Question

Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 64 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 8.9 and a standard deviation of 1.5. What is the 90% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Enter your answers accurate to 4 decimal places.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 8.9

sample standard deviation = s = 1.5

sample size = n = 64

Degrees of freedom = df = n - 1 = 64 - 1 = 63

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

t/2,df = t_0.05,63 = 1.669

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

= 1.669 * (1.5 / 64)

Margin of error = E = 0.3129

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

- E < < + E

8.9 - 0.3129 < < 8.9 + 0.3129

( 8.5871 < < 9.2129 )