Question

A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 428 gram setting. It is believed that the machine is underfilling or overfilling the bags. A 13 bag sample had a mean of 424 grams with a variance of 169. Assume the population is normally distributed. A level of significance of 0.01 will be used. Specify the type of hypothesis test.

Answer 1

Given:

Population mean (µ) = 428
Sample size(n) = 13
Sample mean = xbar = 424.

Sample variance =s^2 = 169
Sample standard deviation = S = 13

The claim statement is, "the machine is underfilling or overfilling the bag".
That is .

Hence, the claim statement goes under the alternative hypothesis.

Hence, the Null hypothesis is, H0: µ = 428
And the alternative hypothesis is,

Here, we have the sample standard deviation. So, we need to use the T-test for testing.
Now, let's find the test statistic.

That is, test statistic T = -1.1094
Degrees of freedom = n -1 = 13 - 1 = 12 .

Now, let's find the P-value.
For finding the P-value, we can use a technology like excel/Ti84 calculator or T table.
The following excel command is used to find the P-value.

= T.DIST.2T(Absolute value of Test statistic, Degrees of freedom)
= T.DIST.2T(1.1094 , 12 )
You will get, P-value = 0.2889955

We are given: Level of significance α = 0.01.

Following is the decision rule for making the decision.
If, P-value > α, then we fail to reject the null hypothesis.
If, P-value <= α, then we reject the null hypothesis.
Here, given α = 0.01
And P-value = 0.289
That is P-value > α.
Hence, we fail to reject the null hypothesis.

That is, there is not sufficient evidence to support the claim that the machine is underfilling or overfilling the bags.