Use the confidence level and Sample data to find a confidence interval for estimating the population mean. Round your answer to the same number of decimal places as the sample mean.

Of the packages received by a Parcel Service, 37 packages were randomly selected. The sample has a mean weight of 17.0 pounds and a standard deviation of 3.3 lb. What is the 95% confidence interval for the true mean weight, μ, of all packages received by the Parcel Service?

A) 15.7 lb<μ<18.3 lb

B) 15.9 lb<μ<18.1 lb

C) 15.6 lb<μ<18.4 lb

D) 16.1 lb<μ<17.9 lb

Question

Marlee Horne

Mathematics

19 September, 23:33

Use the confidence level and Sample data to find a confidence interval for estimating the population mean. Round your answer to the same number of decimal places as the sample mean.

Of the packages received by a Parcel Service, 37 packages were randomly selected. The sample has a mean weight of 17.0 pounds and a standard deviation of 3.3 lb. What is the 95% confidence interval for the true mean weight, μ, of all packages received by the Parcel Service?

A) 15.7 lb<μ<18.3 lb

B) 15.9 lb<μ<18.1 lb

C) 15.6 lb<μ<18.4 lb

D) 16.1 lb<μ<17.9 lb

+2

Answers (1)

Know the Answer?

Landon Henson · Answer 1

B) 15.9 lb < μ < 18.1 lb

Step-by-step explanation:

Confidence interval = mean ± margin of error

CI = μ ± ME

The mean is μ = 17.0.

Margin of error = critical value * standard error

ME = CV * SE

n > 30, so we can approximate CV with a normal distribution. At P = 95%, CV = 1.96.

SE = σ / √n

SE = 3.3 / √37

SE = 0.54

So the margin of error is:

ME = 1.96 * 0.54

ME = 1.1

So the confidence interval is:

CI = 17.0 ± 1.1

CI = (15.9, 18.1)