Question

N Distance Driven Average 25265 Std Dev 29.49301 Variance869.8378 M Height Average Std Dev 3.475409 Variance 12.07847 69.09

Answer 1

a ) We have to find 95% confidence interval on the variance in the height of an evening MBA student.

Confidence interval for variance is given by ,

$\frac{(n-1)s^2}{\chi^2 _{\alpha /2,df}} < \sigma ^2 <\frac{(n-1)s^2}{\chi^2 _{1-\alpha /2,df}}$

Where , n is sample size = 50

s² is sample variance= 12.0785

Degrees of freedom = df = n - 1 = 49

confidence level = 95% ,

significance level =$\alpha$ = 1 - 0.95 = 0.05 , $\alpha$/2 = 0.025 ,

$\chi^2 _{\alpha /2 ,df}$   is right tail critical value of chi-square distribution at given confidence level.

$\chi^2 _{\alpha /2,df}$ = $\chi^2 _{0.025,49}$ = 70.2224              { Using Excel , =CHIINV( 0.025,49) = 70.2224 }

$\chi^2 _{1-\frac{\alpha }{2},df}$ is left tail critical value of chi-square distribution at given confidence level.

$\chi^2 _{1-\frac{\alpha }{2},df}$ = $\chi^2 _{0.975,49}$ = 31.5549            { using Excel , =CHIINV(0.975 , 49 ) = 31.5549}

So 95% confidence interval for variance is ,

$\frac{(50-1)12.0784}{70.2224} < \sigma ^2 <\frac{(50-1)12.0784}{31.5549}$

$8.4281 < \sigma ^2 <18.7559$

95% confidence interval on the variance in the height of an evening MBA student is ( 8.4281 , 18.7559 )

b )

We have to find 90% confidence interval on the standard deviation in the distance driven to campus

Confidence interval for standard deviation is given by,

$\sqrt{\frac{(n-1)s^2}{\chi^2 _{\alpha /2,df}} }<\sigma < \sqrt{\frac{(n-1)s^2}{\chi^2 _{1-\alpha /2,df}}}$

Where , n is sample size = 50

s² is sample variance= 869.8378

Degrees of freedom = df = n - 1 = 49

confidence level = 90% ,

significance level =$\alpha$ = 1 - 0.90 = 0.1 , $\alpha$/2 = 0.05

$\chi^2 _{\alpha /2,df}$ = $\chi^2 _{0.05,49 }$ =  66.3386                      { Using Excel , =CHIINV(0.05,49) = 66.3387 }

$\chi^2 _{1-\alpha /2,df}$ = $\chi^2 _{0.95,49}$ = 33.9303                 { Using Excel , =CHIINV(0.95,49) = 33.9303 }

90% Confidence interval for standard deviation is,

$\sqrt{\frac{(50-1)869.8378}{66.3386} }<\sigma<\sqrt{\frac{(50-1)869.8378}{33.9303}}$

$25.3474 <\sigma<35.4424$

90% confidence interval on the standard deviation in the distance driven to campus is ( 25.3474 , 35.4424 )