Question

A company produces product with a mean weight of 10 and a standard deviation of 0.200. A new process supposedly will produce products with the same mean and a smaller standard deviation. A sample of 20 products produced by the new method has a sample standard deviation of 0.126. At a significance level of 10%, is it appropriate to conclude that the new process is less variable than the old?

Answer 1

`F=(s^2_1)/(s^2_2)=(0.2^2)/(0.126^2)=2.520`

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have
`n_1 -1 =10-1=9` and for the denominator we have
`n_2 -1 =20-1=19` and the F statistic have 9 degrees of freedom for the numerator and 19 for the denominator. And the P value is given by:

Now we can calculate the p value with this probability:

`p_v =P(F_(9,19)>2.520)=0.043`

Using a significance level of 5% we see that the p value is lower than this value and we have enough evidence to reject the null hypothesis and we can conclude that the variation for the new process is lower than the new one.

Explanation:

Information given

`n_1 = 10` represent the sampe size old

`n_2 =20` represent the sample size new

`s_1 = 0.2` represent the sample deviation for old

`s_2 = 0.126` represent the sample deviation for new

The statistic is given by:

`F=(s^2_1)/(s^2_2)`

Hypothesis to test

We want to test if the new process is less variable than the old, so the system of hypothesis are:

H0:
`\sigma^2_1 \leq \sigma^2_2`

H1:
`\sigma^2_1 >\sigma^2_2`

The statistic is given by:

`F=(s^2_1)/(s^2_2)=(0.2^2)/(0.126^2)=2.520`

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have
`n_1 -1 =10-1=9` and for the denominator we have
`n_2 -1 =20-1=19` and the F statistic have 9 degrees of freedom for the numerator and 19 for the denominator. And the P value is given by:

Now we can calculate the p value with this probability:

`p_v =P(F_(9,19)>2.520)=0.043`

Using a significance level of 5% we see that the p value is lower than this value and we have enough evidence to reject the null hypothesis and we can conclude that the variation for the new process is lower than the new one.