Question

A manufacturer of salad dressings uses machines to dispense liquid ingredients into bottles that move along a filling line.
The machine that dispenses dressing is working properly when 8 ounces are dispensed. The standard deviation of the process is 0.15 ounces.
A sample of 48 bottles is selected periodically, and the filling line is stopped if there is evidence that the mean amount dispensed is different from 8 ounces.

Suppose that the mean amount dispensed in a particular sample of 48 bottles is 7.983 ounces.

Calculate the P-Value.

Answer 1

Answer:

P-value = 0.4324

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 8 ounces

Sample mean, $\bar{x}$ = 7.983 ounces

Sample size, n = 48

Population standard deviation, σ = 0.15 ounces

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 8\text{ ounces}\\H_A: \mu \neq 8\text{ ounces}$

We use Two-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

$z_{stat} = \displaystyle\frac{7.983 - 8}{\frac{0.15}{\sqrt{48}} } = -0.7851$

Now, we calculate the p-value with the help of standard normal z table.

P-value = 0.4324