Question

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 419 gram setting. It is believed that the machine is underfilling the bags. A 15 bag sample had a mean of 409 grams with a standard deviation of 22. A level of significance of 0.05 will be used. Assume the population distribution is approximately normal. Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.

Answer 1

Answer:

We reject the null hypothesis, and accept the alternative hypothesis.

So, it means that the machine is under filling the bags

Step-by-step explanation:

Given

Null Hypothesis

$H_0:\mu = 419$ --- works correctly at 419g

Alternate Hypothesis

$H_0:\mu < 419$ --- under fills the bag

$n =15$ --- the sample

$\bar x = 409$ --- the mean

$\sigma = 22$ --- the standard deviation

$\alpha = 0.05$ --- the significance level

Required

What is the decision rule

Since n is less than 30, we make use of t test

$t = \frac{\bar x - \mu}{\sigma/\sqrt n}$

$t = \frac{409 - 419}{22/\sqrt{15}}$

$t = \frac{-10}{22/ 3.87}$

$t = \frac{-10}{5.58}$

$t = -1.792$

Calculate df

$df = n-1$

$df = 15-1$

$df = 14$

Calculate the p value from the Student's t-distribution  at df = 14 and $t = -1.792$

$p =0.047$  

Given that: $\alpha = 0.05$

Since $p < \alpha$

So, it means that the machine is under filling the bags