Question

According to the recent survey conducted by Sunny Travel Agency, a married

couple spends an average of $500 per day while on vacation. Suppose a sample of 64 couples

vacationing at the Resort XYZ resulted in a sample mean of $490 per day and a sample standard

deviation of $100. Develop a 95% confidence interval estimate of the mean amount spent per day by

a couple of couples vacationing at the Resort XYZ.

Answer 1

Solution:

Note that, Population standard deviation() is unknown. So we use t distribution.

Our aim is to construct 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025

Also, d.f = n - 1 = 64 - 1 = 63

    =    =  _0.025,63 = 1.998

( use t table or t calculator to find this value..)

The margin of error is given by

E =  /2,d.f. * ( / n )

= 1.998 * (100 / 64 )

= 24.975

Now , confidence interval for mean() is given by:

( - E ) <   <  ( + E)

(490 - 24.975)   <   <  (490 + 24.975)

465.025 <   <  514.975

Required 95% confidence interval is (465.025 , 514.975)