Question

Problem (20 points, 5 points for each part . The modules of rupture (MOR) for a particular grade of pencil lead is known to have a standard deviation of 250 psi. (1) A random sample of 16 pencil leads yielded a sample mean of 6490. Construct a 95% confidence interval (two-sided) for the true mean MOR. (2) A random sample of25 pencil leads yielded a sample mean of6520. Construct a 98% upper (3) Find the sample size required to estimate the true mean MOR to within 100 using a 99% (4) What did you assume to do these analyses? How do you check these assumptions? (one -sided) confidence interval for the true mean MOR. confidence interval.

Answer 1

1)

Level of Significance ,    α =    0.05
population std dev ,    σ =    250
Sample Size ,   n =    16
Sample Mean,    x̅ =   6490

z value=   z α/2=   1.9600   [Excel formula =NORMSINV(α/2) ]
          
Standard Error , SE =   σ/√n =   62.5000  
margin of error ,   E=Z*SE =   122.498  

confidence interval is           
Interval Lower Limit=   x̅ - E =    6367.50  
Interval Upper Limit=   x̅ + E =    6612.50

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2)

Level of Significance ,    α =    0.02
population std dev ,    σ =    250
Sample Size ,   n =    25
Sample Mean,    x̅ =   6520

z value=      2.0537   [Excel formula =NORMSINV(α) ]

Standard Error , SE =   σ/√n =   50.0000
margin of error ,   E=Z*SE =   102.687

upper one sided confidence interval is      

Interval Upper Limit=   x̅ + E =    6622.69

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3)

Standard Deviation ,   σ =    250  
sampling error ,    E =   100  
Confidence Level ,   CL=   99%  
          
alpha =   1-CL =   1%  
Z value =    Zα/2 =    2.5758   [excel formula =normsinv(α/2)]
          
Sample Size,n =    (Z*σ / E)² =   41.4681  
          
          
So,Sample Size needed=       42  

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4)

Assumptions:

Samples used should be randomly distributed.

Samples used should be independent of each other.

population from which samples are taken is normally distributed.