In an initial survey designed to estimate the percentage of time air-express car

Question

In an initial survey designed to estimate the percentage of time air-express cargo loaders are idle, an analyst found that loaders were idle in 7 of the 50 observations.

What is the estimated percentage of idle time? (Omit the "%" sign in your response.)

Based on the initial results, approximately how many observations would you require to estimate the actual percentage of idle time to within 9 percent with a confidence of 95 percent? (Do not round intermediate calculations. Round up your final answer to the next whole number.)

In an initial survey designed to estimate the percentage of time air-express cargo loaders are idle, an analyst found that loaders were idle in 7 of the 50 observations.

Explanation / Answer

(a)the estimated percentage of idle time

p=7/50

=0.14

(b) given error E=0.09

we have E= Z(a/2) (SE(p))

E^2= Z^2(0.05/2)(SE(p))

where SE(p) =sqrt((p(1-p)/n))

(SE(p))^2 = 0.14(1-0.14)/n

(SE(p))^2 = 0.1204/n

there fore E^2= Z^2(0.05/2)*(0.1204)/n

n=(1.96)^2*(0.1204)/(0.09)^2

n=57.102

therefore sample size is n=57

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