Question

Collette is self-employed, selling cosmetics at home parties. She wants to estimate the average amount a client spends per year at these parties. A random sample of 16 receipts had a mean of x = $340.70 with a standard deviation of s = $60.15. Find a 90% confidence interval for the mean amount μ spent by all clients. Assume x has a distribution that is approximately normal.

a.) What is the critical value of the t variable for the given confidence level?

b.) What is the maximal margin of error? $____

c.) The lower endpoint of the 90% confidence interval for μ is $____

d.) The upper endpoint of the 90% confidence interval μ is $____

*Round all your answers to the nearest hundredth

Answer 1

Solution =

Given that,

n = 16

= 340.70   

s = 60.15

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

Our aim is to construct 90% confidence interval.   

c = 0.90

= 1- c = 1- 0.90 = 0.10

  /2 = 0.10 2 = 0.05

Also, d.f = n - 1 = 15  

    =    =  0.05,15 = 1.75

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

The margin of error is given by

E =  /2,d.f. * ( / n)

= 1.753 * ( 60.15/ 16 )

= 26.36

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

( - E ) <   <  ( + E)

( 340.70 - 26.361 )   <   <  ( 340.70 + 26.361 )

314.34 <   < 367.06

Required 90% confidence interval is ( 314.34 , 367.06)

Answer :

a.) the critical value = 1.75

b.) What is the maximal margin of error? $ 26.36

c.) The lower endpoint of the 90% confidence interval for μ is $ 314.34

d.) The upper endpoint of the 90% confidence interval μ is $ 367.06