Question

Potato chip bags are labeled as containing 9 ounces of potato chips. To determine the accuracy of this label, a simple random sample of 37 bags was taken. The sample mean was 8.73 ounces and the sample standard deviation was 0.18 ounces. Construct a 98 %
confidence interval for the population mean weight of bags of potato chips.

a) Give the critical value, ??.

b) Compute the standard error, ?? ̅.

c) Calculate the maximal margin of error, ?.

d) Give the 98 % confidence interval for the population mean weight of bags of potato chips using one complete sentence.

e) Based on the confidence interval you constructed, are the labels likely correct?

Answer 1

a)

sample mean, xbar = 8.73
sample standard deviation, s = 0.18
sample size, n = 37
degrees of freedom, df = n - 1 = 36

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, tc = t(α/2, df) = 2.434

tc= 2.434

b)

standard error = 0.18/sqrt(37) = 0.0296

c)

ME = tc * s/sqrt(n)
ME = 2.434 * 0.18/sqrt(37)
ME = 0.072

d)

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (8.73 - 2.434 * 0.18/sqrt(37) , 8.73 + 2.434 * 0.18/sqrt(37))
CI = (8.66 , 8.8)

e)

yes, because confidence interval does not contain 9