Question

Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 58 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 12.5 and a standard deviation of 1.7. What is the 90% 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,

= 12.5

s = 1.7

n = 58

Degrees of freedom = df = n - 1 = 58 - 1 = 57

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,57 =1.672

Margin of error = E = t_/2,df * (s /n)

= 1.672 * (1.7 / 58)

= 0.4

Margin of error = 0.50

The 90% confidence interval estimate of the population mean is,

- E <  < + E

12.5 - 0.4 < < 12.5 + 0.4

12.1 < < 12.9

(12.1, 12.9 )