A machine that is programmed to package 5.26 pounds of cereal is being tested fo

Question

A machine that is programmed to package 5.26 pounds of cereal is being tested for its accuracy. In a sample of 100 cereal boxes, the sample mean filling weight is calculated as 5.26 pounds. It can be assumed that filling weights are normally distributed with a population standard deviation of 0.08 pound. Use Table 1. a-1. Identify the relevant parameter of interest for these quantitative data. The parameter of interest is the proportion filling weight of all cereal packages. The parameter of interest is the average filling weight of all cereal packages. a-2. 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.) Point estimate Margin of error b-1. Calculate the 95% confidence interval. (Use rounded margin of error. Round your 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 5.26. Yes, since the confidence interval does not contain the target filling weight of 5.26. No, since the confidence interval contains the target filling weight of 5.26. No, since the confidence interval does not contain the target filling weight of 5.26. 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? (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.) Sample size

Explanation / Answer

a-1.

We are interested in the mean (average),

OPTION B: The parameter of interest is the average filling weight of all cereal packages. [ANSWER]

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a-2.

The point estimate is the mean,

X = point estimate = 5.26 [ANSWER]

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Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    0.08          
n = sample size =    100          
              
Thus,              
Margin of Error E =    0.015679712   [ANSWER, MARGIN OF ERROR]      

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b-1.

Here,

Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    5.26          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    0.08          
n = sample size =    100          
              
Thus,              

Lower bound =    5.244320288          
Upper bound =    5.275679712          
              
Thus, the confidence interval is              
              
(   5.244320288   ,   5.275679712   ) [ANSWER]

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b-2.

As 5.26 is inside the interval,

OPTION C: No, since the confidence interval contains the target filling weight of 5.26. [ANSWER]

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c.

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.025  
      
Using a table/technology,      
      
z(alpha/2) =    1.959963985  
      
Also,      
      
s = sample standard deviation =    0.08  
E = margin of error =    0.01  
      
Thus,      
      
n =    245.8533645  
      
Rounding up,      
      
n =    246   [ANSWER]

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