Question

4. Different cereals are randomly selected and the sugar content (grams of sugar per gram of cereal) is obtained for each cereal. The results are given below:

0.03,   0.24, 0.3, 0.47, 0.43, 0.07, 0.47, 0.13, 0.44, 0.39, 0.48, 0.17, 0.13, 0.09, 0.45, 0.43, 0.21, 0.20, 0.26, 0.29, 0.27, 0.33, 0.21, 0.22, 0.19, 0.20,

Do the data support the claim that the average sugar content per gram of cereal does not exceed 0.27 grams? Use 95% confidence first and then use 99% confidence. If the final conclusion is different for the two confidence levels explain the difference. What is the 99% confidence interval for mean sugar content per gram of cereal?

Answer 1

95% CI
sample mean, xbar = 0.2744
sample standard deviation, s = 0.1385
sample size, n = 25
degrees of freedom, df = n - 1 = 24

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.064

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (0.2744 - 2.064 * 0.1385/sqrt(25) , 0.2744 + 2.064 * 0.1385/sqrt(25))
CI = (0.2172 , 0.3316)

99% CI,
Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 2.797

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (0.2744 - 2.797 * 0.1385/sqrt(25) , 0.2744 + 2.797 * 0.1385/sqrt(25))
CI = (0.1969 , 0.3519)

For both the CI there is not significant evidence to conclude that the average sugar content per gram of cereal does not exceed 0.27 grams