| A random sample of 20 nominally measured of the length of a plastic ball is taken and the lengths are measured precisely. The measurements, in mm, are as follows: 2.02 1.94 2.09 1.95 1.98 2.00 2.03 2.04 2.08 2.07 1.99 1.96 1.99 1.95 1.99 1.99 2.03 2.05 2.01 2.03 From the data, we calculate EX = 40.19, x² = 80.7977 Assuming that the lengths are normally distributed with unknown mean u, and unknown variance 82 i find a 90% confidence interval for the variance o?. Interpret your results. i. find a 90% confidence interval for the standard deviation o. Interpret your results.

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s| A random sample of 20 nominally measured of the length of a plastic ball is taken and
the lengths are measured precisely. The measurements, in mm, are as follows:
2.02 1.94 2.09 1.95 1.98 2.00 2.03 2.04 2.08 2.07
1.99 1.96 1.99 1.95 1.99 1.99 2.03 2.05 2.01 2.03
From the data, we calculate E X = 40.19, Ex² = 80.7977
Assuming that the lengths are normally distributed with unknown mean u, and unknown variance 82
i. find a 90% confidence interval for the variance o?. Interpret your results.
find a 90% confidence interval for the standard deviation a. Interpret your results.
ii.
Transcribed Image Text:s| A random sample of 20 nominally measured of the length of a plastic ball is taken and the lengths are measured precisely. The measurements, in mm, are as follows: 2.02 1.94 2.09 1.95 1.98 2.00 2.03 2.04 2.08 2.07 1.99 1.96 1.99 1.95 1.99 1.99 2.03 2.05 2.01 2.03 From the data, we calculate E X = 40.19, Ex² = 80.7977 Assuming that the lengths are normally distributed with unknown mean u, and unknown variance 82 i. find a 90% confidence interval for the variance o?. Interpret your results. find a 90% confidence interval for the standard deviation a. Interpret your results. ii.
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