16. Use the T-Interval Procedure: Chips Ahoy! There was a random sample of 42 18

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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