A manufacturer of snack mix would like to know whether its bag filling machine works correctly at the 450 gram setting. It is believed that the machine is overfilling the bags. A 8 bag sample had a mean of 460 grams with a variance of 196. 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) = 450 $
The Sample Mean $ (\bar{x}) = 460 $
The Sample Variance $ (s^2) = 196 $
The Sample Size $ (n) = 8 $

$\therefore$ The Sample Standard Deviation $ (s) = \sqrt{196} = 14 $
The Significance Level $ (\alpha) = 0.02 $

Solution:
The null and alternative hypothesis:
$ H_0: \mu = 450 $
$ H_1: \mu > 450 $

The test statistic $ (t): $
$t = \frac{\bar{x} – \mu}{\frac{s}{\sqrt{n}}} $
$ = \frac{460 – 450}{\frac{14}{\sqrt{8}}} $

$ = 2.02 $

The degree of freedom $ (df): $
$ df = n – 1 $
$ = 8 – 1 $
$ = 7 $

The p-value:
$ \text{p-value} = \text{P}(t_{7} > 2.02) $

$ = 0.0415 $

The conclusion:
The p-value is greater than the significance level. Therefore, we fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the bags are overfilled.

Final Answer:
The null and alternative hypothesis:
$ H_0: \mu = 450 $
$ H_1: \mu > 450 $

The test statistic $ (t) = 2.02 $

The p-value $ = 0.0415 $

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