Question

A student wanted to estimate the number of chocolate chips in a commercial brand of cookie. He sampled 100 cookies and found an average of 10.5 chips per cookie. If we assume the standard deviation is 8, what is a 99% confidence interval for the average number of chips per cookie?A. (8.4,12.6)B. (8.9,12.1)C. (5.3,10.7)

Answer 1

The 99% confidence interval is given by (8.4;12.6)

A. (8.4,12.6)

Explanation:

1) Notation and definitions

n=100 represent the sample size

`\bar X= 10.5` represent the sample mean

`\sigma=8` represent the population standard deviation assumed

m represent the margin of error

Confidence =99% or 0.99

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

2) Calculate the critical value zc

On this case we can to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by
`\alpha=1-0.99=0.01` and
`\alpha/2 =0.005`. The degrees of freedom are given by:

We can find the critical values in excel using the following formulas:

"=NORM.INV(0.005,0,1)" for
`z_(\alpha/2)=-2.58`

"=NORM.INV(1-0.005,0,1)" for
`z_(1-\alpha/2)=2.58`

The critical value
`zc=\pm 2.74`

3) Calculate the margin of error (m)

The margin of error for the sample mean is given by this formula:

`m=z_c (\sigma)/(√(n))`

`m=2.58 (8)/(√(100))=2.064`

4) Calculate the confidence interval

The interval for the mean is given by this formula:

`\bar X \pm z_(c) (\sigma)/(√(n))`

And calculating the limits we got:

`10.5 - 2.58 (8)/(√(100))=8.4`

`10.5 + 2.58 (8)/(√(100))=12.6`

The 99% confidence interval is given by (8.4;12.6)

A. (8.4,12.6)