In order to estimate the proportion of office workers who liste n to streamed mu

Question

In order to estimate the

proportion

of office workers who liste

n to streamed

music on a work computer on a regular basis, a sample of 1200 office

workers who work at a computer was taken. Of them, 543 listen to steamed

music on the computer at work. A 99% confidence interval for the

proportion

of all office workers

who listen to streamed music at work is

about:

(a)

(0.453

-

0.001, 0.453+0.001)

(b)

(0.453

-

0.037, 0.453+0.037)

(c)

(0.453

-

0.028, 0.453+0.028)

(d)

(0.453

-

0.044,0.453+0.044)

(e)

(0.453

-

0.033,

453+0.033)

Explanation / Answer

Solution : Option (b) . (0.453 - 0.037 , 0.453 + 0.037)

Explanation :-
  
Given data x = 543 , n = 1200

=> p = x/n = 543/1200 = 0.453

=> q = 1 - p = 0.548

For 99% confidence interval , Z = 2.576

=> The 99% confidence interval of the proportion is (p - Z*sqrt(p*q/n) , p + Z*sqrt(p*q/n))

=> (0.453 - 2.567*sqrt(0.453*0.548/1200) , 0.453 + 2.567*sqrt(0.453*0.548/1200))

=> (0.453 - 0.037 , 0.453 + 0.037)

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