Respuesta :
Answer:
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.
