A machine that fills beverage cans is supposed to put 12 ounces of beverage in each can. The standard deviation of the amount in each can is 0.12 ounce. The machine is overhauled with new Components, and ten cans are filled to determine whether the standard deviation has changed. Assume the fill Amounts to be a random sample from a normal population.
$$\begin{array} { l l l l l }
\hline 12.03 & 11.97 & 12.10 & 11.63 & 11.83 \
11.87 & 12.12 & 11.47 & 11.95 & 11.87 \
\hline
\end{array}$$
Perform a hypothesis test to determine whether the standard deviation differs from 0.12 ounce. Use the $\alpha = 0.01$ level of significance.
A) Reject $H _ { 0 } .$
B) Do not reject $H _ { 0 }$

Question

A machine that fills beverage cans is supposed to put 12 ounces of beverage in each can. The standard deviation of the amount in each can is 0.12 ounce. The machine is overhauled with new Components, and ten cans are filled to determine whether the standard deviation has changed. Assume the fill Amounts to be a random sample from a normal population. 
$$\begin{array} { l l l l l } 
\hline 12.03 & 11.97 & 12.10 & 11.63 & 11.83 \
11.87 & 12.12 & 11.47 & 11.95 & 11.87 \
\hline
\end{array}$$ 
Perform a hypothesis test to determine whether the standard deviation differs from 0.12 ounce. Use the $\alpha = 0.01$ level of significance.
A) Reject $H _ { 0 } .$
B) Do not reject $H _ { 0 }$

Quizplus · Accepted Answer

Step 1: State the hypotheses:

The null hypothesis is that the standard deviation is equal to 0.12 ounce.
The alternative hypothesis is that the standard deviation is not equal to 0.12 ounce.

$H_0: \sigma = 0.12$
$H_1: \sigma \neq 0.12$

Step 2: Calculate the test statistic:

$$s=\sqrt{\frac{\sum_{i=1}^{n}(x_i-\bar{x})^2}{n-1}} = 0.181$$

$$ \chi^2 = \frac{(n-1)s^2}{\sigma^2} = \frac{(10-1)(0.181)^2}{(0.12)^2} = 23.5875$$

Step 3: Determine the p-value:

Since this is a two-tailed test, the p-value is the probability of getting a chi-square statistic greater than or equal to 23.5875 or less than or equal to 2.7001 (the critical value with 9 degrees of freedom at the 0.01 level of significance).

$$p= 2 P(Chi \geq 23.5875)= 2 (1 - P(Chi \leq 2.7001)) = 0.0002 < \alpha$$

Step 4: Make a decision and interpret the results:

Since the p-value is less than the level of significance, we reject the null hypothesis. There is sufficient evidence to conclude that the standard deviation differs from 0.12 ounce. Therefore, we can conclude that the overhaul of the machine with new components has had an effect on the standard deviation of the amount of beverage in each can.