Question

A machine that is programmed to package 3.60 pounds of cereal is being tested for its accuracy. In a sample of 81 cereal boxes, the sample mean filling weight is calculated as 3.68 pounds. The population standard deviation is known to be 0.08 pound. [You may find it useful to reference the z table.]

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

Which one?

• 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 95% 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 95% 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?

• Yes, since the confidence interval does not contain the target filling weight of 3.60.

• No, since the confidence interval does not contain the target filling weight of 3.60.

• Yes, since the confidence interval contains the target filling weight of 3.60.

• No, since the confidence interval contains the target filling weight of 3.60.

c. How large a sample must we take if we want the margin of error to be at most 0.01 pound with 95% 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

Solution :

Given that,

a-1)The parameter of interest is the average filling weight of all cereal packages.

a-2) Point estimate = sample mean = = 3.68

Population standard deviation =    = 0.08

Sample size = n = 81

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_0.025 = 1.96

Margin of error = E = Z/2 * ( /n)

= 1.96 * ( 0.08 /  81 )

= 0.02

b-1) At 95% confidence interval estimate of the population mean is,

  ± E

3.68 ± 0.02   

( 3.66, 3.70 )  

b-2) No, since the confidence interval does not contain the target filling weight of 3.60.

c) margin of error = E = 0.01

sample size = n = [Z_/2* / E] ²

n = [1.96 * 0.08 / 0.01]²

n = 245.86

Sample size = n = 246