Question

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 431 gram setting. It is believed that the machine is underfilling the bags. A 26 bag sample had a mean of 421 grams with a standard deviation of 20. A level of significance of 0.01 will be used. Assume the population distribution is approximately normal. Is there sufficient evidence to support the claim that the bags are underfilled

Answer 1

The p-value of the test is of 0.0086 < 0.01, which means that there is sufficient evidence to support the claim that the bags are underfilled

Explanation:

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 431 gram setting. It is believed that the machine is underfilling the bags.

At the null hypothesis, we test if they are at the correct weight, that is:

`H_0: \mu = 431`

At the alternative hypothesis, we test if they are underfilled, that is:

`H_1: \mu < 431`

The test statistic is:

We have the standard deviation for the sample, so the t-distribution is used to solve this question

`t = (X - \mu)/((s)/(√(n)))`

In which X is the sample mean,
`\mu` is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

431 is tested at the null hypothesis:

This means that
`\mu = 431`

A 26 bag sample had a mean of 421 grams with a standard deviation of 20.

This means that
`n = 26, X = 421, s = 20`

Value of the test statistic:

`t = (X - \mu)/((s)/(√(n)))`

`t = (421 - 431)/((20)/(√(26)))`

`t = -2.55`

P-value of the test and decision:

Using a t-distirbution calculator, the p-value is found using a left-tailed test(test if the mean is less than a value), with 26 - 1 = 25 df and t = -2.55. This p-value is of 0.0086.

The p-value of the test is of 0.0086 < 0.01, which means that there is sufficient evidence to support the claim that the bags are underfilled