Homework Sourse13 A machine that is programmed to package 368 pounds of cer
(13)
A machine that is programmed to package 3.68 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. It can be assumed that filling weights are normally distributed with a population standard deviation of 0.08 pound. Use Table 1 - See below at the very bottom past all questions.
Identify the relevant parameter of interest for these quantitative data.
Compute the point estimate as well as the margin of error with 95% confidence. (Round intermediate calculations to 4 decimal places. Round \"z\" value and final answers to 2 decimal places.)
Calculate the 95% confidence interval. (Use rounded margin of error. Round your answers to 2 decimal places.)
Can we conclude that the packaging machine is operating improperly?
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 4 decimal places. Round \"z\" value to 2 decimal places and round up your final answer to the next whole number.)
Sample size
| A machine that is programmed to package 3.68 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. It can be assumed that filling weights are normally distributed with a population standard deviation of 0.08 pound. Use Table 1 - See below at the very bottom past all questions. |
Solution
(a-1)The parameter of interest is the average filling weight of all cereal packages.
------------------------------------------------------------------------------------------------------------------------
(a-2) Point estimate =3.68
Given a=0.05, Z(0.025) =1.96 (from standard normal table)
So Margin of error =Z*s/vn
=1.96*0.08/sqrt(81)
=0.02
------------------------------------------------------------------------------------------------------------------------
(b-1) The lower bound is
xbar -Z*s/vn = 3.68 -0.02 =3.66
The upper bound is
xbar +Z*s/vn = 3.68 +0.02 =3.7
------------------------------------------------------------------------------------------------------------------------
(b-2)Yes, since the confidence interval contains the target filling weight of 3.68.
------------------------------------------------------------------------------------------------------------------------
(c) n=(Z*s/E)^2
=(1.96*0.08/0.01)^2
=245.8624
Take n=246
