Homework: Homework- Ch-10.1 Score: 0 of 5 pts Instructor-created question Save 7
ID: 3070048 • Letter: H
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Homework: Homework- Ch-10.1 Score: 0 of 5 pts Instructor-created question Save 7 of 11 (5 complete) HW Score: 33.33%, 16.67 of 50 pts Question Help A recent study found that children who watched a cartoon with food advertising ate, on average, 27.6 grams of crackers as compared to an average of 21.1 grams of crackers for children who watched a cartoon without food advertising. Suppose that there were 58 children in each group, and the sample standard deviation for those children who watched the food ad was 8.4 grams and the sample standard deviation for those children who did not watch the food ad was 8.2 grams. Complete parts (a) and (b) below a. Assuming that the population variances are equal and -0.05, is there evidence that the mean amount of crackers eaten was significantly higher for the children who watched food ads? Let population 1 be the weights of crackers eaten by children who watch food ads and let population 2 be the weights of crackers eaten by children who do not watch food ads. What are the correct null and alternative hypotheses? , : 1- 2 20 1 H1 -H20 What is the test statistic? STAT-(Round to two decimal places as needed.) What is the correct conclusion? A. Reject Ho . There is sufficient evidence that the mean amount of crackers eaten was significantly higher for the children who watched food ads B. Do not reject Ho. There is sufficient evidence that the mean amount of crackers eaten was significantly higher for the children who watched food ads. C. Do not reject Ho. There is insufficient evidence that the mean amount of crackers eaten was significantly higher for the children who watched food ads 0 D. Reject Ho. There is insufficient evidence that the mean amount of crackers eaten was significantly higher for the children who watched food ads Click to select your answer(s) and then click Check AnswerExplanation / Answer
a) Option D
H0: mu1 - mu2 <= 0
H1: mu1 - mu2 > 0
n1 = 58
n2 = 58
x1bar = 27.6000
x2bar = 21.1000
sigma1 (s1) = 8.4000
sigma2 (s2) = 8.2000
Sp = sqrt(((n1-1)*s1^2+(n2-1)*s2^2)/(n1+n2-2))
Sp = sqrt((57*8.4^2 + 57*8.2^2)/(58 + 58 - 2)
Sp = 8.3006
Test statistic,
t = (27.6 - 21.1)/(8.3006*sqrt(1/58 + 1/58))
t = 4.217
p-value = 0.00002
As p-value is less than the significance level of 0.05, we reject the null hypothesis.
Option A
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