6.2 Tests on a normal population

We assume in this section that the population rv $X$ has distribution $\mathcal{N}(\mu,\sigma^2),$ where both $\mu$ and $\sigma^2$ are unknown. We will test hypotheses about one of the two parameters from a srs $(X_1,\ldots,X_n)$ of $X.$ The sampling distributions obtained in Section 2.2 are key for obtaining the critical regions of the forthcoming tests.

6.2.1 Tests about the mean

One-sided right test

We want to test the hypothesis⁷⁷

\[\begin{align*} H_0:\mu=\mu_0 \quad \text{vs.}\quad H_1:\mu > \mu_0. \end{align*}\]

As seen in the previous section, we must fix first the bound $\alpha$ for the probability of Type I error, that is,

\[\begin{align} \mathbb{P}(\text{Reject $H_0$}|\text{$H_0$ true})\leq \alpha.\tag{6.1} \end{align}\]

To obtain the critical region, we need a statistic $T(X_1,\ldots,X_n).$ If we take $\bar{X},$ whose distribution is $\mathcal{N}(\mu_0,\sigma^2/n),$ then we should reject the null hypothesis for large values of $\bar{X}.$ Therefore, the critical region has the form

\[\begin{align*} C=\{(x_1,\ldots,x_n)'\in \mathbb{R}^n: \bar{X}>k\}, \end{align*}\]

for a given value $k.$ We can compute the constant $k$ in such a way that verifies (6.1), that is, such that

\[\begin{align} \mathbb{P}(\bar{X}>k|\mu=\mu_0)\leq \alpha.\tag{6.2} \end{align}\]

But there are infinitely many values $k$ that satisfy the above relation. For example, if we consider a value $k_1$ that is not the smallest one, then there will be another value $k_2<k_1$ such that

\[\begin{align*} \mathbb{P}(\bar{X}>k_1|\mu=\mu_0)\leq \mathbb{P}(\bar{X}>k_2|\mu=\mu_0)\leq \alpha. \end{align*}\]

Then, it will happen that the probability of Type II error of the test with critical region $\{\bar{X}>k_1\}$ will be larger than the one for the test with critical region $\{\bar{X}>k_2\}.$ Therefore, among the tests with critical region of the type $\{\bar{X}>k\}$ that verify (6.2), the most efficient is the one with smallest $k.$

However, recall that in this case it is not possible to determine the smallest $k$ that verifies (6.2) since the distribution of $\bar{X}$ is partially unknown ($\sigma^2$ is unknown). But if we estimate $\sigma^2$ with $S'^2,$ then a test statistic is

\[\begin{align} T(X_1,\ldots,X_n)=\frac{\bar{X}-\mu_0}{S'/\sqrt{n}}\stackrel{H_0}{\sim} t_{n-1}.\tag{6.3} \end{align}\]

Determining the critical region from $T$ is simple, since the range of $T$ is $\mathbb{R}$ and therefore the critical region is an interval of $\mathbb{R}.$ The null hypothesis is to be rejected for large values of $T,$ hence the critical region is of the form $\{T(x_1,\ldots,x_n)>k\}.$ We must select the smallest $k$ such that

\[\begin{align*} \mathbb{P}(T>k|\mu=\mu_0)\leq \alpha. \end{align*}\]

From here, it is deduced that $k=t_{n-1;\alpha}.$ Therefore, the critical region is

\[\begin{align*} C &=\{(x_1,\ldots,x_n)'\in\mathbb{R}^n: T(x_1,\ldots,x_n)>t_{n-1;\alpha}\} \\ &=\left\{(x_1,\ldots,x_n)'\in\mathbb{R}^n: \frac{\bar{X}-\mu_0}{S'/\sqrt{n}}>t_{n-1;\alpha} \right\} \\ &=\left\{(x_1,\ldots,x_n)'\in\mathbb{R}^n: \bar{X}>\mu_0+t_{n-1;\alpha}\frac{S'}{\sqrt{n}}\right\}. \end{align*}\]

Observe that the quantity $t_{n-1;\alpha}\frac{S'}{\sqrt{n}}$ can be regarded as the “uncertainty tolerance” that is added to the no-uncertainty rule that would reject $H_0:\mu=\mu_0$ if $\bar{X}>\mu_0.$

