SOLUTION: A random sample of 49 shoppers showed that they spend an average of $23.45 per visit at the Saturday Mornings Bookstore. The standard deviation of the sample was $2.80. What is the

Algebra ->  Statistics  -> Confidence-intervals -> SOLUTION: A random sample of 49 shoppers showed that they spend an average of $23.45 per visit at the Saturday Mornings Bookstore. The standard deviation of the sample was $2.80. What is the      Log On


   

Question 1121781: A random sample of 49 shoppers showed that they spend an average of $23.45 per visit at the Saturday Mornings Bookstore. The standard deviation of the sample was $2.80. What is the 90% confidence interval of the true mean?
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
CI half-interval is t0.95, df=49*s/sqrt(n)
t=1.67
half-interval is 1.67*2.8/sqrt(49)=0.668
interval is $23.45+/-0.67
($22.78, $24.12)