A sample of 30 boxes of Fruit Loops was chosen and each box was weighed. The mea

Question

A sample of 30 boxes of Fruit Loops was chosen and each box was weighed. The mean weight of the sample was 22 ounces with a sample deviation of 0.4 ounces. a. Construct a 90% confidence interval for the mean weight of the cereal boxes. b. Construct a 98% confidence interval for the mean weight of the cereal boxes. c. Explain the difference between the answers for (a) and (b). d. Imagine that your sample was 50. Construct a 90% confidence interval. e. Explain the difference between the answers for (a) and (d). f. What do the 90% and 98% confidence levels mean? Be specific. Hint: It does not mean that we are 90% or 98% sure that the interval contains the actual population mean. That is close, but is not precise enough.

Explanation / Answer

mean = 22 , s = 0.4 , n =30
a) z value at 90% CI = 1.645

CI = mean +/- z * (s/sqrt(n))
= 22 +/- 1.645 * ( 0.4 /sqrt(30))
= 21.879 to 22.120

b) z value at 98% CI = 2.33

CI = mean +/- z * (s/sqrt(n))
= 22 +/- 2.33 * ( 0.4 /sqrt(30))
= 21.829 to 22.170

c)

for answers a) and b) confidence interval are differnt so answers ar differnt.

d)
n =50
z value at 90% CI = 1.645

CI = mean +/- z * (s/sqrt(n))
= 22 +/- 1.645 * ( 0.4 /sqrt(50))
= 21.868 to 22.132

e) for answer a) sample size is 30 and for answer d) sample size is 50 .

f) The selection of a confidence level for an interval determines the probability that the confidence interval produced will contain the true parameter value

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