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Recall the Spice Girls Emporium example. A list of useful information is given below. n = 36 The sample mean income is $41,100 The population standard deviation is estimated to be $4,500 What if we wanted to change our level of confidence to be 99%? What would our new margin of error be? Your answer should be given as an integer.

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Answer: Margin of error = 1932

Step-by-step explanation: Margin of Error is the amount of variation a survey's results have. In other words, it is understood as the measure of variation one can see if the same survey was taken multiple times.

Margin of error is calculated as [tex]z\frac{\sigma}{\sqrt{n}}[/tex]

z is z-score related to the percentage of confidence, in this z = 2.576

σ is population standard deviation

n is how many individuals are there in the sample or population

With a new level of confidence of 99%:

ME = [tex]2.576.\frac{4500}{\sqrt{36}}[/tex]

ME = 2.576(750)

ME = 1932

The new margin of error would be 1932.

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