Respuesta :
Answer:
Confidence interval variance [21.297 ; 64.493]
Confidence interval standard deviation;
4.615, 8.031
Step-by-step explanation:
Given :
Variance, s² = 34.34
Standard deviation, s = 5.86
Sample size, n = 27
Degree of freedom, df = 27 - 1 = 26
Using the relation for the confidence interval :
[s²(n - 1) / X²α/2, n-1] ; [s²(n - 1) / X²1-α/2, n-1]
From the chi distribition table :
X²α/2, n-1 = 41.923 ; X²1-α/2, n-1 = 13.844
Hence,
[34.34*26 / 41.923] ; [34.34*26 / 13.844]
[21.297 ; 64.493]
The 95% confidence interval for the population variance is :
21.297 < σ² < 64.493
Standard deviation is the square root of variance, hence,
The 95% confidence interval for the population standard deviation is :
4.615 < σ < 8.031
The population variance of 95% confidence interval is [tex]4.615<\sigma<8.031[/tex] and this can be determined by using the given data.
Given :
- Variance = 34.34
- Standard Deviation = 5.86
- Sample Size = 27
- 95% confidence interval
First, determine the degree of freedom.
[tex]df = 27-1=26[/tex]
The determine the confidence interval using the below formula:
[tex]\left(\dfrac{s^2(n-1)}{X^2_{\alpha/2,(n-1) }}\right);\left(\dfrac{s^2(n-1)}{X^2_{(1-\alpha)/2,(n-1) }}\right)[/tex]
Now, using the chi distribution the above expression becomes:
[tex]\left(\dfrac{34.34\times 26}{41.923 }}\right);\left(\dfrac{34.34\times 26}{13.844 }}\right)[/tex]
Simplify the above expression.
(21.297 ) ; (64.493)
So, the population variance of 95% confidence interval is:
[tex]\begin{aligned}\\21.297&<\sigma^2<64.493\\4.615&<\sigma<8.031\\\end{alingned}[/tex]
For more information, refer to the link given below:
https://brainly.com/question/7635845
