Construct the indicated confidence intervals for (a) the population variance&nbsp;σ2&nbsp;and (b) the population standard deviation&nbsp;σ&nbsp;Assume that the population in the sample has a normal distribution.c=0.95,s2=11.56,n=30

Determine the critical values using table 6 with df=n-1=30-1=29&nbsp;x{1−0.025}2=x{0.975}2=16.047&nbsp;Following are the perimeters of the confidence interval for the standard deviation:n−1x{α2}2{˙s=30−145.722{˙11.56≈2.708&nbsp;n−1x12−{α2}{˙s=30−116.047{˙11.56≈4.571&nbsp;2. The boundaries of the confidence interval for the variance is then the square of the boundaries of the confidence intervals for the standard deviation:&nbsp;(2.7082,4.5712)=(7333,50.894)

Get the solution to "Construct the indicated confidence intervals for (a) the..."

Chardonnay Felix

Answered question

2021-05-29

Construct the indicated confidence intervals for (a) the population variance
$σ^{2}$
and (b) the population standard deviation
$σ$
Assume that the population in the sample has a normal distribution.
$c = 0.95, s^{2} = 11.56, n = 30$

Answer & Explanation

Derrick

Skilled2021-05-30Added 94 answers

Determine the critical values using table 6 with df=n-1=30-1=29
$x_{{1 - 0.025}}^{2} = x_{{0.975}}^{2} = 16.047$
Following are the perimeters of the confidence interval for the standard deviation:
$\sqrt{\frac{n - 1}{x_{{\frac{α}{2}}}^{2}}} \dot{{} s = \sqrt{\frac{30 - 1}{45.722}} \dot{{} \sqrt{11.56} \approx 2.708$
$\sqrt{\frac{n - 1}{x_{1}^{2} - {\frac{α}{2}}}} \dot{{} s = \sqrt{\frac{30 - 1}{16.047}} \dot{{} \sqrt{11.56} \approx 4.571$
2. The boundaries of the confidence interval for the variance is then the square of the boundaries of the confidence intervals for the standard deviation:
$({2.708}^{2}, {4.571}^{2}) = (7333, 50.894)$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in College Statistics

Didn't find what you were looking for?