Question

Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 45 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 9.1 and a standard deviation of 3.8. What is the 80% confidence interval for the number of chocolate chips per cookie for Big Chip cookies?

Enter your answers accurate to one decimal place.

< μμ >

Answer 1

Solution :

Given that,

Point estimate = sample mean =     = 9.1

Population standard deviation =     = 3.8

Sample size n =45

At 80% confidence level the z is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 =0.10

Z / 2   = Z0.10 = 1.28 ( Using z table )

Margin of error = E =   Z/2    * ( /n)
= 1.28 * ( 3.8 / 45 )

= 0.73
At 80% confidence interval estimate of the population mean
is,

- E <   < + E

9.1 - 0.73 <   <9.1 + 0.73

8.4 <   < 9.8