Question

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 435.0435.0 gram setting. It is believed that the machine is underfilling the bags. A 4646 bag sample had a mean of 429.0429.0 grams. A level of significance of 0.050.05 will be used. Determine the decision rule. Assume the standard deviation is known to be 24.024.0.

Answer 1

Answer:

The machine is underfilling the bags.        

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 435.0 grams

Sample mean, $\bar{x}$ = 429.0 grams

Sample size, n = 46

Alpha, α = 0.05

Population standard deviation, σ = 24.0 grams

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 435.0\text{ grams}\\H_A: \mu < 435.0\text{ grams}$

We use one-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{429.0 - 435.0}{\frac{24.0}{\sqrt{46}} } = -1.695$

Now, $z_{critical} \text{ at 0.05 level of significance } = -1.64$

Decision Rule:

If the calculated z statistic is less than the critical value of z, we fail to accept the null hypothesis and reject it.

Since,

$z_{stat} < z_{critical}$

We reject the null hypothesis and accept the alternate hypothesis. Thus,  the machine is underfilling the bags.