈ 7.21

Finally, we can construct the confidence interval:

Confidence interval = (
`\bar{x_1} - \bar{x_2}`) ± ME

= 21.33 - 17.2 ± 7.21

= 4.13 ± 7.21

≈ (-3.08, 11.34)

Therefore, the 99% confidence interval for the difference in average time to fall asleep between Medicine X and Medicine Y is approximately (-3.08, 11.34) minutes.