Question

A)

Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 52 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 17.9 and a standard deviation of 2.5. What is the 90% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Enter your answers accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).
< μ <

Answer 1

Solution :

Given that,

n = 52

= 17.9

s = 2.5

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 = 52 - 1 = 51

    =    =  _0.05,51 = 1.675

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

The margin of error is given by

E =  /2,d.f. * ( / n)

= 1.675* ( 2.5/ 52 )

= 0.581

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

( - E ) <   <  ( + E)

( 17.9 - 0.581 )   <   <  ( 17.9 + 0.581 )

17.3 <   < 18.5

Required 90% confidence interval is ( 17.3 , 18.5 )