Question

In: Statistics and Probability

Isabella wants to advertise how many chocolate chips are in each Big Chip cookie at her...

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 < ?

Solutions

Expert Solution

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


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