Simple random sample of size n is drawn from a population that is normally distributed. the sample mean, x overbar, is found to be 106, and the sample standard deviation, s, is found to be 10. (a) construct a 95% confidence interval about mu if the sample size, n, is 24. (b) construct a 95% confidence interval about mu if the sample size, n, is 15. (c) construct a 70% confidence interval about mu if the sample size, n, is 24. (d) could we have computed the confidence intervals in parts () if the population had not been normally distributed?

C.I: 95%            (101.777 < x < 110.223)   n = 24

C.I: 70%            (103.836 < x < 108.164)  n = 24

C.I: 95%            (101.622 < x < 110.378)   n = 15

Yes we can do parts through a-c if population had not been normally distributed. See explanation

Step-by-step explanation:

Given:

- Sample mean x_bar = 106

- Sample standard deviation s = 10

Find:

(a) Construct a 95% confidence interval about mu if the sample size, n, is 24.

(b) Construct a 95% confidence interval about mu if the sample size, n, is 15.

(c) Construct a 70% confidence interval about mu if the sample size, n, is 24.

Solution:

- For sample size n = 24. n < 30 T-score

          DOF = n - 1 = 24 - 1 = 23

          a = 1 - 0.95 = 0.05 ,           t_a/2 = t_0.025 = 2.069

          C.I: ( x_bar - t_0.025*s/sqrt(n) < x < x_bar + t_0.025*s/sqrt(n) )

           ( 106 - 2.069*10/sqrt(24) < x < 106 + 2.069*10/sqrt(24) )    

            C.I: 95%            (101.777 < x < 110.223)

- For sample size n = 24. n < 30 T-score

          DOF = n - 1 = 24 - 1 = 23

          a = 1 - 0.70 = 0.30 ,           t_a/2 = t_0.15 = 1.060

          C.I: ( x_bar - t_0.15*s/sqrt(n) < x < x_bar + t_0.15*s/sqrt(n) )

           ( 106 - 1.06*10/sqrt(24) < x < 106 + 1.06*10/sqrt(24) )    

            C.I: 70%        (103.836 < x < 108.164)

- For sample size n = 15. n < 30 T-score

          DOF = n - 1 = 15 - 1 = 14

          a = 1 - 0.95 = 0.05 ,           t_a/2 = t_0.025 = 2.145

          C.I: ( x_bar - t_0.025*s/sqrt(n) < x < x_bar + t_0.025*s/sqrt(n) )

           ( 106 - 2.145*10/sqrt(24) < x < 106 + 2.145*10/sqrt(24) )    

            C.I: 95%            (101.622 < x < 110.378)

- The answer is yes.

First, in many cases, the sample is not normally distributed but you can use the central limit theorem to make a normal approximation and obtain an asymptotical confidence interval.

But you can also find exact confidence intervals even if the data is not normal. You need to find a pivotal quantity, i.e. a statistic whose distribution does not depend on the parameter of the underlying distribution.

or example let X1,…,Xn∼Exp(λ). We have λX¯∼Γ(n,n). Thus we have :

                            P( (u_α/2) /X¯ ≤ λ ≤ u_(1−α/2) / X¯) = 1 − α

where u_α/2 and u_1−α/2 are the quantities of Γ(n,n) with levels α/2 and 1−α/2.                            

answer: 2 and 1

step-by-step explanation: got this question