Question

Use technology to construct the confidence intervals for the population variance sigmaσsquared2 and the population standard deviation sigmaσ. Assume the sample is taken from a normally distributed population. equals=0.95​, equals=31​, equals=20

Answer 1

Solution:- Given that information :

95% confidence interval , s = 31, n = 20 , s^2 = 31^2

Degrees of freedom = n-1 = 20-1 = 19

X^2R = X^20.025 = 32.8523 <--- CHIINV(0.025,19)

X^2L = X^20.975 = 8.9065 <--- CHIINV(0.975,19)

=> The confidence interval for the population variance :

: ((n-1)*s^2)/X^2R < σ^2 < ((n-1)*s^2)/X^2L

= ((19*961/32.8523) < σ^2 < ((19*961/8.9065)

= 555.7906 < σ^2 < 2050.0758

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=> the confidence interval for the population standard deviation:

: sqrt((n-1)*s^2)/X^2R) < σ< sqrt((n-1)*s^2)/X^2L)

sqrt(555.7906)< σ < sqrt(2050.0758)

= 23.5752 < σ < 45.2778