Question

Emma 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 8.3 and a standard deviation of 2.3. What is the 99% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Assume the data is from a normally distributed population. Round answers to 3 decimal places where possible. < μ <

Answer 1

The 99% confidence interval is

7.558 - 9.042

Step-by-step explanation:

The formula for the confidence interval is:

`Confidence\text{ }interval=\bar{X}\pm(\sigma)/(√(n))`

Where:

X is the mean

σ is the standard deviation

z is the z-score for the confidence interval

n is the sample size.

Also, the interval has:

`Upper\text{ }limit=\bar{X}+(\sigma)/(√(n))`
`Lower\text{ }limit=\bar{X}-(\sigma)/(√(n))`

Then, in this case,

The sample size is n = 64

The mean is X = 8.3

The z-score for a 99% confidence interval is z = 2.58

The standard deviation is σ = 2.3

Then:

`Lower\text{ }limit=8.3-2.58\cdot(2.3)/(√(64))=9.04175`
`Upper\text{ }limit=8.3+2.58\cdot(2.3)/(√(64))=7.55825`

Thus, the confidence interval, rounded to 3 decimals is

7.558 - 9.042