Question

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. Note that the correct answer will be evaluated based on the full-precision result you would obtain using Excel.

Answer 1

Answer:

The standard error for the new sample size is of 23.4.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Interpretation:

From this, we can gather that the standard error is inversely proportional to the square root of the sample size, that is, for example, if the sample size is multiplied by 4, the standard error is divided by the square root of 4, which is 2.

Standard error of 52.4, sample size multiplied by 5. What is the standard error for the new sample size?

The standard error of 52.4 divided by the square root of 5. So

$s = \frac{52.4}{\sqrt{5}} = 23.4$

The standard error for the new sample size is of 23.4.