Question

Inference for one mean

Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 43 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 15.2 and a standard deviation of 1.3. What is the 95% confidence interval for the number of chocolate chips per cookie for Big Chip cookies?

Enter your answers accurate to one decimal place:

_< μ < _

Answer 1

Solution :

Given that,

= 15.2

s =1.3

n =43

Degrees of freedom = df = n - 1 = 43- 1 = 42

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,42 = 2.018 ( using student t table)

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

=2.018 * (1.3 / 43)

= 0.4

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

- E + E

15.2 - 0.4 15.2 + 0.4

14.8 15.6