Question

50:19

All else being equal, if you cut the sample size in half, how does this affect the margin of error when using the sample to make a

statistical inference about the mean of the normally distributed population from which it was drawn?

ME.

The margin of error is multiplied by /05

The margin of error is multiplied by 12

The margin of error is multiplied by 05,

The margin of error is multiplied by 2.

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Answer 1

Answer:

The margin of error is multiplied by 1.41, which is 1 divided by the square root of 5.

Step-by-step explanation:

The margin of error is:

$M = z*\frac{\sigma}{\sqrt{n}}$

In which z is related to the confidence level, $\sigma$ is the standard deviation of the population and n is the size of the sample.

The margin of error is inverse proportional to the square root of the sample size.

Then

Sample size n:

$M = z*\frac{\sigma}{\sqrt{n}}$

Modified(half the sample size):

$M_{M} = z*\frac{\sigma}{\sqrt{0.5n}}$

Ratio

$\frac{M_{M}}{M} = \frac{z*\frac{\sigma}{\sqrt{0.5n}}}{z*\frac{\sigma}{\sqrt{n}}} = \frac{\sqrt{n}}{\sqrt{0.5n}} = \frac{\sqrt{n}}{\sqrt{0.5}*\sqrt{n}} = \frac{1}{\sqrt{0.5}} = 1.41$

The margin of error is multiplied by 1.41, which is 1 divided by the square root of 5.