Question

A machine that is programmed to package 3.90 pounds of cereal is being tested for its accuracy. In a sample of 64 cereal boxes, the sample mean filling weight is calculated as 3.95 pounds. The population standard deviation is known to be 0.14 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 average filling weight of all cereal packages.
The parameter of interest is the proportion filling weight of all cereal packages.

a-2. Compute its point estimate as well as the margin of error with 90% 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 90% 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 contains the target filling weight of 3.90.
No, since the confidence interval does not contain the target filling weight of 3.90.
Yes, since the confidence interval contains the target filling weight of 3.90.
Yes, since the confidence interval does not contain the target filling weight of 3.90.

c. How large a sample must we take if we want the margin of error to be at most 0.02 pound with 90% 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:

n = sample size = 64

Sample mean =

The population standard deviation is known to be 0.14 pound

thus

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

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

Part a-2. Compute its point estimate as well as the margin of error with 90% confidence.

Point estimate is sample mean:

Margin of Error is:

Z_c is z critical value for c = 90% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.90) / 2 = 1.90 / 2 = 0.9500

Look in z table for Area = 0.9500 or its closest area and find corresponding z value.

Area 0.9500 is in between 0.9495 and 0.9505 and both the area are at same distance from 0.9500

Thus we look for both area and find both z values

Thus Area 0.9495 corresponds to 1.64 and 0.9505 corresponds to 1.65

Thus average of both z values is : ( 1.64+1.65) / 2 = 1.645

Thus Z_c = 1.645

Thus

Part b-1. Calculate the 90% confidence interval.

Part 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 3.90.

Part c. How large a sample must we take if we want the margin of error to be at most 0.02 pound with 90% confidence?

E = 0.02