Question

Question 2: The American Journal of Clinical Nutrition published the results of a study of the dietary intake of sugar by children.

In the trial a sample of 657 randomly selected American children between the ages of 5 and 12 was chosen, and for each child a seven day food diary was filled in. An analysis of the data showed a sample mean of 134.3 g of sugar per day with a standard deviation of 48.1 g. Using the results from this sample, estimate the 68%, 95%, and 99% confidence intervals for the mean daily sugar intake of all American children (i.e., the population mean).

Answer 1

Solution:-

Given that

n = 657

= 134.3

= 48.1

At 68% confidence level,

= 1 - 68% = 1 - 0.68 = 0.32

Margin of error = E =

E =

E = 1.87

At 68% confidence interval,

134.3 - 1.87 < < 134.3 + 1.87

132.43 < < 136.17

(132.43, 136.17)

The 68% confidence interval is (132.43, 136.17)

At 95% confidence level,

= 1 - 95% = 1 - 0.95 = 0.05

Margin of error = E =

E =

E = 3.68

A 95% confidence interval is,

134.3 - 3.68 < 134.3 + 3.68

130.62 < 137.98

(130.62, 137.98)

The 95% confidence interval is, (130.62, 137.98)

At 99% confidence interval,

= 1 - 99% = 1 - 0.99 = 0.01

Margin of error E =

E =

E = 4.83

A 99% confidence interval is,

134.3 - 4.83 < < 134.3 + 4.83

129.47 < < 139.13

(129.47, 139.13)

The 99% confidence interval is (129.47, 139.13)