A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 408 gram setting. It is believed that the machine is overfilling the bags. A 13 bag sample had a mean of 416 grams with a variance of 169. Assume the population is normally distributed. Is there sufficient evidence at the 0.01 level that the bags are overfilled?

Answer:
Given:
The Hypothesized Mean $(\mu )=408$
The Sample Mean $(\overline{x})=416$
The Sample Variance $(s^2)=169$
The Sample Size $(n)=13$

$\therefore$ The Sample Standard Deviation $(s)=\sqrt{169}=13$
The Significance Level $(\alpha )=0.01$

Solution:
The null and alternative hypothesis:
$H_0:\mu =408$
$H_1:\mu >408$

The test statistic $(t):$
$t=\frac{\overline{x}–\mu }{\frac{s}{\sqrt{n}}}$
$=\frac{416–408}{\frac{13}{\sqrt{13}}}$

$=2.219$

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

The p-value:
$\text{p-value}=\text{P}(t_{12}>2.219)$

$=0.0233$

The conclusion:
The p-value is greater than the significance level. Therefore, we fail to reject the null hypothesis. There is not sufficient evidence at the 0.01 level to support the claim that the bags are overfilled.

Final Answer:
The null and alternative hypothesis:
$H_0:\mu =408$
$H_1:\mu >408$

The test statistic $(t)=2.219$

The p-value $=0.0233$

The conclusion:
The p-value is greater than the significance level. Therefore, we fail to reject the null hypothesis. There is not sufficient evidence at the 0.01 level to support the claim that the bags are overfilled.