Answer:
No, there is not sufficient evidence to support the claim that the bags are underfilled
Step-by-step explanation:
A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 434 gram setting.
This means that the null hypothesis is:
![H_{0}: \mu = 434](https://tex.z-dn.net/?f=H_%7B0%7D%3A%20%5Cmu%20%3D%20434)
It is believed that the machine is underfilling the bags.
This means that the alternate hypothesis is:
![H_{a}: \mu < 434](https://tex.z-dn.net/?f=H_%7Ba%7D%3A%20%5Cmu%20%3C%20434)
The test statistic is:
![z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
In which X is the sample mean,
is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
434 is tested at the null hypothesis:
This means that ![\mu = 434](https://tex.z-dn.net/?f=%5Cmu%20%3D%20434)
A 9 bag sample had a mean of 431 grams with a variance of 144.
This means that ![X = 431, n = 9, \sigma = \sqrt{144} = 12](https://tex.z-dn.net/?f=X%20%3D%20431%2C%20n%20%3D%209%2C%20%5Csigma%20%3D%20%5Csqrt%7B144%7D%20%3D%2012)
Value of the test-statistic:
![z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
![z = \frac{431 - 434}{\frac{12}{\sqrt{9}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B431%20-%20434%7D%7B%5Cfrac%7B12%7D%7B%5Csqrt%7B9%7D%7D%7D)
![z = -0.75](https://tex.z-dn.net/?f=z%20%3D%20-0.75)
P-value of the test:
The pvalue of the test is the pvalue of z = -0.75, which is 0.2266
0.2266 > 0.01, which means that there is not sufficient evidence to support the claim that the bags are underfilled.