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 (a) - (c) 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.