A soft-drink dispensing machine is said to be out of control if the variance of the contents exceeds 1.5 dl. If a random sample of 25 drinks from this machine has a variance of 2.03 dl, does this machine indicate at the 0.05 level of significance that the machine is out of control?

#### Solution

Given that the sample size $n = 25$ and sample standard deviation $s = 2.03$.

**Hypothesis Problem**

The hypothesis testing problem is

$H_0 : \sigma = 1.5$ against $H_1 : \sigma > 1.5$ ($\text{right-tailed}$)

**Test Statistic**

The test statistic for testing above hypothesis testing problem is

$$

\chi^2 =\frac{(n-1)s^2}{\sigma^2}

$$

**Level of Significance**

The significance level is $\alpha = 0.05$.

**Critical Value **

As the alternative hypothesis is **$\text{right-tailed}$**, the critical value of $\chi^2$ at $24$ degrees of freedom $\text{ is }$ $\text{36.415}$ (from $\chi^2$ statistical table).

The rejection region (i.e. critical region) is $\chi^2 > 36.415$.

**Test Statistic**

The test statistic under the null hypothesis is

$$

\begin{aligned}

\chi^2 &=\frac{(n-1)s^2}{\sigma^2_0}\

& = \frac{(25-1)*(2.03)^2}{(2.25)}\

&= 43.956

\end{aligned}

$$

**Decision**

**Critical value approach**

The test statistic is $\chi^2 =43.956$ which falls $inside$ the critical region, we **$\text{reject}$** the null hypothesis.

**$p$-value Approach**

This is a **$\text{right-tailed}$** test, so the p-value is the area to the left of the test statistic ($\chi^2=43.956$) is p-value = $0.0077$.

The p-value is $0.0077$ which is **$\text{less than}$** the significance level of $\alpha = 0.05$, we **$\text{reject}$** the null hypothesis.

Thus at 0.05 level of significance we conclude that the machine is out of control.

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators