7.58 A company that produces cell phones shipped 1,000 to a retail distributor.

Question

7.58 A company that produces cell phones shipped 1,000 to a retail distributor. The company assured the distributor that they are confident that 98% of the cell phones are operable, that is, non-defective. The retailer took a random sample of n=50 from the 1,000 cell phones and found that 46 of the phones were operable, ready to be sold.

(a) Obtain a point estimate of the true proportion of operable cell phones: p^ = .

(b) Obtain an approximate 95% confidence interval of the true population proportion, p, of operable components. ( , ) (give answer to three decimal places)

Explanation / Answer

a)

Note that              
              
p^ = point estimate of the population proportion = x / n = 46/50 =   0.92   [ANSWER]

*******************************

b)      
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.038366652          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
Margin of error = z(alpha/2)*sp =    0.075197256          
lower bound = p^ - z(alpha/2) * sp =   0.844802744          
upper bound = p^ + z(alpha/2) * sp =    0.995197256          
              
Thus, the confidence interval is              
              
(   0.844802744   ,   0.995197256   ) [ANSWER]

**************************************

Hi! I didn't use finite population correction factor here. If you use that in your class, please resubmit this question indicating that you need to use finite population correction (FPC) for this problem. That way we can continue helping you! Thanks!

Related Questions

Navigate

• Browse (All)
• Browse 7
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.