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Question

A rivet is to be inserted into a hole. A random sample of n = 15 parts is selected, and the hole diameter is measured. The sample standard deviation of the hole diameter measurements is s = 0.008 millimeters. Construct a 99% lower confidence bound for σ².

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Answered 2 years ago
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If s2s^2 is the sample variance from a random sample of nn observations from a normal distribution with unknown variance σ2\sigma^2, then a $\textbf{100(1α)%100(1 - \alpha)\% lower confidence bounds on σ2\sigma^2}$ is

(n1)s2χα,n12σ2.\begin{align} \frac{(n-1)s^2}{\chi^2_{\alpha,n-1}}\leq\sigma^2 . \end{align}

where χα,n12\chi^2_{\alpha,n-1} is the lower 100α100\alpha percentage points of the chi-square distribution with n1n - 1 degrees of freedom. We know that confidence level is 99%99\%, degrees of freedom is n1=151=14n-1= 15-1=14 and sample standard deviation is s=0.008s=0.008, than

α=10.99=0.01χ0.01,142=29.14.\begin{align*} \alpha=1-0.99=0.01\quad\Rightarrow\quad \chi^2_{0.01,14}= 29.14. \end{align*}

Now from Equation (1) we get

14×0.008229.14σ2.\frac{14\times0.008^2}{29.14}\leq\sigma^2.

Therefore a lower confidence bound for σ2\sigma^2

0.0000307σ2.\boxed{0.0000307\leq\sigma^2.}

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