Question

A. Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 47 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 10.3 and a standard deviation of 1.2. What is the 99% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Enter your answers accurate to 4 decimal places.
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B. You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately σ=34.4σ=34.4. You would like to be 95% confident that your estimate is within 2.5 of the true population mean. How large of a sample size is required?

Use a z* value accurate to TWO places for this problem. (Not z = 2)

n =

C. You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately σ=45.6σ=45.6. You would like to be 90% confident that your estimate is within 0.2 of the true population mean. How large of a sample size is required?

As in the reading, in your calculations:
--Use z = 1.645 for a 90% confidence interval
--Use z = 2 for a 95% confidence interval
--Use z = 2.576 for a 99% confidence interval.


n =

Answer 1

A) At 99% confidence interval the critical value is z_0.005 = 2.576

The 99% confidence interval is

+/- z_0.005 * s/

= 10.3 +/- 2.576 * 1.2/

= 10.3 +/- 0.4509

= 9.8491, 10.7509

B) At 95% confidence interval the critical value is z_0.025 = 1.96

Margin of error = 2.5

or, z_0.025 * / = 2.5

or, 1.96 * 34.4/ = 2.5

or, n = (1.96 * 34.4/2.5)^2

or, n = 728

C) At 90% confidence interval the critical value is z_0.05 = 1.645

Margin of error = 0.2

or, z_0.05 * / = 0.2

or, 1.645 * 45.6/ = 0.2

or, n = (1.645 * 45.6/0.2)^2

or, n = 140671