Question

A recent study found that children who watched a cartoon with food advertising​ ate, on​ average, 28.6 grams of crackers as compared to an average of 18.8 grams of crackers for children who watched a cartoon without food advertising. Suppose that there were 61 children in each​ group, and the sample standard deviation for those children who watched the food ad was 8.7 grams and the sample standard deviation for those children who did not watch the food ad was 7.7 grams. Complete parts​ (a) and​ (b) below.

b. Assuming that the population variances are​ equal, construct 95% confidence interval estimate of the difference μ1−μ2 between the mean amount of crackers eaten by the children who watch and do not watch the food ad.

___≤ μ1−μ2 ≤ ___ (Round to two decimal places as needed.)

Answer 1

Solution:

Confidence interval for difference between two population means is given as below:

Confidence interval = (X1bar – X2bar) ± t*sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

We are given

n1 = 61

n2 = 61

df = n1 + n2 – 2 = 61 + 61 – 2 = 120

Confidence level = 95%

Critical t value = 1.9799

(by using t-table)

X1bar = 28.6

X2bar = 18.8

S1 = 8.7

S2 = 7.7

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = [(61 – 1)*8.7^2 + (61 – 1)*7.7^2]/(61 + 61 – 2)

S_p² = 67.4900

(X1bar – X2bar) = 28.6 – 18.8 = 9.8

t*sqrt[S_p²*((1/n1)+(1/n2))] = 1.9799*sqrt[67.4900*((1/61)+(1/61))]

t*sqrt[S_p²*((1/n1)+(1/n2))] = 1.9799*1.4875

t*sqrt[S_p²*((1/n1)+(1/n2))] = 2.9452

Confidence interval = (X1bar – X2bar) ± t*sqrt[S_p²*((1/n1)+(1/n2))]

Confidence interval = 9.8 ± 2.9452

Lower limit = 9.8 - 2.9452 = 6.8548

Upper limit = 9.8 + 2.9452 = 12.7452

Confidence interval = (6.8548, 12.7452)

6.85 ≤ µ1 - µ2 ≤ 12.75