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AI HL · AHL 8

Statistics 2

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AHL 8.5

The sample mean and its standard error

AHL 8.5

The sample mean and its standard error

Target: I can say why the sample mean \(\bar{X}\) is itself a random variable. (+4 more)

By the end of this section

Key points

The sample mean

  • \(\bar{X}=\dfrac{X_1+X_2+\dots+X_n}{n}\) is a random variable: it changes from sample to sample.
  • \(E(\bar{X})=E(X)\) and \(\mathrm{Var}(\bar{X})=\dfrac{\mathrm{Var}(X)}{n}\).
  • Divide the variance by \(n\). Divide the standard deviation by \(\sqrt{n}\).
  • The standard error is the standard deviation of \(\bar{X}\): \(SE=\dfrac{\sigma}{\sqrt{n}}\).

What it tells you

  • The variance of the sample mean answers: if I took many samples of size \(n\), how much would their means vary?
  • A larger sample gives means that cluster closer to \(\mu\), so the variance is smaller.
  • To halve the standard error you need four times as many observations.
  • \(n\) is the number of observations in the sample. It is not the number of trials in a binomial \(X\).

Finding \(E(X)\) and \(\mathrm{Var}(X)\) first

  • \(X\sim N(\mu,\sigma^2)\): the mean is \(\mu\) and the variance is \(\sigma^2\). The second number is the variance.
  • \(X\sim B(m,p)\): \(E(X)=mp\) and \(\mathrm{Var}(X)=mp(1-p)\).
  • From a table: \(E(X)=\sum xP(X=x)\). The probabilities must sum to 1, which is how you find an unknown \(k\).
  • \(E(aX+b)=aE(X)+b\) and \(\mathrm{Var}(aX+b)=a^2\mathrm{Var}(X)\), as in the last lesson.

Explore: the sample mean

Choose the shape of a population and a sample size n. Take samples, and each sample mean is added to the histogram. The curve is N(μ, σ²/n).

Samples taken
Mean of the sample means
μ, the population mean
Standard deviation of the sample means
σ/√n, the standard error

Take a sample to start.

  1. With n = 1, take 100 samples. What does the histogram look like? Why?
  2. Choose the two-humped population and make n larger. At what size does the histogram first look like the curve?
  3. Compare the mean of the sample means with μ, and their standard deviation with σ/√n. Do they agree better after 10 samples or after 1000?
  4. To halve the standard error, how many times larger must n be? Check it with the slider.

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Mean of a sample of jars

The mass of a jar of honey has mean 340 g and variance 22.5 g². A sample of 25 jars is taken and the mean mass \(\bar{X}\) is found. Find \(E(\bar{X})\) and \(\mathrm{Var}(\bar{X})\).

Working
  1. Name the sample mean

    \(\bar{X}=\dfrac{X_1+X_2+\dots+X_{25}}{25}\)

    \(\bar{X}\) is a random variable: a different sample gives a different mean.

  2. The expected value

    \(E(\bar{X})=E(X)=340\)

    The sample mean has the same expected value as one observation.

  3. The variance

    \(\mathrm{Var}(\bar{X})=\dfrac{22.5}{25}=0.9\)

    Divide the variance of \(X\) by the sample size.

Video walkthrough coming
A normal variable and the standard error

The height of a certain plant is modelled by \(X\sim N(68,\,14.4)\), in cm. The mean height \(\bar{X}\) of a sample of 12 plants is found. Find \(E(\bar{X})\), \(\mathrm{Var}(\bar{X})\) and the standard error of the mean.

Working
  1. Read the model

    \(E(X)=68\), \(\mathrm{Var}(X)=14.4\), \(n=12\)

    The second number in \(N(\mu,\sigma^2)\) is the variance, not the standard deviation.

  2. Mean and variance

    \(E(\bar{X})=68\), \(\mathrm{Var}(\bar{X})=\dfrac{14.4}{12}=1.2\)

    Same rules as before.

  3. Standard error

    \(SE=\sqrt{1.2}=1.10\) cm

    The standard error is the standard deviation of \(\bar{X}\): \(\sigma/\sqrt{n}=\sqrt{14.4}/\sqrt{12}\).

Video walkthrough coming
A binomial variable

\(X\sim B(20,\,0.3)\). The mean \(\bar{X}\) of a sample of 15 observations of \(X\) is found. Find \(E(\bar{X})\) and \(\mathrm{Var}(\bar{X})\).

Working
  1. Mean and variance of X first

    \(E(X)=np=20(0.3)=6\), \(\mathrm{Var}(X)=np(1-p)=20(0.3)(0.7)=4.2\)

    The 20 is the number of trials in \(X\). The 15 is the number of observations in the sample. Keep them apart.

  2. The sample mean

    \(E(\bar{X})=6\), \(\mathrm{Var}(\bar{X})=\dfrac{4.2}{15}=0.28\)

    Now use the sample-mean rules with \(n=15\).

Video walkthrough coming
Finding k

The random variable \(W\) has \(P(W=w)=\dfrac{k}{w}\) for \(w=2,\,3,\,4,\,6\). Find the value of \(k\).

Working
  1. Write the probabilities

    \(P(W=2)=\tfrac k2,\ P(W=3)=\tfrac k3,\ P(W=4)=\tfrac k4,\ P(W=6)=\tfrac k6\)

    Substitute each value of \(w\).

  2. They sum to 1

    \(k\left(\tfrac12+\tfrac13+\tfrac14+\tfrac16\right)=1\), so \(\tfrac{5}{4}k=1\)

    \(\tfrac12+\tfrac13+\tfrac14+\tfrac16=\tfrac{6+4+3+2}{12}=\tfrac{15}{12}\).

  3. Solve

    \(k=0.8\)

    Check: the four probabilities are 0.4, 0.267, 0.2, 0.133, and they total 1.

Video walkthrough coming
From a table to a transformation

The table shows the probability distribution of \(X\). The random variable \(Y\) is given by \(Y=4X-3\). Find \(E(X)\) and \(E(Y)\). Given that \(\mathrm{Var}(X)=3.79\), find \(\mathrm{Var}(Y)\).

X2468P(X = x)0.150.350.30.2
Working
  1. E(X) from the table

    \(E(X)=2(0.15)+4(0.35)+6(0.3)+8(0.2)=5.1\)

    \(E(X)=\sum xP(X=x)\): multiply each value by its probability, then add.

  2. The mean of Y

    \(E(Y)=4E(X)-3=4(5.1)-3=17.4\)

    Use \(E(aX+b)=aE(X)+b\).

  3. The variance of Y

    \(\mathrm{Var}(Y)=4^2(3.79)=60.64\)

    Use \(\mathrm{Var}(aX+b)=a^2\mathrm{Var}(X)\). The \(-3\) drops out.

Video walkthrough coming