Question

A machine that is programmed to package 1.20 pounds of cereal is being tested for its accuracy In a sample of 36 cereal boxes, the sample mean filling weight is calculated as 1.22 pounds. The population standard deviation is known to be 0.06 pound. Identify the relevant parameter of interest for these quantitative data. The parameter of interest is the proportion filling weight for all cereal packages. The parameter of interest is the average filling weight for all cereal packages.

A-1. Identify the relevant parameter of interest for these quantitative data.
A) The parameter of interest is the average filling weight of all cereal packages.
B) 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 95% confidence. (Round intermediate calculations to 4 decimal places.
B-1. Calculate the 95% confidence interval.
B-2. Can we conclude that the packaging machine is operating improperly?
A) Yes, since the confidence interval contains the target filling weight of 1.20.
B) Yes, since the confidence interval does not contain the target filling weight of 1.20.
C) No, since the confidence interval contains the target filling weight of 1.20.
D) No, since the confidence interval does not contain the target filling weight of 1.20.
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?

Answer 1

A) The parameter of interest is the average filling weight of all cereal packages. ;

Point estimate = 1.22

Margin of Error = 0.0196

Confidence interval = (1.2004, 1.2396)

C) No, since the confidence interval contains the target filling weight of 1.20.

Explanation:

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

The point estimate = sample mean = 1.22

The margin of error with 95% confidence :

Zcritical * σ/sqrt(n)

Zcritical at 95% confidence = 1.96

Margin of Error = 1.96 * 0.06/sqrt(36)

Margin of Error : 1.96 * 0.01

Margin of Error = 0.0196

Confidence interval :

xbar ± margin of error

1.22 ± 0.0196

Lower boundary = 1.22 - 0.0196 = 1.2004

Upper boundary = 1.22 + 0.0196 = 1.2396

(1.2004, 1.2396)

C) No, since the confidence interval contains the target filling weight of 1.20.

Since, the internal contains 1.20, then we can conclude that it is filling properly.