Question

The owner of a local phone store wanted to determine how much customers are willing to spend on the purchase of a new phone. In a random sample of 8 phones purchased that day, the sample mean was $383.299 and the standard deviation was $24.0154. Calculate a 90% confidence interval to estimate the average price customers are willing to pay per phone. Question 6 options: 1) ( -367.213 , 399.385 ) 2) ( 374.808 , 391.79 ) 3) ( 367.213 , 399.385 ) 4) ( 367.506 , 399.092 ) 5) ( 381.404 , 385.194 )

Answer 1

Solution:

Given that,

   = 383.299 ....... Sample mean

s = 24.0154 ......Sample standard deviation

n = 8 ....... Sample size

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 = 8 - 1 = 7

    =    =  0.05, 7   = 1.895

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

The margin of error is given by

E =  /2,d.f. * ( / n)

= 1.895 * (24.0154 / 8 )

= 16.086

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

( - E ) <   <  ( + E)

(383.299 - 16.086)   <   <  (383.299  + 16.086)

367.213 <   < 399.385

Required interval is (367.213 , 399.385)