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Suppose your marketing colleague used a known population mean and standard deviation to compute the standard error as 52.4 for samples of a particular size. You don't know the particular sample size but your colleague told you that the sample size is greater than 70. Your boss asks what the standard error would be if you quintuple (5x) the sample size. What is the standard error for the new sample size? Please round your answer to the nearest tenth.

User Baski
by
6.8k points

1 Answer

5 votes

Answer:

The new standard error is
e_(new) = 23.4

Explanation:

From the question we are told that

The standard error is
e = 52.4

Generally the standard error is mathematically represented as


e = (6)/(√(n) )

Where n is the sample size

for the original standard error we have


52.4 = (6)/(√(n) )

Now sample size is quintuple


e_(new) = (6)/(√(5 * n) )


but \ \ 52.4 = (6)/(√(n) )

So
e_(new) = (52.4)/(√(5) )


e_(new) = 23.4

User TheDavidBarton
by
7.0k points
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