Question

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 418 gram setting. It is believed that the machine is underfilling the bags. A 9 bag sample had a mean of 413 grams with a standard deviation of 20. A level of significance of 0.1 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

No. At a significance level of 0.1, there is not enough evidence to support the claim that the bags are underfilled (population mean significantly less than 418 g.)

Explanation:

This is a hypothesis test for the population mean.

The claim is that the bags are underfilled (population mean significantly less than 418 g.)

Then, the null and alternative hypothesis are:

`H_0: \mu=418\\\\H_a:\mu< 418`

The significance level is 0.1.

The sample has a size n=9.

The sample mean is M=413.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=20.

The estimated standard error of the mean is computed using the formula:

`s_M=(s)/(√(n))=(20)/(√(9))=6.6667`

Then, we can calculate the t-statistic as:

`t=(M-\mu)/(s/√(n))=(413-418)/(6.6667)=(-5)/(6.6667)=-0.75`

The degrees of freedom for this sample size are:

`df=n-1=9-1=8`

This test is a left-tailed test, with 8 degrees of freedom and t=-0.75, so the P-value for this test is calculated as (using a t-table):

`\text{P-value}=P(t<-0.75)=0.237`

As the P-value (0.237) is bigger than the significance level (0.1), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.1, there is not enough evidence to support the claim that the bags are underfilled (population mean significantly less than 418 g.)