Sample means

Sampling distribution of sample means

Suppose we want to estimate the population mean $\mu$. We take a sample from the population and compute the sample mean. To judge how reliable that sample mean is, we need to understand how much sample means tend to vary from sample to sample. In general, a larger sample size $n$ leads to less variation in sample means.

Sidenote

Important to know!

Start with a population with mean $\mu$ and standard deviation $\sigma$.

• The mean of the set of sample means, ${\mu }_{\bar{x}}$, equals the population mean $\mu$.
• The standard deviation of the set of sample means is approximately $\frac{\sigma }{\sqrt{n}}$, where $\sigma$ is the population standard deviation and $n$ is the sample size.

The sampling distribution of $\bar{x}$ has mean $\mu$ and standard deviation $\frac{\sigma }{\sqrt{n}}$. If the population is approximately normal, or if the central limit theorem applies (sample size $n\ge 30$), then the set of sample means is approximately normally distributed.

Practice problems

Example 1:

Suppose that the battery life for cell phones is normally distributed with a mean of $30$ hours and a standard deviation of $5$ hours. A quality inspector tests $16$ phones. What is the distribution of the average battery life from these $16$ phones?

We are given $\mu =30$ hours, $\sigma =5$ hours, and $n=16$. Compute the standard error:

$$\begin{array}{rl}{\sigma }_{\bar{x}}& =\frac{\sigma }{\sqrt{n}}\\ & =\frac{5}{\sqrt{16}}\\ & =\frac{5}{4}\\ & =1.25\text{ hours}\end{array}$$

Solution:

Example 2:

Suppose that teenagers in the United States spend an average of $7$ hours on computers each day with a standard deviation of $2.5$ hours. What is the probability that a sample of $36$ teenagers will average between $6.5$ and $7.5$ hours?

We are given $\mu =7$ hours, $\sigma =2.5$ hours, and $n=36$.

First, find the standard error:

$$\begin{array}{rl}\text{SE}={\sigma }_{\bar{x}}& =\frac{\sigma }{\sqrt{n}}\\ & =\frac{2.5}{\sqrt{36}}\\ & =\frac{2.5}{6}\\ & \approx 0.4167\text{ hours}\end{array}$$

Next, convert the endpoints to z-scores:

$$\begin{array}{rl}{z}_1& =\frac{6.5-7}{0.4167}\\ & =-1.20\end{array}\begin{array}{rl}{z}_2& =\frac{7.5-7}{0.4167}\\ & =1.20\end{array}$$

Look up the cumulative probabilities and subtract:

$$\begin{array}{rl}P(-1.20<Z<1.20)& =P(Z<1.20)-P(Z<-1.20)\\ & =0.8849-0.1151\\ & =0.7698\end{array}$$

Solution:

Comparing two population means

Many real-life questions involve comparing two population means. For example:

• Do students who graduate from school A earn more income than students from school B?
• Does tech company A offer higher salaries on average to entry-level graduates than tech company B?
• Is the life expectancy of people working in office jobs greater than that of people in jobs requiring physical labor?
• Does the average American earn more than the average Canadian?

In these situations, we focus on the sampling distribution of the difference in sample means: we repeatedly take a sample from each population, compute each sample mean, and subtract them.

Consider two populations with means ${\mu }_1$ and ${\mu }_2$ and standard deviations ${\sigma }_1$ and ${\sigma }_2$. Take samples of sizes $n_1$ and $n_2$ and find the difference ${\bar{x}}_1-{\bar{x}}_2$ for each sample pairing.

• The mean of the set of differences equals the difference in population means:

$$\begin{array}{rl}{\mu }_{{\bar{X}}_1-{\bar{X}}_2}& ={\mu }_1-{\mu }_2\end{array}$$

$$\begin{array}{rl}E[{\bar{X}}_1-{\bar{X}}_2]& ={\mu }_1-{\mu }_2\end{array}$$

• The standard deviation of the set of differences is approximately:

$$\begin{array}{rl}{\sigma }_{{\bar{X}}_1-{\bar{X}}_2}& =\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}\end{array}$$

• If the populations are normal, or if the sample sizes are at least $30$, the set of all differences of sample means is approximately normally distributed.

Practice problem

Example 3:

Two brands of lightbulbs are tested to see which lasts longer. Brand A has a mean lifespan of $1200$ hours with a standard deviation of $100$ hours, and Brand B has a mean of $1100$ hours with a standard deviation of $120$ hours. If we sample $50$ bulbs from each brand, what is the probability that the sample difference in average lifespan is at least $70$ hours?

We are given ${\mu }_1=1200$, ${\mu }_2=1100$, ${\sigma }_1=100$, ${\sigma }_2=120$, $n_1=n_2=50$. Since both sample sizes are at least $30$, the sampling distribution of ${\bar{X}}_1-{\bar{X}}_2$ is approximately normal.

Find the standard deviation of the differences:

$$\begin{array}{rl}{\sigma }_{{\bar{X}}_1-{\bar{X}}_2}& =\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}\\ & =\sqrt{\frac{100^2}{50}+\frac{120^2}{50}}\\ & =\sqrt{200+288}\\ & =\sqrt{488}\\ & \approx 22.09\end{array}$$

The mean of the sampling distribution is the difference in population means:

$$\begin{array}{rl}{\mu }_1-{\mu }_2& =1200-1100\\ & =100\end{array}$$

Compute the z-score for a sample difference of $70$ hours:

$$\begin{array}{rl}z& =\frac{70-100}{22.09}\\ & =\frac{-30}{22.09}\\ & \approx -1.358\end{array}$$

Find the probability of getting a difference of at least $70$ hours:

$$\begin{array}{rl}P(Z>-1.358)& =1-P(Z<-1.358)\\ & =1-0.0871\\ & =0.9129\end{array}$$

Solution: