What ar etwo estimates for 68÷503
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Answer:
Step-by-step explanation:
Given that confidence level is 95% and sample mean is within 10 minutes of the population mean. i.e. margin of error = 10
Std deviation of population= 211 minutes
Margin of error = Z critical * std error
Std error = sigma/sq rt n =
Hence we have
\frac{211}{\sqrt{n} }
The minimum sample size is 1710
a) There may not be 1,809 computer users to survey