Question

A machine that is programmed to package 5.60 pounds of cereal is being tested for its accuracy. In a sample of 100 cereal boxes, the sample mean filling weight is calculated as 5.69 pounds. The population standard deviation is known to be 0.09 pound. [You may find it useful to reference the z table.]

a-1. Identify the relevant parameter of interest for these quantitative data.

• The parameter of interest is the proportion filling weight of all cereal packages.

• The parameter of interest is the average filling weight of all cereal packages.

a-2. Compute its point estimate as well as the margin of error with 99% confidence. (Round intermediate calculations to at least 4 decimal places. Round "z" value to 3 decimal places and final answers to 2 decimal places.)

b-1. Calculate the 99% confidence interval. (Use rounded margin of error. Round your final answers to 2 decimal places.)

b-2. Can we conclude that the packaging machine is operating improperly?

• No, since the confidence interval does not contain the target filling weight of 5.60.

• Yes, since the confidence interval contains the target filling weight of 5.60.

• No, since the confidence interval contains the target filling weight of 5.60.

• Yes, since the confidence interval does not contain the target filling weight of 5.60.

c. How large a sample must we take if we want the margin of error to be at most 0.01 pound with 99% confidence? (Round intermediate calculations to at least 4 decimal places. Round "z" value to 3 decimal places and round up your final answer to the next whole number.)

Answer 1

a-1. Identify the relevant parameter of interest for these quantitative data.

• Answer: The parameter of interest is the average filling weight of all cereal packages.

a-2. Compute its point estimate as well as the margin of error with 99% confidence.

Point estimate = Xbar = 5.69

Margin of error = E = Z*σ/sqrt(n)

We are given

Confidence level = 99%

Critical Z value = 2.576

(by using z-table)

Sample size = n = 100

Population standard deviation = σ = 0.09

Margin of error = E = 2.576*0.09/sqrt(100) = 0.023184

Margin of error = E = 0.02

b-1. Calculate the 99% confidence interval.

Lower limit = Xbar – E = 5.69 – 0.02 = 5.67

Upper limit = Xbar + E = 5.69 + 0.02 = 5.71

b-2. Can we conclude that the packaging machine is operating improperly?

Yes, since the confidence interval does not contain the target filling weight of 5.60.

c. How large a sample must we take if we want the margin of error to be at most 0.01 pound with 99% confidence?

n = (Z*σ/E)^2

E = 0.01

Z = 2.576

σ = 0.09

n = (2.576*0.09/0.01)^2

n = 537.4979

n = 538

Required sample size = lk538