Question

A sample of 15 small bags of Skittles was selected. The mean weight of the sample of bags was 2 ounces and the standard deviation was 0.12 ounces. The population standard deviation is known to be 0.1 ounces. (Assume the population distribution of bag weights is normal)

a) Construct a 95% confidence interval estimating the true mean weight of the candy bags. (4 decimal places

b) What is the margin of error for this confidence interval?

c) Interpret your confidence interval.

Answer 1

a) we have n = 15 , xbar = 2 ,   = 0.1

The 95% confidence interval estimating the true mean weight of the candy bags is

xbar - E <   < xbar + E

Where E = Z​​​​​​a/2*( /√n)

For a = 0.05 , Z​​​​​​a/2 = Z​​​​​​0.025 = 1.96

E = 1.96*(0.1/√15) = 0.0506

2 - 0.0506 < < 2 + 0.0506

1.9494 < < 2.0506

This is 95% confidence interval

b) margin of error E = 0.0506

c) interpretation of confidence interval

We are 95% confident that true mean weight of the candy bags is lies between 1.9494 and 2.0506