Chapter 5, Problem 5.52LM

The standard deviation (or, as it is usually called , the standard error) of the sampling distribution for the sample mean, $\bar{x}$, 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,

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

1. a. As the sample size is increased, what happens to the standard error of $\bar{x}$ ? Why is this property considered important?
2. b. 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. c. 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, $\bar{x}$ or A, is preferable as an estimator of the population mean? Why?

4. d. Suppose that the population standard deviation σ is equal to 10 and that the sample size is 64. Calculate the standard errors of $\bar{x}$ 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 $\bar{x}$?

To determine

To describe: what happens to the standard error of $\overline{x}$ when sample size increases.

To explain: Why is this property considered important.

Answer to Problem 5.52LM

The standard deviation (standard error) decreases when sample size increases.

The property is important because the estimate variable would be less when the sample size is large.

Explanation of Solution

Calculation:

Sampling distribution of $\overline{x}$:

When a random sample of size n is selected from a population which is normally distributed, then the sampling distribution of $\overline{x}$ is also normally distributed.

Properties:

• The mean of sampling distribution is equal to the mean of the population.

  ${\mu }_{\overline{x}}=E\left(x\right)=\mu$

• The standard deviation of the sampling distribution is equal to the ratio of the population standard deviation and square root of sample size.

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

• When the sample size n is or more of the population size N then the $\frac{\sigma }{\sqrt{n}}$ must be multiplied by the finite correction factor $\sqrt{\frac{\left(N-n\right)}{\left(N-1\right)}}$.
• For most sampling situations, the correction factor is close to 1 and is ignored.
• The standard deviation ${\sigma }_{\overline{x}}$ sometimes referred as the standard error of the mean.

As the sample size increases the standard deviation (standard error) decreases because the sample size and the standard deviation of $\overline{x}$ are inversely proportional.

This property is important because the estimate variable would be less when the sample size is large.

To determine

To explain: About the statistic as an estimator of a population parameter when the standard error remains constant as n changes.

Explanation of Solution

If the standard error remains constant as n changes then it would not be a good estimator for the parameter. When the standard error is not a function of the sample size then the statistic based on one observation is as good estimator as the statistic based on 1,000 observations.

To determine

To explain: Which of the sample statistic $\overline{x}\text{or }A$ is preferable as an estimator of the population.

Explanation of Solution

Given info:

An unbiased estimator A of a population mean is a sample statistic with a standard error equal to ${\sigma }_A=\frac{\sigma }{\sqrt[3]{n}}$ is considered.

Justification:

The statistic $\overline{x}$ should be preferred than A as an estimator for the population mean because the standard error of $\overline{x}$ is fewer than the standard error of A.

To determine

To find: The standard errors of $\overline{x}$ and A.

To explain: Whether the assumption of normality is unnecessary for the sampling distribution of $\overline{x}$.

Answer to Problem 5.52LM

The standard error of $\overline{x}$ is 1.25.

The standard error of A is 2.5.

Explanation of Solution

Given info:

The population standard deviation $\sigma$ is 10 and the sample size is 64. Consider the sampling distribution of A is approximately normal.

Calculation:

The standard error of $\overline{x}$ is,

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

Substitute $\sigma$ as 10 and n as 64 in the formula,

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{10}{\sqrt{64}}\\ =\frac{10}{8}\\ =1.25\end{array}$

Thus, the standard error of $\overline{x}$ is 1.25.

Central Limit Theorem:

Central limit theorem states that the sampling distribution of $\overline{x}$ with mean ${\mu }_{\overline{x}}=\mu$ and standard deviation ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$ would be approximately normal as the sample size is sufficiently large.

The normal approximation to the sampling distribution of $\overline{x}$ would be better when the sample size is larger, $n=64$.

Therefore the sampling distribution of $\overline{x}$ is approximately normal with mean ${\mu }_{\overline{x}}=\mu$ and standard deviation ${\sigma }_{\overline{x}}=1.25$.

Empirical rule:

It the distributions of a data set are approximately symmetric or bell shaped then the standard deviations have the following features,

• About 68% of the data falls within one standard deviation of the mean.
• About 95% of the data falls within two standard deviations of the mean.
• About 99.7% of the data falls within three standard deviations of the mean.

One standard deviation from mean is,

$\begin{array}{c}{\mu }_{\overline{x}}\pm {\sigma }_{\overline{x}}=\mu \pm 1.25\\ =\left(\mu -1.25,\mu +1.25\right)\end{array}$

Two standard deviations from mean is,

$\begin{array}{c}{\mu }_{\overline{x}}\pm 2{\sigma }_{\overline{x}}=\mu \pm 2\left(1.25\right)\\ =\mu \pm 2.50\\ =\left(\mu -2.50,\mu +2.50\right)\end{array}$

Three standard deviations from mean is,

$\begin{array}{c}{\mu }_{\overline{x}}\pm 3{\sigma }_{\overline{x}}=\mu \pm 3\left(1.25\right)\\ =\mu \pm 3.75\\ =\left(\mu -3.75,\mu +3.75\right)\end{array}$

From the empirical rule 68% of the data falls within $\left(\mu -1.25,\mu +1.25\right)$, 95% of the data falls within $\left(\mu -2.50,\mu +2.50\right)$, and 99.7% of the data falls within $\left(\mu -3.75,\mu +3.75\right)$.

The standard error of A is,

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

Substitute $\sigma$ as 10 and n as 64 in the formula,

$\begin{array}{c}{\sigma }_A=\frac{\sigma }{\sqrt[3]{n}}\\ =\frac{10}{\sqrt[3]{64}}\\ =\frac{10}{4}\\ =2.5\end{array}$

Thus, the standard error of A is 2.5.

One standard deviation from mean is,

$\begin{array}{c}{\mu }_A\pm {\sigma }_A=\mu \pm 2.50\\ =\left(\mu -2.50,\mu +2.50\right)\end{array}$

Two standard deviations from mean is,

$\begin{array}{c}{\mu }_A\pm 2{\sigma }_A=\mu \pm 2\left(2.50\right)\\ =\mu \pm 5.00\\ =\left(\mu -5.00,\mu +5.00\right)\end{array}$

Three standard deviations from mean is,

$\begin{array}{c}{\mu }_A\pm 3{\sigma }_A=\mu \pm 3\left(2.50\right)\\ =\mu \pm 7.50\\ =\left(\mu -7.50,\mu +7.50\right)\end{array}$

From the empirical rule 68% of the data falls within $\left(\mu -2.50,\mu +2.50\right)$, 95% of the data falls within $\left(\mu -5.00,\mu +5.00\right)$, and 99.7% of the data falls within $\left(\mu -7.50,\mu +7.50\right)$.