Question

A machine that is programmed to package 6.20 pounds of cereal is being tested for its accuracy. In a sample of 36 cereal boxes, the sample mean filling weight is calculated as 6.29 pounds. The population standard deviation is known to be 0.13 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 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.)

Answer 1

Solution:-

1) Yes.

2)
Point estimate = sample mean = = 6.29  

Population standard deviation = = 0.13

Sample size = n =36

At 99% confidence level

= 1-0.99% =1-0.99 =0.01

/2 =0.01/ 2= 0.005

Z/2 = Z0.005 = 2.576

Z/2 = 2.576

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

= 2.576 * (0.13 /  36 )

= 0.0558 = 0.06

b)At 99 % confidence interval estimate of the population mean is,

- E < < + E

6.29 - 0.06 <   < 6.29 + 0.06

6.23<   < 6.35

(6.23 ,6.35 )