Question

A beef cattle nutritionist wants to compare the birth weights of calves from cows that receive two different diets during gestation. He therefore selects 16 pairs of cows, where the cows within each pair have similar characteristics. One cow within each pair is randomly assigned to diet 1, while the other cow in the pair is assigned to diet 2. He obtains the following results:

     Mean difference in birth weights of the pairs of calves = 10 lb

     Standard deviation of the difference in birth weights of the pairs = 8.0 lb

Construct a 95% confidence interval for the true mean difference in birth weights of the calves from cows receiving diet 1 vs. diet 2.

Group of answer choices

(5.738 lb, 14.262 lb)

(6.08 lb, 13.92 lb)

(6.606 lb, 13.394 lb)

(4.97635 lb, 17.02365 lb)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 10

sample standard deviation = s = 8

sample size = n = 16

Degrees of freedom = df = n - 1 = 15

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,15 = 2.131

Margin of error = E = t_/2,df * (s /n)

= 2.131 * (8 / 16)

= 4.262

The 95% confidence interval estimate of the population mean is,

- E < < + E

10 - 4.262 < < 10 + 4.262

5.738 < < 14.262

The answer is,

(5.738 lb , 14.262 lb)