Question

A personnel manager has found that historically the scores on aptitude tests given to applicants for entry- level positions follow a normal distribution with a standard deviation of 32.4 points. A random sample of nine test scores from the current group of applicants had a mean score of 187.9 points.
a. Find an 80% confidence interval for the population mean score of the current group of applicants.
b. Based on these sample results, a statistician found for the population mean a confidence interval extending from 165.8 to 210.0 points. Find the confidence level of this interval.

Answer 1

A personnel manager has found that historically the scores on aptitude tests given to applicants for entry-level positions follow a normal distribution with a standard deviation of 32.4 points. A random sample of nine test scores from the current group of applicants had a mean score of 187.9 points.

a. Find an 80% confidence interval for the population mean score of the current group of applicants.

b. Based on these sample results, a statistician found for the population mean a confidence interval extending from 165.8 to 210.0 points. Find the confidence level of this interval.

n= 9

a= 187.9

d=32.4

a. Find an 80% confidence interval for the population mean score of the current group of applicants.

P(a-t<x<a+t)="0.8" p(a-t<x<a+t)="P(-t<x-a<t)" =="" =f(^t="" =central="" f(^t="" limit="" p(-22.1<x="" t="" t<="" t)="2F(^t" x-a<t)="P(-t<x-a<t)" x-a<t)="0.8" x-a<t)="1.28" x-t="0.8" x-t<x="0.9" x<210.0)="u" www="F" x<22.1="0.9" www.="" x<10.0)="0.5034" www.assignmentexpert.com="">

interval is {187.9-41.472, 187.9+41.472}

b. Based on these sample results, a statistician found for the population mean a confidence interval extending from 165.8 to 210.0 points. Find the confidence level of this interval.

P(165.8 <x<210.0 )="u" confidence="" level="">

P(-22.1<x-187.9<22.1)=u

P(^22.1 / d < ^2 - u / d < ^22.1 / d) = Central limit theorem=

=F(^22.1 / d) - F(^22.1 / d) = 2F(^22.1 / 32.4) - 1 = u

2F(0.68) - 1 = 2*0.7517 - 1 = 0.5034

confidence level= 0.5

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