As quality control chief for Blotto Brewery, Inc., you wish to determine the ave
ID: 3355193 • Letter: A
Question
As quality control chief for Blotto Brewery, Inc., you wish to determine the average calorie content of bottles of Blotto Lite. You randomly select 49 bottles off the production line. They have an average calorie count of 100. The sample standard deviation is 4.9 calories.
a) Compute by hand a 95% confidence interval based on this sample. (Use Minitab to get the t multiplier.) Confirm your confidence interval result with Minitab and give an interpretation of this interval.
b) If we compute a 90% confidence interval based on this sample, will the margin of error be larger or smaller than the margin of error for our 95% interval? Briefly explain your answer.
Explanation / Answer
Answer to the question is as follows:
a. The mean calorie content is 100 calorie
The sample deviation is 4.9 calories
The t value for df = n-1 = 48 is 1.677
So, 95% CI is given by:
mean + /- t*s/sqrt(n)
= 100 +/- 1.677*4.9/sqrt(49) = 98.82 to 101.174
b. The 90% confidence interval requires less than sample size than the 95% CI. The CI is given by :
t is lower for 90% CI, so MOE is also less, thereby CI for 90% is also less than that of 95%
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