One-sided left test

If the null hypothesis to test is

\[\begin{align*} H_0:\mu=\mu_0 \quad \text{vs.}\quad H_1:\mu < \mu_0, \end{align*}\]

then the test statistic is exactly the same as before, but in this case we must reject the null hypothesis for small values of $T.$ Therefore, the critical region has the form $\{T<k\}.$ Fixing a significance level $\alpha$ and selecting the larger $k$ that verifies the relation

\[\begin{align*} \mathbb{P}(T<k|\mu=\mu_0)\leq \alpha \end{align*}\]

we get that the value of $k$ is $k=-t_{n-1;\alpha}$ and the critical region is

\[\begin{align*} C=\{(x_1,\ldots,x_n)'\in\mathbb{R}^n : T(x_1,\ldots,x_n)<-t_{n-1;\alpha}\}. \end{align*}\]

Two-sided test

Now we want to test the hypothesis

\[\begin{align*} H_0:\mu=\mu_0 \quad \text{vs.}\quad H_1:\mu\neq \mu_0. \end{align*}\]

For the same test statistic (6.3), now we will reject for large absolute values of $T$ that will indicate deviation from $H_0.$ That is, the critical region has the form

\[\begin{align*} \{(x_1,\ldots,x_n)'\in\mathbb{R}^n: T(x_1,\ldots,x_n)\in(-\infty,k_1)\cup (k_2,\infty)\}. \end{align*}\]

Now we must determine the value of the two constants $k_1$ and $k_2$ in such a way that

\[\begin{align*} \mathbb{P}(T\in(-\infty,k_1)\cup (k_2,\infty)|\mu=\mu_0)\leq \alpha. \end{align*}\]

Evenly splitting $\alpha/2$ to both tails, since the distribution of $T$ under $H_0$ is $t_{n-1}$ and is symmetric, gives $k_2=t_{n-1;\alpha/2}$ and $k_1=-t_{n-1;\alpha/2}.$ Then, the critical region is

\[\begin{align*} C=\{(x_1,\ldots,x_n)'\in\mathbb{R}^n : |T(x_1,\ldots,x_n)|>t_{n-1;\alpha/2}\}. \end{align*}\]

Example 6.5 Eight bullets made of a new type of gunpowder were fired in a gun, and their initial speeds in m/s were measured:

\[\begin{align*} 910,\quad 921,\quad 918,\quad 896,\quad 909,\quad 905,\quad 895,\quad 903. \end{align*}\]

The producer of the gunpowder claims that the new gunpowder delivers an average initial speed above $915$ m/s. Is the sample providing evidence against such claim, at the significance level $\alpha=0.025$? Assume that the initial speeds follow a normal distribution.

We want to test the hypothesis

\[\begin{align*} H_0:\mu=915\quad \text{vs.}\quad H_1:\mu<915. \end{align*}\]

The critical region is

\[\begin{align*} C=\{T<-t_{7;0.025}\approx-2.365\}. \end{align*}\]

The observed value of the statistic is

\[\begin{align*} T=\frac{\bar{X}-\mu_0}{S'/\sqrt{n}}\approx\frac{907.125-915}{9.3723/\sqrt{8}}\approx -2.377<-2.365. \end{align*}\]

Then, the observed initial speeds are sufficiently low to question the claim of the gunpowder producer.

Recall that $t_{7;0.025}$ can be computed as

qt(0.025, df = 7, lower.tail = FALSE)
## [1] 2.364624

The R function t.test() implements the (one-sample) test of $H_0:\mu=\mu_0$ against different alternatives. The main arguments of the function are as follows:

t.test(x, alternative = c("two.sided", "less", "greater"), mu = 0, ...)

