Question

Isabella wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 60 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 14.3 and a standard deviation of 1.4. What is the 95% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Assume the data is from a normally distributed population. Round answers to 3 decimal places where possible.

? < μμ < ?

Answer 1

Solution :

Given that,

Point estimate = sample mean =     = 14.3

Population standard deviation =     = 1.4

Sample size n =60

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z / 2   = Z0.025 = 1.96   ( Using z table )

Margin of error = E =   Z / 2     * ( /n)
= 1.96* ( 1.4/ 60 )

= 0.354
At 95% confidence interval estimate of the population mean
is,

- E <   < + E

14.3 - 0.354 <   < 14.3+ 0.354

13.946 <   < 14.654