Question

Noah wants to advertise how many chocolate chips are in each Big Chip cookie at his bakery. He randomly selects a sample of 43 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 19.5 and a standard deviation of 2.1. 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.

Answer 1

Soiution

Given that,

Point estimate = sample mean = = 19.5

Population standard deviation =    = 2.1

Sample size = n = 43

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 = 0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96

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

= 1.96 * (2.1 /  43 )

= 0.628

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

- E < < + E

19.5 - 0.628 <   < 19.5 + 0.628

18.872 <   < 20.128

( 18.872 , 20.128 )