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Sholpan [36]
3 years ago
14

A simple random sample of size n=250 individuals who are currently employed is asked if they work at home at least once per week

. Of the 250 employed individuals ​surveyed,42 responded that they did work at home at least once per week. Construct a​ 99% confidence interval for the population proportion of employed individuals who work at home at least once per week.
The lower bound is nothing.

(Round to three decimal places as​ needed.)
Mathematics
1 answer:
Travka [436]3 years ago
4 0

Answer:

solution is given below

Step-by-step explanation:

A simple random sample size n = 250 . Of the 250 employed individuals ​surveyed,42 responded that they did work at home at least once per week.

Construct 99% confidence interval for population

For proportion : 42 / 250 = 1/10 = 0.16

Mean = 2.5 * sqrt [ 0.1 * 0.9  / 250]

     = 2.5 * 0.01

    = 0.47

Construction of hypothesis:

0.10 - 0.047 < p < 0.10 + 0.047

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2 1/2r + 1 + 2r = 9 1/10
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This should help you here are the steps of your problem :)


2*(1/2)*r+2*r+1 = 9*(1/10) // - 9*(1/10)


2*(1/2)*r+2*r-(9*(1/10))+1 = 0


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r = -1/30


So your answer would be 1/30

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