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Moving to another question will save this response Question 4 of 8 > Question 4 5 points Saved In a random sam ple of 80 LCD monitors selected from a production line, only 56 meet the specifications. Find an 86% confidence interval for the true percentage of LCD monitors produced by this production line that meet the specifications. a. [62.4%, 77.6%) b.[22.6%, 33.4%) O c. 22.4%, 37.6%) Od. [21.8%, 42.1%) > Moving to another question will save this response. Question 4 of 8 >

Explanation / Answer

z for 86% confidence is 1.48 (1.0.86=.14 and 14/2=.07 and therefore .07 in the z chart is closest to the z score 1.48)

p is 56/80=0.7

the lower limit is p-z*sqrt(p*(1-p)/N)=0.7-1.48*sqrt(0.7*0.3/80)=0.62417

upper limit is p+z*sqrt(p*(1-p)/N)=0.7+1.48*sqrt(0.7*0.3/80)=0.7758

thus the correct answer is A

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