Tests of center

Two tests of scale

1. Chi-square test of variance

2. t-test of variance

Chi-square test of variance

Population: \(Y \sim\) some population distribution

Sample: n i.i.d from population, \(Y_1, \ldots, Y_n\)

Parameter: Population variance \(\sigma^2 = Var(Y)\)

Sample variance

The sample variance, \[ s^2 = \frac{1}{n-1}\sum_{i= 1}^n \left(Y_i - \overline{Y} \right)^2 \]

is an unbiased and consistent estimate of \(\sigma^2\).

Furthermore, if \(Y \sim N(\mu, \sigma)\), it can be shown the sampling distribution of \(s^2\) is a scaled Chi-square distribution: \[ (n-1)\frac{s^2}{\sigma^2} \sim \chi^2_{(n-1)} \]

Using the sampling distribution to formulate a test

Assume the population distribution is \(N(\mu, \sigma^2)\).

Consider the null hypothesis \(H_0: \sigma^2 = \sigma^2_0\).

Let the test statistic be: \[ X(\sigma_0^2) = (n-1)\frac{s^2}{\sigma_0^2} \] What’s the distribution of the test statistic if the null hypothesis is true?

Rejection regions

For a test at level \(\alpha\):

• \(H_A: \sigma^2 > \sigma_0^2\): Reject \(H_0\) if \(X(\sigma_0^2) >\)
• \(H_A: \sigma^2 < \sigma_0^2\): Reject \(H_0\) if \(X(\sigma_0^2) <\)
• \(H_A: \sigma^2 \ne \sigma_0^2\): Reject \(H_0\) if \(X(\sigma_0^2) >\)
  or \(X(\sigma_0^2) <\)

p-values: Your turn

Shade the area for the p-value when \(X(\sigma_0) = 20\), with \(H_A: \sigma^2 > \sigma_0^2\)

Shade the area for the p-value when \(X(\sigma_0) = 20\), with \(H_A: \sigma^2 < \sigma_0^2\)

p-values: Your turn

Shade the area for the p-value when \(X(\sigma_0) = 20\), with \(H_A: \sigma^2 \ne \sigma_0^2\)

p-values: In general

p-value is:

• \(H_A: \sigma^2 > \sigma_0^2\):
• \(H_A: \sigma^2 < \sigma_0^2\):
• \(H_A: \sigma^2 \ne \sigma_0^2\):

where \(X \sim \chi^2_{(n-1)}\)

In R: \(P(X \le x)\) = pchisq(x, df = n - 1)

Confidence interval

From inverting test statistic.

\((1 - \alpha)100\)% confidence interval

\[ \left( \frac{s^2(n - 1)}{\chi^2_{(n-1)}(1 - \alpha/2)} , \frac{s^2(n - 1)}{\chi^2_{(n-1)}(\alpha/2)}\right) \]

What if the population isn’t Normal

(from Sarah Emerson’s slides F2016)

t-test of variance

An alternative to the Chi-square test of variance, based on considering a transformed response:

\[ Z_i = \left( Y_i - \overline{Y}\right)^2 \quad i = 1, \ldots, n \]

What is \(E(Z)\), are the \(Z_i\) independent?

A CLT for the sample variance

The \(Z_i\) aren’t independent but are weakly dependent, turns out there is a CLT for this case, as long as \(Z\) has finite fourth moment:

\[ \frac{\overline{Z} - E(Z)}{\sqrt{Var(Z)/n}} \rightarrow_d N(0, 1) \]

Substitute in for \(E(Z)\) \[ \frac{\overline{Z} - \frac{n - 1}{n}\sigma^2}{\sqrt{Var(Z)/n}} \rightarrow_d N(0, 1) \]

Leads to a t-test

We don’t know the population variance of the \(Z\) (transformed \(Y\)), so substitute sample estimate for it.

Under null hypothesis \(H_0: \sigma^2 = \sigma_0^2\)

\[ t(\sigma_0^2) = \frac{\overline{Z} - \frac{n-1}{n} \sigma_0^2 }{\sqrt{s_{Z}^2/n}} \, \dot \sim \, t_{(n-1)} \]

So to test the null \(H_0: \sigma^2 = \sigma_0^2\), do a t-test on \(Z_i = \left(Y_i - \overline{Y}\right)^2\) with the null hypothesis \(H_0: \mu_Z = \frac{n-1}{n}\sigma_0^2\).

Performance of t-test of variance

Charlotte’s simulations, rejection rate of \(H_0\) for \(\alpha = 0.05\).