Question

2. The file FastFood contains the amount that a sample of 15 customers spent for lunch ($) at a fast-food restaurants:

7.42 6.29 5.83 6.50 8.34 9.51 7.10 6.80 5.90 4.89 6.50 5.52 7.90 8.30 9.60

a. Construct 95% confidence interval estimate for the population mean amount spent for lunch ($) at a fast-food restaurant assuming a normal distribution.

b. Interpret the interval constructed in (a).

Answer 1

Solution: Here , the given things are

5n=15 and give the observations. alpha =0.05

First find mean and standard deviation from the given data using excel

From excel we can calculate the mean ( ) = 7.0933 and standard deviation (s)=1.406

a) 95% confidence interval is calculated as follows

=      *

Calculate the for =0.05 and n=15 in t table. Because sample size less than 30 and mean and standard deviation not given

   = = 2.145

  =      *

=7.0933 2.145*

=7.0933 0.7786

95 % confidence interval (6.3146 , 7.8719) stimate for the population mean amount spent for lunch ($) at a fast-food restaurant assuming a normal distribution.

The confidence interval is 6.3146 ? µ ? 7.8719

b) This means that with 95% confidence, you conclude that the mean amount spent for lunch at fast food restaurants is between $6.3146 and $7.8719