A manufacturer uses production method to produce steel rods. A random sample of 17 steel rods resulted in lengths with a standard deviation

Question

A manufacturer uses production method to produce steel rods. A random sample of 17 steel rods resulted in lengths with a standard deviation of 4.7 cm. at the 0.10 significance level, test the claim that the new production method has lengths with a standard deviation different from 3.5 cm, which was the standard deviation for the old method.

Use the traditional and confidence level methods to test this claim.

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Serenity 4 days 2021-09-07T22:13:58+00:00 1 Answer 0

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    2021-09-07T22:15:05+00:00

    Answer:

    \chi^2 =\frac{17-1}{12.15} 22.09 =15.87

    Traditional method

    We can find a critical value in the chi square distribution with df =16 who accumulates \alpha/2 =0.05 of the area on each tail and we got:

    \chi^2_{crit}= 7.962

    \chi^2_{crit}=26.296

    Since the calculated value is between the two critical values we don’t have enough evidence to conclude that the true deviation is NOT significantly different from 3.5 cm

    Confidence level method

    We can find the p value like this

    p_v =2*P(\chi^2 >15.87)=0.92

    Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is NOT significantly different from 3.5 cm

    Step-by-step explanation:

    Data provided

    n=17 represent the sample selected

    \alpha=0.1 represent the significance

    s^2 =4.7^2 =22.09 represent the sample variance

    \sigma^2_0 =3.5^2 =12.15 represent the value to check

    Null and alternative hypothesis

    We want to verify if the new production method has lengths with a standard deviation different from 3.5 cm, so the system of hypothesis would be:

    Null Hypothesis: \sigma^2 = 12.15

    Alternative hypothesis: \sigma^2 \neq 12.15

    The statistic for this case is given by:

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

    The degrees of freedom are:

     df =n-1= 17-1=16

    \chi^2 =\frac{17-1}{12.15} 22.09 =15.87

    Traditional method

    We can find a critical value in the chi square distribution with df =16 who accumulates \alpha/2 =0.05 of the area on each tail and we got:

    \chi^2_{crit}= 7.962

    \chi^2_{crit}=26.296

    Since the calculated value is between the two critical values we don’t have enough evidence to conclude that the true deviation is significantly different from 3.5 cm

    Confidence level method

    We can find the p value like this

    p_v =2*P(\chi^2 >15.87)=0.92

    Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is significantly different from 3.5 cm

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