Respuesta :
Answer:
95% confidence interval for the population variance = [56.31 , 224.75].
Step-by-step explanation:
We are given that Scores on an IQ test are normally distributed. A sample of 18 IQ scores had standard deviation s = 10.
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(7.564 < [tex]\chi^{2} __1_7[/tex] < 30.19) = 0.95
P(7.564 < [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] < 30.19) = 0.95
P( [tex]\frac{7.564}{(n-1)s^{2} }[/tex] < [tex]\frac{1}{\sigma^{2} }[/tex] < [tex]\frac{30.19}{(n-1)s^{2} }[/tex] ) = 0.95
P( [tex]\frac{(n-1)s^{2}}{ 30.19}[/tex] < [tex]\sigma^{2}[/tex] < [tex]\frac{(n-1)s^{2}}{ 7.564}[/tex] ) = 0.95
95% Confidence interval for [tex]\sigma^{2}[/tex] = [ [tex]\frac{(n-1)s^{2}}{ 30.19}[/tex] , [tex]\frac{(n-1)s^{2}}{ 7.564}[/tex] ]
= [ [tex]\frac{17* 10^{2}}{ 30.19}[/tex] , [tex]\frac{17*10^{2}}{ 7.564}[/tex] ]
= [ 56.31 , 224.75 ]
Therefore, 95% confidence interval for the population variance, [tex]\sigma^{2}[/tex] is [ 56.31 , 224.75 ] .
The 95 percent confidence interval for the population variance [tex]\sigma ^2[/tex] is (56.31, 224.75)
Given-
Score of the IQ test are normally distributed.
The sample of 18 IQ scores had standard deviation [tex]s[/tex] is 10.
Population variance is [tex]\sigma^2[/tex].
Pivotal quantity or pivot is a function of observations and unobservant parameters and it can be represent with the formula,
[tex]=\dfrac{(n-1)s^2}{\sigma^2} \sim X^2_{n-1}[/tex]
Now From the standards 95 percent confidence interval for the population variation is,
[tex]P(7.564 < X^2_{18-1} < 30.19) = 0.95[/tex]
[tex]P(7.564 < \dfrac{(n-1)s^2}{\sigma^2} < 30.19) = 0.95[/tex]
[tex]P(\dfrac{(n-1)s^2}{7.564} < {\sigma^2} < \dfrac{(n-1)s^2}{30.19}) = 0.95[/tex]
Claculate 95 percent confidence level for,
[tex]\sigma^2=P(\dfrac{(n-1)s^2}{7.564},\dfrac{(n-1)s^2}{30.19})[/tex]
[tex]\sigma^2=P(\dfrac{(17)10^2}{30.9},\dfrac{(17)10^2}{7.564})[/tex]
[tex]\sigma^2=(56.31 , 224.75)[/tex]
Hence, the 95 percent confidence interval for the population variance [tex]\sigma ^2[/tex] is (56.31, 224.75)
For more about the normally distribution follow the link below-
https://brainly.com/question/12421652
