Question

# A sample of 15 small bags of Skittles was selected. The mean weight of the sample...

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?

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

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