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laquitad8693
25.03.2020 •
Mathematics
The weights of 83 randomly selected windshields were found to have a variance of 1.88. Construct the 95% confidence interval for the population variance of the weights of all windshields in this factory. Round your answers to two decimal places.
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Ответ:
95% confidence interval for the population variance = (1.42 , 2.62).
Step-by-step explanation:
We are given that the weights of 83 randomly selected windshields were found to have a variance of 1.88.
So, firstly the pivotal quantity for 95% confidence interval for the population variance is given by;
P.Q. =
~ ![\chi^{2} __n_-_1](/tpl/images/0564/0965/03dd3.png)
where,
= sample variance = 1.88
n = sample of windshields = 83
So, 95% confidence interval for population variance,
is;
P(58.85 <
< 108.9) = 0.95 {As the table of
at 82 degree of freedom
gives critical values of 58.85 & 108.9}
P(58.85 <
< 108.9) = 0.95
P(
<
<
) = 0.95
P(
<
<
) = 0.95
95% confidence interval for
= (
,
)
= (
,
)
= (1.42 , 2.62)
Therefore, 95% confidence interval for the population variance of the weights of all windshields in this factory is (1.42 , 2.62).
Ответ: