Question

Suppose a local manufacturing company claims their production line has a variance of less than 9.0. A quality control engineer decides to test this claim by sampling 35 parts. She finds that the standard deviation of the sample is 2.12. Is this enough evidence at the 1% level of significance, to accept the manufacturing companies claim?

Answer 1

Answer:

$\chi^2 =\frac{35-1}{9} 2.12^2 =16.979$

$p_v =P(\chi^2 <16.979)=0.0065$

If we compare the p value and the significance level provided we see that $p_v <\alpha$ so on this case we have enough evidence in order to reject the null hypothesis at the significance level provided. And that means that the population variance is significantly lower than 9

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

$n=35$ represent the sample size

$\alpha=0.01$ represent the confidence level  

$s =2.12$ represent the sample deviation obtained

$\sigma_0 =3$ represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance specification is lower than 9 and the deviation lower than 3, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \geq 9$

Alternative hypothesis: $\sigma^2 <9$

Calculate the statistic  

For this test we can use the following statistic:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

$\chi^2 =\frac{35-1}{9} 2.12^2 =16.979$

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case df= n-1= 35-1=34. And since is a left tailed test the p value would be given by:

$p_v =P(\chi^2 <16.979)=0.0065$

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(16.979,34,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that $p_v <\alpha$ so on this case we have enough evidence in order to reject the null hypothesis at the significance level provided. And that means that the population variance is significantly lower than 9