A machine that is programmed to package 725 pounds of cereal

A machine that is programmed to package 7.25 pounds of cereal is being tested for its accuracy. In a sample of 25 cereal boxes, the sample mean filling weight is calculated as 7.25 pounds. It can be assumed that filling weights are normally distributed with a population standard deviation of 0.05 pound. Use Table 1.

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.)

Solution

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

as here we are dealing with the weights. we are checking the accuracy of the machine whether it can package 7.25 pounds of cereal.

a-2). the point estimate is the sample mean=7.25 pound

here the margin of error is to be calculated with 95% confidence. so here alpha=0.05 hence alpha/2=0.025

hence z-score is that value of a standard normal distribution under which the probability is 1-alpha/2=0.975

using z-table we find the value as 1.959964

rounding it of to two decimal we get 1.96

hence the margin of error is 0.05/sqrt(25)*1.96=0.0196=0.02 [rounded to two decimal]

b-1) the 95% confidence interval is sample mean-margin or error to sample mean+margin of error

or, 7.25-0.02 to 7.25+0.02   or, 7.23 to 7.27   [answer]

b-2) answer is no because- No, since the confidence interval contains the target filling weight of 7.25.

c) let the sample size be n

the margin of error can be at most 0.01 pound with 95% confidence.

it means 0.05/sqrt(n)*1.96<0.01

or, 0.098/sqrt(n)<0.01 or, 1/sqrt(n)<0.102040816 or, sqrt(n)>9.8 or n>96.04

so we must take the sample to be as large as 97 [answer]