Question

A researcher wants to know if the vitamins will increase the average weight of a cow. She randomly selects 2 cows from each of 18 different breeds of cows. For each breed one cow gets the vitamin, and one cow does not. Assume cow weights are normally distributed. Given the data below, calculate a 90% confidence interval for the difference in the averages between cows on the vitamins and cows not on the vitamins.

For the cows which did not take vitamins: Number of cows: 18 Mean weight: 953 pounds Standard deviation: 40 pounds	For the cows which did take vitamins: Number of cows: 18 Mean weight: 765 pounds Standard deviation: 25 pounds
The standard deviation for the difference in weights was 14.5.

Answer 1

As the two tests are performed on the same breed, we will use paired t test to calculated the confidence interval.

Sample mean of the difference , x̅d = 765 -953 -188

Sample standard deviation of the difference, sd = 14.5

Sample size, n = 18

90% Confidence interval for the difference in the averages between cows on the vitamins and cows not on the vitamins:

At α = 0.10 and df = n-1 = 17, two tailed critical value, t-crit = T.INV.2T(0.1, 17) = 1.740

Lower Bound = x̅d - t-crit*sd/√n = -188 - 1.74 * 14.5/√18 = -193.9454

Upper Bound = x̅d + t-crit*sd/√n = -188 + 1.74 * 14.5/√18 = -182.0546

-193.9454 < µd < -182.0546