Question #288659

A rivet is to be inserted into a hole. If the Standard deviation of hole diameter exceeds


0,01mm, there is an unacceptably high probability that the rivet will not fit. A random sample


of n=15 part is selected and the hole diameter is measured. The sample Standard deviation of


the hole diameter measurements is s=0.008mm. Is there strong evidence to indicate that the


standard deviation of hole diameter exceeds 0.01mm? Use α=0.01

1
Expert's answer
2022-01-24T14:49:56-0500
σ0=0.01 mm,σ02=0.0001 mm2\sigma_0=0.01\ mm, \sigma_0^2=0.0001\ mm^2

s=0.008 mm,s2=0.000064 mm2s=0.008\ mm, s^2=0.000064\ mm^2

The following null and alternative hypotheses need to be tested:

H0:σ20.0001H_0:\sigma^2\leq0.0001

H1:σ2>0.0001H_1:\sigma^2>0.0001

This corresponds to a right-tailed test test, for which a Chi-Square test for one population variance will be used.

Based on the information provided, the significance level is α=0.01,\alpha = 0.01, df=n1=151=14df=n-1=15-1=14 degrees of freedom, and the the rejection region for this right-tailed test is R={χ2:χ2>29.1412}.R = \{\chi^2: \chi^2 >29.1412\}.

The Chi-Squared statistic is computed as follows:


χ2=(n1)s2σ02=(151)(0.000064)0.0001=8.96\chi^2=\dfrac{(n-1)s^2}{\sigma_0^2}=\dfrac{(15-1)(0.000064)}{0.0001}=8.96

Since it is observed that χ2=8.9629.1412=χc2,\chi^2 = 8.96 \le 29.1412= \chi_c^2, it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population variance σ2\sigma^2 is greater than 0.0001,0.0001, at the 0.010.01 significance level.

Therefore, there is not enough evidence to claim that the standard deviation of hole diameter exceeds 0.01 mm0.01\ mm at the 0.010.01 significance level.


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