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A business that repairs major appliances charges a flat fee and hourly amount plus the cost of any needed parts. they know that the time required to repair an appliance has a standard deviation of 0.25 hours. the mean depends on the type of appliance. to estimate the true mean time required to repair side-by-side refrigerators, they will take a random sample of 60 refrigerator repairs. if they want a 99% confidence interval, what will the margin of error in their estimate be?

Respuesta :

jushmk
Error in true mean = +/-Z*[tex] \frac{s}{ \sqrt{n} } [/tex]

Where; Z=2.58 at 99% confidence interval, s= sd = 0.25 hours, n=sample population = 60 refrigerators

Therefore;

Margin of error = +/-2.58*[tex] \frac{0.25}{ \sqrt{60} } [/tex] = 0.0832 hours

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