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.
Point estimate = sample mean = = 8.3
Population standard deviation =
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 = Z0.005 = 2.576 ( Using z table )
Margin of error = E = Z/2* ( /n)
= 2.576 * ( 2.2/ 53)
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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