Question

A sample of 172 babies in the zinc group had a mean birth weight of 3072 grams. A sample of 158 babies in the placebo group had a mean birth weight of 3426 grams. Assume that the population standard deviation for the zinc group is 792 grams, while the population standard deviation for the placebo group is 708 grams. Determine the 95% confidence interval for the true difference between the mean birth weights for "zinc" babies versus "placebo" babies.

Answer 1

The 95% confidence interval for the true difference between the mean birth weights for "zinc" babies versus "placebo" babies is (-515.8566, -192.1434)

Explanation:

Let
`\mu_(1)-\mu_(2)` be the true difference between the mean birth weights for "zinc" babies versus "placebo" babies. We have the sample sizes
`n_(1) = 172` babies and
`n_(2) = 158` babies, the unbiased point estimate for
`\mu_(1)-\mu_(2)` is
`\bar{x}_(1) - \bar{x}_(2)`, i.e., 3072-3426 = -354

The standard error is given by
`\sqrt{(\sigma^(2)_(1))/(n_(1))+(\sigma^(2)_(2))/(n_(2))}`, i.e.,

`\sqrt{((792)^(2))/(172)+((708)^(2))/(158)}` = 82.5799. Then, the endpoints for a 95% confidence interval for
`\mu_(1)-\mu_(2)` is given by

-354-
`(z_(0.05/2))`82.5799 and -354+
`(z_(0.05/2))`82.5799, i.e.,

-354-
`(z_(0.025))`82.5799 and -354+
`(z_(0.025))`82.5799 where
`z_(0.025)` is the 2.5th quantile of the standard normal distribution, i.e., -1.96, so, we have

-354-(1.96)(82.5799) and -354+(1.96)(82.5799), i.e.,

-515.8566 and -192.1434