Question

Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 53 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 8.3 and a standard deviation of 2.2. What is the 99% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Enter your answers accurate to one decimal place.

Answer #1

Solution :

Given that,

Point estimate = sample mean =
= 8.3

Population standard deviation =
= 2.2

Sample size = n =53

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_{/2}
= Z_{0.005} = 2.576 ( Using z table )

Margin of error = E = Z_{/2}*
(
/n)

= 2.576 * ( 2.2/ 53)

= 0.8

At 99% confidence interval estimate of the population mean is,

- E < < + E

8.3-0.8 < <8.3+0.8

7.5< < 9.1

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