An automotive part must be machined to close tolerances to be acceptable to customers. Production specifications call for a maximum variance

Question

An automotive part must be machined to close tolerances to be acceptable to customers. Production specifications call for a maximum variance in the lengths of the parts of .0004. Suppose the sample variance for 30 parts turns out to be s^2=.0005. Use a = .05 to test whether the population variance specification is being violated.

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Mary 1 week 2021-11-19T17:48:53+00:00 1 Answer 0 views 0

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    2021-11-19T17:50:45+00:00

    Answer:

    H0: \sigma \leq 0.0004

    H1: \sigma >0.0004

     t=(30-1) [\frac{0.0224}{0.02}]^2 =36.25

    For this case since we have a right tailed test the p value is given by:

    p_v = P(\Chi^2_{29}>36.25)=0.166

    If we compare the p value and the significance level given we see that p_v >\alpha so then we have enough evidence to FAIL to reject the null hypothesis that the true population variance it’s lower or equal  than 0.0004 at 5% of significance.  

    Step-by-step explanation:

    Previous concepts and notation

    The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test “two-sided test or a one-sided test”.

    n = 30 sample size

    s^2 =0.0005 represent the sample variance

    s= 0.0224 represent the sample deviation

    \sigma_o =0.095 the value that we want to test

    p_v represent the p value for the test

    t represent the statistic

    \alpha=0.05 significance level

    State the null and alternative hypothesis

    On this case we want to check if the population variance is higher than 0.0004 (the specification), so the system of hypothesis are:

    H0: \sigma \leq 0.0004

    H1: \sigma >0.0004

    In order to check the hypothesis we need to calculate th statistic given by the following formula:

     t=(n-1) [\frac{s}{\sigma_o}]^2

    This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

    What is the value of your test statistic?

    Now we have everything to replace into the formula for the statistic and we got:

     t=(30-1) [\frac{0.0224}{0.02}]^2 =36.25

    What is the critical value for the test statistic at an α = 0.05 significance level?

    Since is a right tailed test the critical zone it’s on the right tail of the distribution. On this case we need a quantile on the chi square distribution with df=30-1=19 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.  

    We can calculate the critical value in excel with the following code: “=CHISQ.INV(0.95,29)”. And our critical value would be \Chi^2 =42.56

    Since our calculated value is lower than the critical value we to reject the null hypothesis.

    What is the approximate p-value of the test?

    For this case since we have a right tailed test the p value is given by:

    p_v = P(\Chi^2_{29}>36.25)=0.166

    If we compare the p value and the significance level given we see that p_v >\alpha so then we have enough evidence to FAIL to reject the null hypothesis that the true population variance it’s lower or equal  than 0.0004 at 5% of significance.  

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