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The amount of lateral expansion (mils) was determined for a sample of n = 8 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s = 2.89 miles. Assuming normality, derive a 95% CI for? 2 and for?

User RonQi
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Answer:

95% Confidence interval for the variance:


3.6511\leq \sigma^2\leq 34.5972

95% Confidence interval for the standard deviation:


1.9108\leq \sigma \leq 5.8819

Explanation:

We have to calculate a 95% confidence interval for the standard deviation σ and the variance σ².

The sample, of size n=8, has a standard deviation of s=2.89 miles.

Then, the variance of the sample is


s^2=2.89^2=8.3521

The confidence interval for the variance is:


( (n - 1) s^2)/( \chi_(\alpha/2)^2) \leq \sigma^2 \leq ( (n - 1) s^2)/(\chi_(1-\alpha/2)^2)

The critical values for the Chi-square distribution for a 95% confidence (α=0.05) interval are:


\chi_(0.025)=1.6899\\\\\chi_(0.975)=16.0128

Then, the confidence interval can be calculated as:


( (8 - 1) 8.3521)/( 16.0128) \leq \sigma^2 \leq ( (8 - 1) 8.3521)/(1.6899)\\\\\\3.6511\leq \sigma^2\leq 34.5972

If we calculate the square root for each bound we will have the confidence interval for the standard deviation:


√(3.6511)\leq \sigma\leq √(34.5972)\\\\\\1.9108\leq \sigma \leq 5.8819

User Nuander
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