In: Statistics and Probability
A sample of 6 credit ratings for car loan applicants has a mean of 602 and a variance of 4356. Use the data to construct a 98% confidence interval for the population standard deviation of credit scores for all applicants for car loans. (6 points) State the Best point estimate and Include the written statement
Confidence interval for population standard deviation is given as below:
Sqrt[(n – 1)*S2 / χ2 α/2, n – 1 ] < σ < sqrt[(n – 1)*S2 / χ2 1 - α/2, n – 1 ]
We are given
Confidence level = 98%
Sample size = n = 6
Degrees of freedom = n – 1 = 5
Best Point estimate = Sample standard deviation = S = sqrt(4356) = 66
χ2 α/2, n – 1 = 15.0863
χ2 1 - α/2, n – 1 = 0.5543
(By using chi square table)
Sqrt[(n – 1)*S2 / χ2 α/2, n – 1 ] < σ < sqrt[(n – 1)*S2 / χ2 1 - α/2, n – 1 ]
Sqrt[(6 – 1)*4356/ 15.0863] < σ < sqrt[(6 – 1)*4356 / 0.5543]
37.9960 < σ < 198.2245
Lower limit = 37.9960
Upper limit = 198.2245