A food company is concerned about recent criticism of the sugar content of their
ID: 3150726 • Letter: A
Question
A food company is concerned about recent criticism of the sugar content of their children's cereals. The data show the sugar content (as a percentage of weight) of several national brands of children's and adults' cereals. Complete parts a through d. Full data set Children's cereal: 41, 57, 45.1, 44, 52.5, 48.5, 53.7, 40.8, 42.4, 44.3, 46.1, 44, 37.9, 59.4, 47.5, 50.2, 36.2, 57, 43.1, 31.7 Adults' cereal: 23.8, 27.7, 0.4, 9.4, 1.6, 24.9, 19.8, 14.5, 24.5, 7.2, 8.7, 14.3, 16.3, 10.9, 4.6, 18.5, 0.5, 2.5, 2.2, 6.3, 13.6, 1.9, 2.6, 3, 9.5, 1.2, 17.1, 6.6, 16.5, 13.6 (b) Check the conditions. Which of the following conditions are satisfied for the given data? Select all that apply. Randomization condition Nearly normal condition Independent groups assumption Find the 95% confidence interval for the positive difference in means. The 95% confidence interval is (grams, grams). (Round to two decimal places as needed.)Explanation / Answer
Calculating the means of each group,
X2 = 10.80666667
X1 = 67.945
Calculating the standard deviations of each group,
s2 = 8.229464864
s1 = 98.43073694
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n2 = sample size of group 1 = 30
n1 = sample size of group 2 = 20
Thus, df = n1 + n2 - 2 = 48
Also, sD = 22.06100561
For the 0.95 confidence level, then
alpha/2 = (1 - confidence level)/2 = 0.025
t(alpha/2) = 2.010634758
lower bound = [X1 - X2] - t(alpha/2) * sD = 101.494958
upper bound = [X1 - X2] + t(alpha/2) * sD = 12.78170867
Thus, the confidence interval is
( 12.78170867, 101.494958 ) [ANSWER]
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