A manufacturer receives parts from two suppliers. An SRS of 400 parts from suppl

Question

A manufacturer receives parts from two suppliers. An SRS of 400 parts from supplier 1 finds 20 defective parts. An SRS of 100 parts from supplier 2 finds 10 defective parts. Let p1 and p2 be the proportion of all parts from suppliers 1 and 2, respectively, that are defective. A 98% confidence interval for p1 – p2, the difference in the two proportions is –0.05 ± 0.033. –0.05 ± 0.068. –0.05 ± 0.074. A manufacturer receives parts from two suppliers. An SRS of 400 parts from supplier 1 finds 20 defective parts. An SRS of 100 parts from supplier 2 finds 10 defective parts. Let p1 and p2 be the proportion of all parts from suppliers 1 and 2, respectively, that are defective. A 98% confidence interval for p1 – p2, the difference in the two proportions is –0.05 ± 0.033. –0.05 ± 0.068. –0.05 ± 0.074.

Explanation / Answer

Getting p1^ and p2^,          
          
p1^ = x1/n1 =    0.05      
p2 = x2/n2 =    0.1      

Hence,

p1 - p2 = 0.05 - 0.1 = -0.05.
Also, the standard error of the difference is          
          
sd = sqrt[ p1 (1 - p1) / n1 + p2 (1 - p2) / n2] =    0.031917863      
          
For the   98%   confidence level, then  
          
alpha/2 = (1 - confidence level)/2 =    0.01      

z(alpha/2) =    2.326347874      
          
Thus, the margin of error is

E = z(alpha/2)*sd = 0.074252054

Hence

OPTION C: -0.05 +/- 0.074 [ANSWER]

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