Question

The average amount of Canning in 12-ounce cans at a major canning manufacturer's Quality Control Department it is being studied. 1 in 50 (1-50) systematic sampling is applied on the production line and the following data are obtained: were collected. The mass size is 1800 boxes.

Amount of Canning in the Box(ons)

12.00	11.97	12.01	12.03	12.01	11.80
11.91	11.98	12.03	11.98	12.00	11.83
11.87	12.01	11.98	11.87	11.90	11.88
12.05	11.87	11.91	11.93	11.94	11.89
11.75	11.93	11.95	11.97	11.93	12.05
11.85	11.98	11.87	12.05	12.02	12.04

a) Create a 95% confidence interval for the average amount of Canning in boxes.

b) To do a similar study, you can use the values obtained in (a) as preliminary information.
What should be the number of boxes that need to be examined to estimate the population average with Bound (B)=0.03 units? Which (1-k) systematic sampling scheme would you recommend in this case?

Answer 1

a)

Level of Significance ,    α =    0.05
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   0.0762
Sample Size ,   n =    36
Sample Mean,    x̅ = ΣX/n =    11.9456

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   35          
't value='   tα/2=   2.0301   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   0.0762   / √   36   =   0.012702
margin of error , E=t*SE =   2.0301   *   0.01270   =   0.025786
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    11.95   -   0.025786   =   11.919769
Interval Upper Limit = x̅ + E =    11.95   -   0.025786   =   11.971342
95%   confidence interval is (   11.92   < µ <   11.97   )

b)

Standard Deviation ,   σ =    0.076211902                  
sampling error ,    E =   0.03                  
Confidence Level ,   CL=   95%                  
                          
alpha =   1-CL =   5%                  
Z value =    Zα/2 =    1.960   [excel formula =normsinv(α/2)]              
                          
Sample Size,n = (Z * σ / E )² = (   1.960   *   0.076211902   /   0.03   ) ² =   24.791
                          
                          
So,Sample Size needed=       25                  

1800/25 = 72

Hence (1-72) should be the technique.

Thanks in advance!

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