Question

The thickness of 81 randomly selected aluminum sheets were found to have a variance of 3.23. construct the 98% confidence interval for the population variance of the thickness of all aluminum sheets in this factory. Assume that the population is normally distributed. Round your answer to two decimal places.

I have been working on these problems for 8 hours and im still not getting it. They are due tomorrow. Thank you.

Answer #1

Confidence interval for population variance is given as below:

[(n – 1)*S^{2} / χ^{2}_{α/2, n}_{–
1} ] < σ < [(n – 1)*S^{2} / χ^{2}_{1
-}_{α/2, n}_{– 1} ]

We are given

Confidence level = 98%

Sample size = n = 81

Degrees of freedom = n – 1 = 80

Sample Variance = S^2 = 3.23

χ^{2}_{α}_{/2, n – 1} = 112.3288

χ^{2}_{1 -}_{α/2, n}_{– 1} =
53.5401

(By using chi square table)

[(n – 1)*S^{2} / χ^{2}_{α/2, n}_{–
1} ] < σ^2 < [(n – 1)*S^{2} /
χ^{2}_{1 -}_{α/2, n}_{– 1} ]

[(81 – 1)* 3.23 / 112.3288] < σ^2 < [(81 – 1)* 3.23/ 53.5401]

2.3004 < σ^2 < 4.8263

Lower limit = 2.30

Upper limit = 4.83

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1Lower Bound ？
2Upper Bound？

1) A researcher reports that when groups of four children are
randomly selected from a population of couples meeting certain
criteria, the probability distribution for the number of girls, ?,
is given in the following table where ? is a positive constant.
x
?(?)
?(? ≤ ?)
0
0.4
?
1
?
?
2
3? − 0.2
?
3
2?
?
4
0.2
?
a) Find ?(2).
b) Complete the column of cumulative probability. (Please rewrite
the table in your...

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