Question

Use technology to construct the​ 80%, 90%, and​ 95% confidence intervals for the population mean. Interpret the results.

A random sample of

3030

sandwiches from a fast food restaurant has a mean of

1048.51048.5

milligrams of sodium and a standard deviation of

340.6340.6

milligrams of sodium.

The​ 80% confidence interval is

left parenthesis nothing comma nothing right parenthesis,.

Answer 1

mean = x' = 1048.5

standard deviation = s = 340.6

sample size = n = 30

a) 80% confidence interval

=> 80% confidence interval z score = 1.28

Lower limit = x' - z*s/sqrt(n) = 1048.5 - 1.28*340.6/sqrt(30) = 968.90

Upper limit = x' + z*s/sqrt(n) = 1048.5 + 1.28*340.6/sqrt(30) = 1128.10

Thus, 80% confidence interval is (968.90, 1128.10)

b) 90% confidence interval

=> 90% confidence interval z score = 1.64

Lower limit = x' - z*s/sqrt(n) = 1048.5 - 1.64*340.6/sqrt(30) = 946.52

Upper limit = x' + z*s/sqrt(n) = 1048.5 + 1.64*340.6/sqrt(30) = 1150.48

Thus, 90% confidence interval is (946.52, 1150.48)

c) 95% confidence interval

=> 95% confidence interval z score = 1.96

Lower limit = x' - z*s/sqrt(n) = 1048.5 - 1.96*340.6/sqrt(30) = 926.62

Upper limit = x' + z*s/sqrt(n) = 1048.5 + 1.96*340.6/sqrt(30) = 1170.38

Thus, 95% confidence interval is (926.62, 1170.38)