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Test the breaking strength of a sample of wood fibers

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The breaking strength of a random sample of n = 20 bundles of wood fibers has a sample mean x-bar = 436.5 and sample standard deviation s = 11.90.

Construct 90%, 95%, and 99% t-interval for the average breaking strength µ.

Do you think that the average breaking strength is equal to 450.0? Please explain me why or why not.

Also, how many additional data observations should be obtained to construct a 99% t-intervals for the average breaking strength with a length no longer than 10.0? Recall that the length of interval is the difference between upper bound and lower bound of interval

Also, construct a 95% confidence interval for the variance.

It is claimed that the standard deviation of the wood fibers would not exceed 12.01 with 95%% confidence level. Would the given statistics and the t-intervals constructed in the beginning if the problem be able to confirm this claim? Why or why not?

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Solution Summary

The expert tests the breaking strength of a sample of wood fibers. The 99% t-intervals are constructed.

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Confidence interval is equal to: ¯x±t^* s/√n

So for the 90% confidence interval: 436.5±1.729 11.90/√20=(431.9,441.1)

For the 95% confidence level: 436.5±2.093 11.90/√20=(430.9,442.06)

For the 99% confidence level: 436.5±2.861 11.90/√20=(428.89,444.11)

It is very unlikely that the ...

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