A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 405 gram setting. It is believed that the machine is overfilling the bags. A 15 bag sample had a mean of 412 grams with a variance of 169. 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 $(\mu )=405$
The Sample Mean $(\overline{x})=412$
The Sample Variance $(s^2)=169$
The Sample Size $(n)=15$

$\therefore$ The Sample Standard Deviation $(s)=\sqrt{169}=13$
The Significance Level $(\alpha )=0.05$

Solution:
The null and alternative hypothesis:
$H_0:\mu =405$
$H_1:\mu >405$

The test statistic $(t):$
$t=\frac{\overline{x}–\mu }{\frac{s}{\sqrt{n}}}$
$=\frac{412–405}{\frac{13}{\sqrt{15}}}$

$=2.085$

The degree of freedom $(df):$
$df=n–1$
$=15–1$
$=14$

The p-value:
$\text{p-value}=\text{P}(t_{14}>2.085)$

$=0.0279$

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:
$H_0:\mu =405$
$H_1:\mu >405$

The test statistic $(t)=2.085$

The p-value $=0.0279$

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