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The price for 2 tickets with online service charges, large popcorn and 2 medium soft drinks at a sample of 6 theater chains: $36.15, $31.00, $35.05, $40.25, $33.75, $43.00

a) Construct a 95% confidence interval estimate for the population mean the price for 2 tickets with online service charges, large popcorn and 2 medium soft drinks, assuming a normal distribution.

mean of sample = 36.53
n = 6
standard deviation = 19.26866 ( square root of ((36.15 - 36.53)^2 + (31 - 36.53)^2 + (35.05 - 36.53)^2 + (40.25 - 36.53)^2 + (33.75 - 36.53)^2 + (43 - 36.53)^2) / 6-1)
standard error of sample mean = 7.8664
critical value of t = 0.025
degrees of freedom = 5

95% C.I. = 36.53 +/- (critical value of t = 0.025)(7.8664)
95% C.I. = 36.53 +/- (2.571)(7.8664)
95% C.I. = 36.53 +/- 20.2245

My answer is wrong and the interval estimate is between 31.9267 and 41.1399. I think my error is solving for the standard deviation. How do you solve for it? Please provide a step-by-step solution. Thank you.

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