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 1.03 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 = 39.19, Ex² = 77.7377 Assuming that the lengths are normally distributed with unknown mean u, and unknown variance 82 i. find a 90% confidence interval for the standard deviation o. Interpret your results. find a 90% confidence interval for the variance o?. Interpret your results. ii.
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 1.03 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 = 39.19, Ex² = 77.7377 Assuming that the lengths are normally distributed with unknown mean u, and unknown variance 82 i. find a 90% confidence interval for the standard deviation o. Interpret your results. find a 90% confidence interval for the variance o?. Interpret your results. ii.
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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 1.03 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 = 39.19, Ex² = 77.7377
Assuming that the lengths are normally distributed with unknown mean µ, and unknown variance 82
i.
find a 90% confidence interval for the variance o?. Interpret your results.
ii.
find a 90% confidence interval for the standard deviation o . Interpret your results.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F30285abe-251a-4a54-a897-3270a220efcf%2F7b66ee89-b095-40ed-bb82-64eae143a53d%2Fla4cd8_processed.jpeg&w=3840&q=75)
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 1.03 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 = 39.19, Ex² = 77.7377
Assuming that the lengths are normally distributed with unknown mean µ, and unknown variance 82
i.
find a 90% confidence interval for the variance o?. Interpret your results.
ii.
find a 90% confidence interval for the standard deviation o . Interpret your results.
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