Kalamazoo College
Determine the confidence interval for the mean, given the following information:
x̄=$220
s=56
Confidence Level= 95% Confidence
Standard Error = s/√n = 56/√13 = 15.53Degrees of Freedom = n-1 = 13-1 = 12t-table value = 2.179x̄ ± t(s.e.)220 + (2.179)(15.53) = 220 + 33.83987 = 253.83987220 – (2.179)(15.53) = 220 – 33.83987 = 186.16013ANSWERInterval: 186.16-253.84
Standard Error = s/√n = 56/√13 = 15.53
ReplyDeleteDegrees of Freedom = n-1 = 13-1 = 12
t-table value = 2.179
x̄ ± t(s.e.)
220 + (2.179)(15.53) = 220 + 33.83987 = 253.83987
220 – (2.179)(15.53) = 220 – 33.83987 = 186.16013
ANSWER
Interval: 186.16-253.84