Question

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 450 gram setting. It is believed that the machine is underfilling the bags. A 16 bag sample had a mean of 441 grams with a variance of 256. Assume the population is normally distributed. A level of significance of 0.05 will be used. Find the P-value of the test statistic. You may write the P-value as a range using interval notation, or as a decimal value rounded to four decimal places.

Answer 1

Answer:

Test statistic = -2.25

P-value = 0.0199

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 450 gram

Sample mean, $\bar{x}$ = 441 grams

Sample size, n = 16

Alpha, α = 0.05

Sample variance = 256

$\sigma^2 = 256\\\sigma = \sqrt{256} = 16$

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 450\text{ grams}\\H_A: \mu < 450\text{ grams}$

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

Formula:

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

Putting all the values, we have

$t_{stat} = \displaystyle\frac{441 - 450}{\frac{16}{\sqrt{16}} } = -2.25$

Now,

Degree of freedom =

$=n-1 = 16-1=15$

We can calculate the p-value from the table as:

P-value = 0.0199

Conclusion:

Since the p-value is smaller than the significance level we fail to accept the null hypothesis and reject it.

Thus, there is enough evidence to support the claim that the machine is under filling the bags .