Suppose that $X_1,X_2,…,Xn$ is a random sample from the normal distribution with mean $\mu$ and variance $\sigma^2$.

Derive a test of size/level $\alpha$ for

\[H_0:\sigma^2\ge\sigma_0^2\quad\text{vs.}\quad H_1:\sigma^2\lt\sigma_0^2\] Step One:

Choose a statistic/estimator for $\sigma^2$.

\[S^2=\frac{\sum_{i=1}^{n}(X_i-\overline{X})^2}{n-1}\] Step Two:

Give the form of the test.

Reject $H_0$, in favor of $H_1$, if

\[S^2\lt c\]

for some $c$ to be determined.

Step Three:

Find $c$ using $\alpha$.

\[\begin{align} \alpha&=\text{max}P(\text{Type I Error})\\ &=\underset{\sigma^2\ge\sigma_0^2}{\text{max}}P(\text{Reject }H_0;\sigma^2)\\ &=\underset{\sigma^2\ge\sigma_0^2}{\text{max}}P(S^2\lt c;\sigma^2)\\ &= P\Bigg(\underbrace{\frac{(n-1)S^2}{\sigma^2}}_{\color{red}{\chi^2(n-1)}}\lt\frac{(n-1)c}{\sigma^2};\sigma^2\Bigg)\\ &= P\Bigg(W\lt\frac{(n-1)c}{\sigma^2};\sigma^2\Bigg)\\ &\text{Where }W\sim\chi^2(n-1). \end{align}\]

We have

\[\alpha=\underset{\sigma^2\ge\sigma_0^2}{\text{max}}\underbrace{P\Bigg(W\lt\underbrace{\frac{(n-1)c}{\sigma^2}}_{\color{red}{\text{decreasing in }\sigma^2}}\Bigg)}_{\color{red}{\text{decreasing in }\sigma^2}}\\\]

We know the probability decreasing in $\sigma^2$, to maximize with all the $\sigma^2\ge\sigma_0^2$, we just plug in $\sigma^2=\sigma_0^2$.

\[\begin{align} \alpha&=P\Bigg(W\lt\frac{(n-1)c}{\sigma_0^2}\Bigg)\\ \end{align}\]

png

\[\Downarrow\\ \frac{(n-1)c}{\sigma_0^2}=\chi^2_{1-\alpha,n-1}\] Step Four:

Conclusion.

Reject $H_0$, in favor of $H_1$, if

\[\color{red}{ S^2\lt\frac{\sigma_0^2\chi^2_{1-\alpha,n-1}}{n-1}}\]

Example

A lawn care company has developed and wants to patent a new herbicide applicator spray nozzle.

For safety reasons, they need to ensure that the application is consistent and not highly variable.

The company selected a random sample of 10 nozzles and measured the application rate of the herbicide in gallons per acre.

The measurements were recorded as

\[0.213, 0.185, 0.207, 0.163, 0.179\\ 0.161, 0.208, 0.210, 0.188, 0.195\\\]

Assuming that the application rates are normally distributed, test the following hypotheses at level $0.04$.

\[H_0:\sigma^2=0.01\quad H_1:\sigma^2>0.01\]

Reject $H_0$, in favor of $H_1$, if $S^2\gt c$.

\[\begin{align} \alpha&=P(S^2\gt c;\sigma^2=0.01)\\ &=P\Bigg(\frac{(n-1)S^2}{\sigma^2}\gt\frac{9c}{0.01};\sigma^2=0.01\Bigg)\\ &=P\Bigg(W\gt\frac{9c}{0.01}\Bigg)\quad(\text{where }W\sim\chi^2(9))\\ \end{align}\]

So

\[0.04=P\Bigg(W\gt\frac{9c}{0.01}\Bigg)\\ \huge{\Downarrow}\\ \frac{9c}{0.01}=\chi^2_{0.04, 9} = 17.61\]

In R:

qchisq(1-0.04, 9)

x <- c(0.213, 0.185, 0.207, 0.163, 0.179, 0.161, 0.208, 0.210, 0.188, 0.195)

# Compute variance
sigma_sq = var(x)

# Or
sigma_sq = (sum(x^2)-(sum(x)^2)/10)/9

sigma_sq

0.000364322222222219

Reject $H_0$, in favor of $H_1$, if $S^2\gt c$.

\[\begin{align} &c = (17.61)(0.01)/9\approx0.0196\\ &s^2= 0.000364 \end{align}\]

Fail to reject $H_0$, in favor of $H_1$, at level $0.04$. There is not sufficient evidence in the data to suggest that $\sigma^2\gt0.01$.

# install.packages("EnvStats")
# library(EnvStats)

varTest(x, alternative = "greater", conf.level = 1-0.04, 
    sigma.squared = 0.01, data.name = NULL)

$statistic
Chi-Squared 
    0.32789 

$parameters
df 
 9 

$p.value
[1] 0.9999951

$estimate
    variance 
0.0003643222 

$null.value
variance 
    0.01 

$alternative
[1] "greater"

$method
[1] "Chi-Squared Test on Variance"

$data.name
[1] "x"

$conf.int
         LCL          UCL 
0.0001862136          Inf 
attr(,"conf.level")
[1] 0.96

attr(,"class")
[1] "htestEnvStats"