Respuesta :
Answer:
95% confidence interval estimate for the standard deviation = [4.78 , 8.06]
Step-by-step explanation:
We are given that the manager of the service department of a local car dealership has noted that the service times of a sample of 30 new automobiles has a standard deviation of 6 minutes, i.e. n = 30 and s = 6.
The pivotal quantity for 95% confidence interval for the population variance is given by;
[tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] ~ [tex]\chi^{2}__n_-_1[/tex]
95% confidence interval for the population variance is;
P(16.05 < [tex]\chi^{2}__2_9[/tex] < 45.72) = 0.95
P(16.05 < [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] < 45.72) = 0.95
P( [tex]\frac{16.05}{(n-1)s^{2} }[/tex] < [tex]\frac{1 }{\sigma^{2} }[/tex] < [tex]\frac{45.72}{(n-1)s^{2} }[/tex] ) = 0.95
P( [tex]\frac{(n-1)s^{2}}{ 45.72}[/tex] < [tex]\sigma^{2}[/tex] <[tex]\frac{(n-1)s^{2}}{ 16.05}[/tex] ) = 0.95
95% Confidence interval for [tex]\sigma^{2}[/tex] = [ [tex]\frac{(n-1)s^{2}}{ 45.72}[/tex] , [tex]\frac{(n-1)s^{2}}{ 16.05}[/tex] ]
= [ [tex]\frac{(30-1)6^{2}}{ 45.72}[/tex] , [tex]\frac{(30-1)6^{2}}{ 16.05}[/tex] ]
= [ 22.835 , 65.047 ]
So, 95% Confidence interval for [tex]\sigma[/tex] = [ [tex]\sqrt{ 22.835} ,\sqrt{ 65.047}[/tex] ] = [ 4.78 , 8.06 ]
Therefore, 95% confidence interval for the standard deviation of the service times for all their new automobiles is [ 4.78 , 8.06 ] .
The Lower and Upper confidence limits are "4.78 and 8.07", and the further calculation of the confidence interval limits can be defined as follows:
Given:
Standard deviation [tex] \sigma=6 \ minutes [/tex]
Sample size[tex] n=30 [/tex]
Confidence interval=[tex] 95\%[/tex]
To find:
confidence interval limits=?
Solution:
[tex] \sigma=6 \ minutes [/tex]
[tex] n=30 [/tex]
C.I=[tex] 95\%[/tex]
Calculating freedom of degree [tex]=n-1= 30-1=29 [/tex]
Calculating Level of significance[tex] =1-0.95=0.05[/tex]
Using the Chi-square for calculating the critical value of lower tail and higher tail:
Using Excel command
lower value:[tex]=CHIINV(1-(\frac{0.05}{2}),29) =16.05[/tex]
higher value:[tex]=CHIINV((\frac{0.05}{2}),29) =45.72[/tex]
Using the C.I value with standard deviation:
[tex]\to \sqrt{\frac{(n 1)S^2}{χ_1^2 -(\frac{\alpha}{2})}} < \sigma < \sqrt{\frac{(n 1)S^2}{χ^2_(\frac{\alpha}{2})}} \\\\ \to \sqrt{\frac{(30-1)(6)^2}{(45.72)}} < \sigma <\sqrt{ \frac{(30-1)(6)^2}{16.05}}\\\\ \to (4.78 , 8.07) \\\\[/tex]
Therefore, the Lower and Upper confidence limits are "4.78 and 8.07".
Learn more:
brainly.com/question/2396419
