A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 500 gram setting. It is believed that the machine is overfilling the bags. A 12 bag sample had a mean of 510 grams with a variance of 144. Assume the population is normally distributed. Is there sufficient evidence at the 0.05 level that the bags are overfilled?

Answer:
Given:
The Hypothesized Mean $ (\mu) = 500 $
The Sample Mean $ (\bar{x}) = 510 $
The Sample Variance $ (s^2) = 144 $

$\therefore$ The Sample Standard Deviation $ (s) = \sqrt{144} = 12 $
The Significance Level $ (\alpha) = 0.05 $

Solution:
The null and alternative hypothesis:
$ H_0: \mu = 500 $
$ H_1: \mu > 500 $

The test statistic $ (t): $
$t = \frac{\bar{x} – \mu}{\frac{s}{\sqrt{n}}} $
$ = \frac{510 – 500}{\frac{12}{\sqrt{12}}} $

$ = 2.887 $

The degree of freedom $ (df): $
$ df = n – 1 $
$ = 12 – 1 $
$ = 11 $

The p-value:
$ \text{p-value} = \text{P}(t_{11} > 2.887) $

$ = 0.0074 $

The conclusion:
The p-value is less than the significance level. Therefore, we reject the null hypothesis. There is sufficient evidence to support the claim that the bags are overfilled.

Final Answer:
The null and alternative hypothesis:
$ H_0: \mu = 500 $
$ H_1: \mu > 500 $

The test statistic $ (t) = 2.886 $

The p-value $ = 0.0076 $

The conclusion:
The p-value is less than the significance level. Therefore, we reject the null hypothesis. There is sufficient evidence to support the claim that the bags are overfilled.