A manufacturer of candy would like to know whether its bag filling machine works correctly at the 300 gram setting. It is believed that the machine is overfilling the bags. A 13 bag sample had a mean of 308 grams with a variance of 81. Assume the population is normally distributed. Is there sufficient evidence at the 0.02 level that the bags are overfilled?

Answer:
Given:
The Hypothesized Mean $ (\mu) = 300 $
The Sample Mean $ (\bar{x}) = 308 $
The Sample Variance $ (s^2) = 81 $
The Sample Size $ (n) = 13 $

$\therefore$ The Sample Standard Deviation $ (s) = \sqrt{81} = 9 $
The Significance Level $ (\alpha) = 0.02 $

Solution:
The null and alternative hypothesis:
$ H_0: \mu = 300 $
$ H_1: \mu > 300 $

The test statistic $ (t): $
$t = \frac{\bar{x} – \mu}{\frac{s}{\sqrt{n}}} $
$ = \frac{308 – 300}{\frac{9}{\sqrt{13}}} $

$ = 3.205 $

The degree of freedom $ (df): $
$ df = n – 1 $
$ = 13 – 1 $
$ = 12 $

The p-value:
$ \text{p-value} = \text{P}(t_{12} > 3.205) $

$ = 0.0038 $

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 = 300 $
$ H_1: \mu > 300 $

The test statistic $ (t) = 3.205 $

The p-value $ = 0.0038 $

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.