16. Use the T-Interval Procedure: Chips Ahoy! There was a random sample of 42 18
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Question
16. Use the T-Interval Procedure: Chips Ahoy! There was a random sample of 42 18-ounce bags of Chips Ahoy! cookies that found that the mean was 1261.6 chips per bag with a standard deviation of 117.6 chips per bag. a. Determine a 95% confidence interval for the mean number of chips per bag for all 18-ounce bags of Chips Ahoy! Cookies, and interpret your result in words. b. Can you conclude that the average 18-ounce bag of Chips Ahoy! cookies contains at least 1,000 chocolate chips? Explain your answer.Explanation / Answer
a)
Standard error of the mean = SEM = S/N = 18.146
t(, N-1) = 2.02
Confidence interval = m +/- (t(, N-1)*SEM)
Mean = 1261.6
Lower bound: 1224.95
Upper bound: 1298.25
b)
For 99.9 confidence interval
Standard error of the mean = SEM = S/N = 18.146
t(, N-1) = 3.544
Confidence interval = m +/- (t(, N-1)*SEM)
Mean = 1261.6
Lower bound: 1197.29
Upper bound: 1325.91
there are chances,which are less than 0.01 that the a bag might contain less than 1000 chocolate chips
as this a normal distibution. the event is highly unprobable but its possible
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