Question

A machine that puts corn flakes into boxes is adjusted to put an average of 15.6 ounces into each box, with standard deviation of 0.26 ounce. If a random sample of 15 boxes gave a sample standard deviation of 0.39 ounce, do these data support the claim that the variance has increased and the machine needs to be brought back into adjustment

Answer 1

Answer:

$\chi^2 =\frac{15-1}{0.0676} 0.1521 =31.5$

$p_v =P(\chi^2_{14} >31.5)=0.0047$

If we compare the p value and the significance level assumed of 0.05 we see that $p_v$ 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 higher than 0.26^2=0.0676.

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"

$n=15$ represent the sample size

$\alpha$ represent the confidence level  

$s^2 =0.39^2 =0.1521$ represent the sample variance obtained

$\sigma^2_0 =0.26^2 =0.0676$ 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 violated, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \leq 0.0676$

Alternative hypothesis: $\sigma^2 >0.0676$

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.

$\chi^2 =\frac{15-1}{0.0676} 0.1521 =31.5$

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=15-1=14. And since is a right tailed test the p value would be given by:

$p_v =P(\chi^2_{14} >31.5)=0.0047$

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

"=1-CHISQ.DIST(31.5,14,TRUE)"

Conclusion

If we compare the p value and the significance level assumed of 0.05 we see that $p_v$ 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 higher than 0.26^2=0.0676.