Question: The standard deviation (or, as it is usually called, the standard error) of the sampling distribution for the sample mean, $\overline{x}$ , is equal to the standard deviation of the population from which the sample was selected, divided by the square root of the sample size. That is

${\sigma }_{\overline{X}}=\frac{\sigma }{\sqrt{n}}$

1. As the sample size is increased, what happens to the standard error of? Why is this property considered important?
2. Suppose a sample statistic has a standard error that is not a function of the sample size. In other words, the standard error remains constant as n changes. What would this imply about the statistic as an estimator of a population parameter?
3. Suppose another unbiased estimator (call it A) of the population mean is a sample statistic with a standard error equal to

${\sigma }_A=\frac{\sigma }{\sqrt[3]{n}}$

Which of the sample statistics,$\overline{x}$or A, is preferable as an estimator of the population mean? Why?

1. Suppose that the population standard deviation $\sigma$is equal to 10 and that the sample size is 64. Calculate the standard errors of $\overline{x}$and A. Assuming that the sampling distribution of A is approximately normal, interpret the standard errors. Why is the assumption of (approximate) normality unnecessary for the sampling distribution of$\overline{x}$?

The standard error of is $\frac{\sigma }{\sqrt{n}}$

If the sample size increases, then the standard error will decrease as the sample size is in the denominator.

This property is considered to be important since the sample mean will be closer to the population mean.

The standard error remains constant as the sample size changes.

Then statistics is not a good estimator of a population parameter in this case.

Hence, we can say that statistic for 100 observations remains the same as the statistic for 200 observations

Given that,

The standard error of an unbiased estimator of $\overline{x}$ is ${\sigma }_{\overline{X}}=\frac{\sigma }{\sqrt{n}}$

The standard error of the unbiased estimator of A is ${\sigma }_A=\frac{\sigma }{\sqrt[3]{n}}$

So, $\overline{x}$is preferable as an estimator of the population mean.

*The standard error of $\overline{x}$is smaller than the standard error of A as the square root of sample size is in the denominator.

The standard deviation is 10.

The sample size is 64.

The standard error of$\overline{x}$ is computed as

$$\begin{gathered} {\sigma }_{\overline{X}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{10}{\sqrt{64}} \end{gathered}$$

And the standard error of A is computed as

$$\begin{gathered} {\sigma }_A=\frac{\sigma }{\sqrt[3]{n}} \\ =\frac{10}{\sqrt[3]{64}} \\ =2.5 \end{gathered}$$

Therefore, the standard error of $\overline{x}$ and A are 1.25 and 2.5.

Interpretation of ${\sigma }_{\overline{X}}$

According to empirical rules,

All the values of will fall within 3 standard deviations

$$\begin{gathered} \Rightarrow \left(\mu -3\sigma ,\mu -3\sigma \right)\backslash \mathrm{hfill} \\ \Rightarrow \left(\mu -3\times 1.25, \mu -3\times 1.25\right)\backslash \mathrm{hfill} \\ \Rightarrow \left(\mu -3.75,\mu +3.75\right)\backslash \mathrm{hfill} \end{gathered}$$

Interpretation of ${\sigma }_A$

According to empirical rules,

All the values of A will fall within 3 standard deviations. i.e.

$$\begin{gathered} \Rightarrow \left(\mu -3\sigma ,\mu -3\sigma \right)\backslash \mathrm{hfill} \\ \Rightarrow \left(\mu -3\times 2.5, \mu -3\times 2.5\right)\backslash \mathrm{hfill} \\ \Rightarrow \left(\mu -7.5,\mu +7.5\right)\backslash \mathrm{hfill} \end{gathered}$$

According to CLT, if n is large, then it follows the normal distribution with mean$\mu$ and standard deviation$\frac{\sigma }{\sqrt{n}}$ .

Therefore, the assumption of normality is unnecessary for the sampling distribution of$\overline{x}$.