## 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:

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

1. 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.↩︎