Question 620500

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Hi,
In order to test a new production method, 36 employees were selected randomly to try the new method. 
The mean was 75.79 parts per hour with a standard deviation of 28.25 parts per hour.
What is the sample size needed to obtain a margin of error 14.01 with a 90% confidence interval?
SE = 14.01 ={{{ 1.645(28.25/sqrt(n))}}}
    {{{sqrt(n) = 1.645*28.25/14.01 = 3.317}}}   n = 11