To study p, the proportion of parts on hand that have type A seals, a random sam

Question

To study p, the proportion of parts on hand that have type A seals, a random sample of n=50 parts on hand are selected and 31 of them have type A seals.
(a) Give the value of an unbiased estimate of p.
(b) Is n large enough to use the large sample confidence interval for p? Whatever your answer, compute the large sample 98% confidence interval for p.

(c) Using the preliminary estimate given here, how large must n be so that and p do not differ by more than .03 units with 99% confidence.
(d) If one did not have a preliminary estimate of p, how large must n be so that and p do not differ by more than .03 units with 99% confidence.

(e) State the hypotheses used to test if more than half of the parts on hand have type A seals. With = 0.05, what is the conclusion for this test? State your conclusion in the terminology of this example; not just reject some hypothesis or do not reject a hypothesis.
(f) What is the p-value for the test in part (e)?

Explanation / Answer

a)

Note that              
              
p^ = point estimate of the population proportion = x / n = 31/50 =   0.62   [ANSWER]

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b)

yes, as the number of successes (31) and the number of failures (50-31) are both greater than 15.   [ANSWER]

  
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.068644009          
              
Now, for the critical z,              
alpha/2 =   0.01          
Thus, z(alpha/2) =    2.326347874          
Thus,              
Margin of error = z(alpha/2)*sp =    0.159689845          
lower bound = p^ - z(alpha/2) * sp =   0.460310155          
upper bound = p^ + z(alpha/2) * sp =    0.779689845          
              
Thus, the confidence interval is              
              
(   0.460310155   ,   0.779689845   ) [ANSWER]

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c)

Note that      
      
n = z(alpha/2)^2 p (1 - p) / E^2      
      
where      
      
alpha/2 =    0.005  
       
      
Using a table/technology,      
      
z(alpha/2) =    2.575829304  
      
Also,      
      
E =    0.03  
p =    0.62  
      
Thus,      
      
n =    1736.868488  
      
Rounding up,      
      
n =    1737   [ANSWER]

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d)

Note that      
      
n = z(alpha/2)^2 p (1 - p) / E^2      
      
where      
      
alpha/2 =    0.005  
As there is no previous estimate for p, we set p = 0.5.      
      
Using a table/technology,      
      
z(alpha/2) =    2.575829304  
      
Also,      
      
E =    0.03  
p =    0.5  
      
Thus,      
      
n =    1843.026834  
      
Rounding up,      
      
n =    1844   [ANSWER]

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