# Solved: 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

#### ByDr. Raju Chaudhari

Feb 23, 2021

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.