When trying to hire managers and executives companies someti

When trying to hire managers and executives, companies sometimes verify the academic credentials described by the applicants. One company that performs these checks summarized their findings for a six-month period. Of the 86 applicants whose credentials were checked, 25 lied about having a degree. (a) Find the proportion of applicants who lied about having a degree, and find the standard error. P = (b) Consider these data to be a random sample of credentials from a large collection of similar applicants. Give a 90% confidence interval for the true proportion of applicants who lie about having a degree. lower bound upper bound A poll of 818 adults aged 18 or older asked about purchases that they intended to make for the upcoming holiday season. One of the questions asked about what kind of gift they intended to buy for the person on whom they would spend the most. Clothing was the first choice of 485 people. Give a 90% confidence interval for the proportion of people in this population who intend to buy clothing as their first choice. A grower believes that one in five of his citrus trees are infected with the citrus red mite. How large a sample should be taken if the grower wishes to estimate the proportion of his trees that are infected with citrus red mite to within 0.09 with probability 0.95? (Round your answer up to the nearest whole number.) trees

Solution

(4) (a) phat=25/86 =0.2906977

SE= sqrt(p*(1-p)/n) =sqrt(0.2906977*(1-0.2906977)/86) =0.04896517

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(b) Given a=0.1, Z(0.05) = 1.645 (from standard normal table)

So the lower bound is

p -Z*sqrt(p*(1-p)/n)= 0.2906977 -1.645*0.04896517 =0.21015

So the upper bound is

p +Z*sqrt(p*(1-p)/n)=0.2906977 +1.645*0.04896517 =0.3712454

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(5) p=485/818 =0.5929095

So the lower bound is

p -Z*sqrt(p*(1-p)/n)= 0.5929095 -1.645*sqrt(0.5929095*(1-0.5929095)/818) =0.5646523

So the upper bound is

p +Z*sqrt(p*(1-p)/n)=0.5929095 +1.645*sqrt(0.5929095*(1-0.5929095)/818) =0.6211667

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(6) p=1/5=0.2

Given a=0.05, Z(0.025) = 1.96 (from standard normal table)

So n=(Z/E)^2*p*(1-p)

=(1.96/0.09)^2*0.2*0.8

=75.88346

Take n=76