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.

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okpalawalter8 okpalawalter8 · Answer 1 · 2020-06-16T05:07:39+02:00

Answer:

The new standard error is [tex]e_{new} = 23.4[/tex]

Step-by-step explanation:

From the question we are told that

The standard error is [tex]e = 52.4[/tex]

Generally the standard error is mathematically represented as

[tex]e = \frac{6}{\sqrt{n} }[/tex]

Where n is the sample size

for the original standard error we have

[tex]52.4 = \frac{6}{\sqrt{n} }[/tex]

Now sample size is quintuple

[tex]e_{new} = \frac{6}{\sqrt{5 * n} }[/tex]

[tex]but \ \ 52.4 = \frac{6}{\sqrt{n} }[/tex]

So [tex]e_{new} = \frac{52.4}{\sqrt{5} }[/tex]

[tex]e_{new} = 23.4[/tex]