A machine that is programmed to package 1.78 pounds of cereal is being tested fo

Question

A machine that is programmed to package 1.78 pounds of cereal is being tested for its accuracy. In a sample of 25 cereal boxes, the sample mean filling weight is calculated as 1.78 pounds. The population standard deviation is known to be 0.04 pound.

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. Compute its point estimate as well as the margin of error with 99% confidence. (Round intermediate calculations to 4 decimal places. Round "z-value" to 3 decimal places and final answers to 2 decimal places.) Point estimate Margin of error b-1. Calculate the 99% confidence interval. (Use rounded margin of error. Round your final answers to 2 decimal places.) Confidence interval to b-2.

Can we conclude that the packaging machine is operating improperly?

-Yes, since the confidence interval contains the target filling weight of 1.78. Yes, since the confidence interval does not contain the target filling weight of 1.78.

-No, since the confidence interval contains the target filling weight of 1.78. No, since the confidence interval does not contain the target filling weight of 1.78. c. How large a sample must we take if we want the margin of error to be at most 0.01 pound with 99% confidence? (Round intermediate calculations to 4 decimal places. Round "z-value" to 3 decimal places and round up your final answer to the next whole number.) Sample size

Explanation / Answer

TRADITIONAL METHOD

given that,

standard deviation, =0.04

sample mean, x =1.78

population size (n)=25

I.

stanadard error = sd/ sqrt(n)

where,

sd = population standard deviation

n = population size

stanadard error = ( 0.04/ sqrt ( 25) )

= 0.008

II.

margin of error = Z a/2 * (stanadard error)

where,

Za/2 = Z-table value

level of significance, = 0.01

from standard normal table, two tailed z /2 =2.576

since our test is two-tailed

value of z table is 2.576

margin of error = 2.576 * 0.008

= 0.021

III.

CI = x ± margin of error

confidence interval = [ 1.78 ± 0.021 ]

= [ 1.759,1.801 ]

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DIRECT METHOD

given that,

standard deviation, =0.04

sample mean, x =1.78

population size (n)=25

level of significance, = 0.01

from standard normal table, two tailed z /2 =2.576

since our test is two-tailed

value of z table is 2.576

we use CI = x ± Z a/2 * (sd/ Sqrt(n))

where,

x = mean

sd = standard deviation

a = 1 - (confidence level/100)

Za/2 = Z-table value

CI = confidence interval

confidence interval = [ 1.78 ± Z a/2 ( 0.04/ Sqrt ( 25) ) ]

= [ 1.78 - 2.576 * (0.008) , 1.78 + 2.576 * (0.008) ]

= [ 1.759,1.801 ]

-----------------------------------------------------------------------------------------------

interpretations:

1. we are 99% sure that the interval [1.759 , 1.801 ] contains the true population mean

2. if a large number of samples are collected, and a confidence interval is created

for each sample, 99% of these intervals will contains the true population mean

[ANSWERS]

parameter is, the average filling weight of all cereal packages

best point of estimate = mean = 1.78

margin of error = 0.021

-Yes, since the confidence interval contains the target filling weight of 1.78

c.

Compute Sample Size  

n = (Z a/2 * S.D / ME ) ^2

Z/2 at 0.01% LOS is = 2.576 ( From Standard Normal Table )

Standard Deviation ( S.D) = 0.04

ME =0.01

n = ( 2.576*0.04/0.01) ^2

= (0.103/0.01 ) ^2

= 106.1724 ~ 107

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