The argument mu stands for $\mu_0.$ The table below shows the encoding of the alternative argument:

`alternative`	`"two.sided"`	`"less"`	`"greater"`
$H_1$	$\mu\neq\mu_0$	$\mu<\mu_0$	$\mu>\mu_0$

Example 6.6 The t.test() solution to Example 6.5 is very simple:

# Apply t.test() with H1: mu < 915
t.test(x = c(910, 921, 918, 896, 909, 905, 895, 903), mu = 915,
       alternative = "less")
## 
##  One Sample t-test
## 
## data:  c(910, 921, 918, 896, 909, 905, 895, 903)
## t = -2.3766, df = 7, p-value = 0.02456
## alternative hypothesis: true mean is less than 915
## 95 percent confidence interval:
##      -Inf 913.4029
## sample estimates:
## mean of x 
##   907.125

That the reported $p$-value is smaller than $\alpha=0.025$ indicates rejection of $H_0,$ as seen in Section 6.5.

6.2.2 Tests about the variance

We study in this section the tests for the following hypotheses:

$H_0:\sigma^2=\sigma_0^2$ vs. $H_1:\sigma^2>\sigma_0^2;$
$H_0:\sigma^2=\sigma_0^2$ vs. $H_1:\sigma^2<\sigma_0^2;$
$H_0:\sigma^2=\sigma_0^2$ vs. $H_1:\sigma^2\neq\sigma_0^2.$

An estimator of $\sigma^2$ is the sample quasivariance, $S'^2,$ for whom we perfectly know its distribution under the null hypothesis,

\[\begin{align*} U=\frac{(n-1)S'^2}{\sigma_0^2}\stackrel{H_0}\sim \chi_{n-1}^2. \end{align*}\]

Therefore, $U$ is a test statistic for the testing problems a, b, and c.

When testing case a, if the null hypothesis is not true, $U$ will tend to have large values. Therefore, the critical region is of the form $\{U>k\}.$ For obtaining the best value $k,$ we select the significance level $\alpha$ and take the smallest $k$ such that

\[\begin{align*} \mathbb{P}(U>k|\sigma^2=\sigma_0^2)\leq \alpha. \end{align*}\]

Then, the best choice is $k=\chi_{n-1;\alpha}^2,$ so the critical region is

\[\begin{align*} C_a=\{(x_1,\ldots,x_n)'\in\mathbb{R}^n:U(x_1,\ldots,x_n)>\chi_{n-1;\alpha}^2\}. \end{align*}\]

With an analogous reasoning, the critical region for b follows:

\[\begin{align*} C_b=\{(x_1,\ldots,x_n)'\in\mathbb{R}^n:U(x_1,\ldots,x_n)<\chi_{n-1;1-\alpha}^2\}. \end{align*}\]

The critical region for c arises from splitting evenly the probability $\alpha$ in the critical regions of a and b:

\[\begin{align*} C_c=\{(x_1,\ldots,x_n)'\in\mathbb{R}^n:U(x_1,\ldots,x_n)>\chi_{n-1;\alpha/2}^2 \ \text{or}\ U(x_1,\ldots,x_n)<\chi_{n-1;1-\alpha/2}^2\}. \end{align*}\]

Example 6.7 A company claims that the diameter of one of the parts of an engine it produces has a production variance not larger than $0.1290$ squared millimeters. A srs of $10$ pieces revealed $S'^2=0.1935.$ Assuming that the measurements of the diameter follow a normal distribution, test at a significance level $\alpha=0.05$

\[\begin{align*} H_0:\sigma^2=0.1290\quad \text{vs.} \quad H_1:\sigma^2>0.1290. \end{align*}\]

The test statistic is

\[\begin{align*} U=\frac{(n-1)S'^2}{\sigma_0^2}=\frac{9\times 0.1935}{0.1290}=13.5 \end{align*}\]

and the rejection region is

\[\begin{align*} C=\{U>\chi_{9;0.05}^2\approx16.919\}. \end{align*}\]

Since $U=13.5<16.919,$ then the data does not provide any evidence against the variance of the diameter being larger than $0.1290$ squared milliliters. Therefore, there is no evidence against the company’s claim.

Unfortunately, there is no function in base R to conduct the (one-sample) variance test.

In some textbooks it is sometimes written $H_0:\mu\leq\mu_0$ vs. $H_1:\mu > \mu_0.$ However, this notation introduces inconsistencies when determining the distribution of the test statistic under $H_0,$ as then several values for $\mu$ are possible. For that reason it is avoided.↩︎