The Sampling Distribution of a Sample Mean

Learning Objectives

1. State the mean and variance of the sampling distribution of the mean
2. Compute the standard error of the mean
3. State the central limit theorem

The sampling distribution of the mean was defined in the section introducing sampling distributions. This section reviews some important properties of the sampling distribution of the mean introduced in the demonstrations in this chapter.

Mean

The mean of the sampling distribution of the mean is the mean of the population from which the scores were sampled. Therefore, if a population has a mean $\mu$, then the mean of the sampling distribution of the mean is also $\mu$. The symbol $\mu_{\mathrm{M}}$ is used to refer to the mean of the sampling distribution of the mean. Therefore, the formula for the mean of the sampling distribution of the mean can be written as:

$\mu_{M}=\mu$

Variance

The variance of the sampling distribution of the mean is computed as follows:

$\sigma_{M}^{2}=\frac{\sigma^{2}}{N}$

That is, the variance of the sampling distribution of the mean is the population variance divided by $\mathrm{N}$, the sample size (the number of scores used to compute a mean). Thus, the larger the sample size, the smaller the variance of the sampling distribution of the mean.

The standard error of the mean is the standard deviation of the sampling distribution of the mean. It is therefore the square root of the variance of the sampling distribution of the mean and can be written as:

$\sigma_{M}=\frac{\sigma}{\sqrt{N}}$

The standard error is represented by a $\sigma$ because it is a standard deviation. The subscript $(M)$ indicates that the standard error in question is the standard error of the mean.

Central Limit Theorem

The central limit theorem states that:

Given a population with a finite mean $\mu$ and a finite non-zero variance $\sigma^{2}$, the sampling distribution of the mean approaches a normal distribution with a mean of $\mu$ and a variance of $\sigma^{2} / N$ as $N$, the sample size, increases.

The expressions for the mean and variance of the sampling distribution of the mean are not new or remarkable. What is remarkable is that regardless of the shape of the parent population, the sampling distribution of the mean approaches a normal distribution as $\mathrm{N}$ increases. If you have used the "Central Limit Theorem Demo," you have already seen this for yourself. As a reminder, Figure 1 shows the results of the simulation for $\mathrm{N}=2$ and $\mathrm{N}=10$. The parent population was a uniform distribution. You can see that the distribution for $\mathrm{N}=2$ is far from a normal distribution. Nonetheless, it does show that the scores are denser in the middle than in the tails. For $\mathrm{N}=10$ the distribution is quite close to a normal distribution. Notice that the means of the two distributions are the same, but that the spread of the distribution for $N=10$ is smaller.

Figure 1. A simulation of a sampling distribution. The parent population is uniform. The blue line under "16" indicates that 16 is the mean. The red line extends from the mean plus and minus one standard deviation.

Figure 2 shows how closely the sampling distribution of the mean approximates a normal distribution even when the parent population is very non-normal. If you look closely you can see that the sampling distributions do have a slight positive skew. The larger the sample size, the closer the sampling distribution of the mean would be to a normal distribution.

Figure 2. A simulation of a sampling distribution. The parent population is very non-normal.

Source: David M. Lane, https://onlinestatbook.com/2/sampling_distributions/samp_dist_mean.html
This work is in the Public Domain.

Question 4 out of 5.
The sampling distribution of the mean, with $N=30$, of a moderately negatively skewed distribution is:

Positively skewed

Negatively skewed

About normal

Question 5 out of 5.
The entire student body of $225$ students took a test. These test scores have a mean of $75$, a standard deviation of $10$, and are slightly positively skewed. If you randomly chose $25$ of these test scores and calculated the mean over and over again, what could be the mean, standard deviation, and skew of this distribution?

Mean = $75$, SD = $10$, Skew = $1.2$

Mean = $75$, SD = $0.67$, Skew = $0.8$

Mean = $80$, SD = $2$, Skew = $-1.2$

Mean = $75$, SD = $2$, Skew = about $0$

Mean = $75$, SD = $0.67$, Skew = about $0$

1. The mean of the sampling distribution of the mean is the mean of the population from which the scores were sampled, in this case $14$.
   
   
2. The variance of the sampling distribution of the mean is the population variance divided by $N$. The population SD is $6$, so the population variance is $36$. $36/9 = 4$
   
   
3. The standard error is the standard deviation of the population divided by the square root of $N$. In this case, $12/4 = 3$
   
   
4. According to the central limit theorem, regardless of the shape of the parent population, the sampling distribution of the mean approaches a normal distribution as $N$ increases. In this case, a sample size of $30$ is sufficiently large to cause the sampling distribution of the mean to look about normal.
   
   
5. Mean = $75$, SD = $2$, Skew = about $0$: This problem is asking about the sampling distribution of the mean: Mean = $75$, SD = $10/sqrt(25)$ = $10/5$ = $2$, Skew = about $0$ because the central limit theorem states that the sampling distribution of the mean would be about normal with a large enough $N$.
   

Learning Objectives

1. State the mean and variance of the sampling distribution of the difference between means
2. Compute the standard error of the difference between means
3. Compute the probability of a difference between means being above a specified value

Statistical analyses are very often concerned with the difference between means. A typical example is an experiment designed to compare the mean of a control group with the mean of an experimental group. Inferential statistics used in the analysis of this type of experiment depend on the sampling distribution of the difference between means.

The sampling distribution of the difference between means can be thought of as the distribution that would result if we repeated the following three steps over and over again: (1) sample $n_{1}$ scores from Population $1$ and $n_{2}$ scores from Population $2,(2)$ compute the means of the two samples ( $M_{1}$ and $M_{2}$ ), and (3) compute the difference between means, $M_{1}-M_{2}$. The distribution of the differences between means is the sampling distribution of the difference between means.

