Question

Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 73 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 9.2 and a standard deviation of 2.5. What is the 98% 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).

Answer 1

Solution :

Given that,

Point estimate = sample mean =   =9.2

Population standard deviation =    = 2.5

Sample size = n =73

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z_/2 = Z_0.01 = 2.326 ( Using z table    )

Margin of error = E = Z_/2    * ( /n)

= 2.326 * ( 2.5/  73 )

= 0.7
At 98% confidence interval estimate of the population mean
is,

- E < < + E

9.2 -0.7 <   < 9.2 + 0.7

8.5 <   < 9.9

( 8.5 , 9.9 )