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

Question

10.4 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.
unbiased estimate of p    = 31 of them have type A seals
sample of n=50 parts on hand are selected
p = 31/50 = 0.38

b.
n*p>5, 50*0.62> 5 => 31>5
n*(1-p)>5, 50*0.38> 5 => 31>5
Can Use Normal Approximation                  
              
c.
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=31
Sample Size(n)=50
Sample proportion = x/n =0.62
Confidence Interval = [ 0.62 ±Z a/2 ( Sqrt ( 0.62*0.38) /50)]
= [ 0.62 - 2.326* Sqrt(0.005) , 0.62 + 2.33* Sqrt(0.005) ]
= [ 0.46,0.78]

d.
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.01 is = 2.576
Sample Proportion = 0.5
ME = 0.03
n = ( 2.576 / 0.03 )^2 * 0.5*0.5
= 1843.271 ~ 1844

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