As you might expect, the mean of the sampling distribution of the difference between means is:

$\mu_{M_{1}-M_{2}}=\mu_{1}-\mu_{2}$

which says that the mean of the distribution of differences between sample means is equal to the difference between population means. For example, say that the mean test score of all 12 -year-olds in a population is $34$ and the mean of 10 -yearolds is 25. If numerous samples were taken from each age group and the mean difference computed each time, the mean of these numerous differences between sample means would be $34-25=9$.

From the variance sum law, we know that:

$\sigma_{M_{1}-M_{2}}^{2}=\sigma_{M_{1}}^{2}+\sigma_{M_{2}}^{2}$

which says that the variance of the sampling distribution of the difference between means is equal to the variance of the sampling distribution of the mean for Population $1$ plus the variance of the sampling distribution of the mean for Population $2$. Recall the formula for the variance of the sampling distribution of the mean:

$\sigma_{M}^{2}=\frac{\sigma^{2}}{N}$

Since we have two populations and two samples sizes, we need to distinguish between the two variances and sample sizes. We do this by using the subscripts $1$ and $2$. Using this convention, we can write the formula for the variance of the sampling distribution of the difference between means as:

$\sigma_{M_{1}-M_{2}}^{2}=\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}$

Since the standard error of a sampling distribution is the standard deviation of the sampling distribution, the standard error of the difference between means is:

$\sigma_{M_{1}-M_{2}}=\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}$

Just to review the notation, the symbol on the left contains a sigma $(\sigma)$, which means it is a standard deviation. The subscripts $M_{1}-M_{2}$ indicate that it is the standard deviation of the sampling distribution of $\mathrm{M}_{1}-\mathrm{M}_{2}$.

Now let's look at an application of this formula. Assume there are two species of green beings on Mars. The mean height of Species $1$ is $32$ while the mean height of Species $2$ is $22$. The variances of the two species are $60$ and $70$, respectively and the heights of both species are normally distributed. You randomly sample $10$ members of Species $1$ and $14$ members of Species $2$. What is the probability that the mean of the $10$ members of Species $1$ will exceed the mean of the $14$ members of Species $2$ by $5$ or more? Without doing any calculations, you probably know that the probability is pretty high since the difference in population means is $10$. But what exactly is the probability?

First, let's determine the sampling distribution of the difference between means. Using the formulas above, the mean is

$\mu_{M_{1}-M_{2}}=32-22=10$

The standard error is:

$\sigma_{M_{1}-M_{2}}=\sqrt{\frac{60}{10}+\frac{70}{14}}=3.317$

The sampling distribution is shown in Figure 1 . Notice that it is normally distributed with a mean of 10 and a standard deviation of $3.317$. The area above $5$ is shaded blue.

Figure 1. The sampling distribution of the difference between means.

The last step is to determine the area that is shaded blue. Using either a $Z$ table or the normal calculator, the area can be determined to be $0.934$. Thus the probability that the mean of the sample from Species $1$ will exceed the mean of the sample from Species $2$ by $5$ or more is $0.934$.

As shown below, the formula for the standard error of the difference between means is much simpler if the sample sizes and the population variances are equal. When the variances and samples sizes are the same, there is no need to use the subscripts $1$ and $2$ to differentiate these terms.

$\sigma_{M_{1}-M_{2}}=\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}=\sqrt{\frac{\sigma^{2}}{n}+\frac{\sigma^{2}}{n}}=\sqrt{\frac{2 \sigma^{2}}{n}}$

This simplified version of the formula can be used for the following problem: The mean height of 15 -year-old boys (in $\mathrm{cm}$ ) is $175$ and the variance is $64$. For girls, the mean is $165$ and the variance is $64$. If eight boys and eight girls were sampled, what is the probability that the mean height of the sample of girls would be higher than the mean height of the sample of boys? In other words, what is the probability that the mean height of girls minus the mean height of boys is greater than $0$?

As before, the problem can be solved in terms of the sampling distribution of the difference between means (girls - boys). The mean of the distribution is $165$ $175=-10$. The standard deviation of the distribution is:

$\sigma_{M_{-}-M_{2}}=\sqrt{\frac{2 \sigma^{2}}{n}}=\sqrt{\frac{(2)(64)}{8}}=4$

A graph of the distribution is shown in Figure 2 . It is clear that it is unlikely that the mean height for girls would be higher than the mean height for boys since in the population boys are quite a bit taller. Nonetheless it is not inconceivable that the girls' mean could be higher than the boys' mean.

Figure 2. Sampling distribution of the difference between mean heights.

A difference between means of $0$ or higher is a difference of $10 / 4=2.5$ standard deviations above the mean of $-10$. The probability of a score $2.5$ or more standard deviations above the mean is $0.0062$.

1. The mean of the distribution of the difference between sample means is equal to the difference between population means. $20 - 15 = 5$
   
   
2. The variance of the sampling distribution of the difference between means is equal to the variance of the sampling distribution of the mean for Pop $1$ plus the variance of the sampling distribution of the mean for Pop $2$. $100/20 + 64/16 = 5 + 4 = 9$
   
   
3. Mean = $10$, SD = $4$, Plug these into the normal calculator and find the area above $6$. You get. $841$. A similar question using this data appears in the text.
   
   
4. Mean = $727 - 532$ = $195$, Var = $12,000/12 + 10,000/14$ = $1,714.3$, SD = $sqrt(1,714.3)$ = $41.404$, Use the normal calculator to calculate the area above $150$ for a distribution with this mean and SD. You get $0.8614$.