A research intern would like to estimate, with 95% confidence, the true proportion of employees who are unsatisfied with their current job. No preliminary estimate is available. The researcher wants the estimate to be within 5% of the population mean. The research intern has a table to use: z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576 They complete the calculation for the minimum sample size as follows.... n=1.962(0.5)(0.5)52≈0.038 Which the intern then rounds up to 1. Which value in the calculation is incorrect?

A First Course in Probability (10th Edition)
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ISBN:9780134753119
Author:Sheldon Ross
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A research intern would like to estimate, with 95% confidence, the true proportion of employees who are unsatisfied with their current job. No preliminary estimate is available. The researcher wants the estimate to be within 5% of the population mean.

The research intern has a table to use:

 

z0.10 z0.05 z0.025 z0.01 z0.005
1.282 1.645 1.960 2.326 2.576

 

They complete the calculation for the minimum sample size as follows....

 

n=1.962(0.5)(0.5)52≈0.038

 

Which the intern then rounds up to 1. Which value in the calculation is incorrect?

 

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this solution is incorrect since the 

confidence level is 90%

margin of error is 3%

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