Confidence interval Question

Question

Measurements of gastric secretion of hydrochloric acid (milliequivalents per hour) in 16 normal subjects and 10 subjects with duodenal ulcer yielded the following results:

Normal subjects: 6.3, 2.0, 2.3, 0.5, 1.9, 3.2, 4.1, 4.0, 6.2, 6.1, 3.5, 1.3, 1.7, 4.5, 6.3, 6.2

Ulcer subjects: 13.7, 20.6, 15.9, 28.4, 29.4, 18.4, 21.1, 3.0, 26.2, 13.0

Construct a 99% confidence interval for the difference in the mean values of the populations. What is the probability that the difference in the population means is greater than 4?

Answer 1

This requires a Welch's t test for 2 independent samples. (AND A LOT OF CALCULATIONS)

The required statistics are standard deviations for each group (s₁ and s₂) and means for each group (xbar₁ and xbar₂). From these we calculate SE which is √( s₁²/n₁. + s₂²/n₂). and t = ¦Xbar1 -Xbar2¦ / SE

finally degrees of freedom (d.f.) is estimated by SE⁴ / ((s₁⁴/n₁²v₁) + (s₂⁴/n₂²v₂))

if group 1 are Normal subjects and group 2 are Ulcer subjects Xbar₁ = 3.75625 s₁ =2.01659 n₁ = 16

and Xbar₂ = 18.9700 s₂ =8.06488 n₂ = 10

SE = 2.59969. t =. 15.21375/2.59969 = 5.85214. d = 15.21375

d.f. = 9.708

First, calculate a confidence interval for alpha = .01.

using a calculator and the inverse cumulative t distribution function for p = .005 & p= .995. with a d.f. = 9.7 we have a t crit of = 3.1906, a MOE of 3.1906*2.59969 or 8.294571

confidence interval CI is then [6.9192 23.5083]

Second, the probability that the |d| is > 4 ( or. 4 < d < -4 )

we calculate the t stat as (10.154389-4 )/ 2.59969 or 2.367355

from this , using the CDF of the t distribution , and using 9.7 d.f. we get. .9597 (or 95.9686%)