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

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

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Piper 2 weeks 2021-09-13T11:28:51+00:00 1 Answer 0

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    2021-09-13T11:30:45+00:00

    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 <\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 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 <\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 higher than 0.26^2=0.0676.

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