Respuesta :
Answer: a. CI for the mean: 17.327 < μ < 26.473
b. CI for variance: 29.7532 ≤ [tex]\sigma^{2}[/tex] ≤ 170.9093
Step-by-step explanation:
a. To construct a 95% confidence interval for the mean:
The given data are:
mean = 21.9
s = 7.7
n = 12
df = 12 - 1 = 11
1 - α = 0.05
[tex]\frac{\alpha}{2}[/tex] = 0.025
t-score = [tex]t_{0.025,11}[/tex] = 2.2001
Note: since the sample population is less than 30, it is used a t-score.
The formula for interval:
mean ± [tex]t.\frac{s}{\sqrt{n} }[/tex]
Substituing values:
21.9 ± 2.200.[tex]\frac{7.7}{\sqrt{12} }[/tex]
21.9 ± 4.573
The interval is: 17.327 < μ < 26.473
b. A 95% confidence interval for the variance:
The given values are:
[tex]s^{2}[/tex] = [tex]7.7^{2}[/tex]
[tex]s^{2}[/tex] = 59.29
α = 0.05
[tex]\frac{\alpha}{2}[/tex] = 0.025
[tex]1-\frac{\alpha}{2}[/tex] = 0.975
[tex]\chi^{2}_{0.025,11}[/tex] = 21.92
[tex]\chi^{2}_{0.975,11}[/tex] = 3.816
Note: To find the values for [tex]\chi^{2}_{\alpha/2,n-1}[/tex] and [tex]\chi^{2}_{1-\alpha/2,n-1}[/tex], look for them at the chi-square table
The formula to calculate interval:
([tex]\frac{(n-1).s^{2}}{\chi^{2}_{\alpha/2,n-1}} , \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2,n-1}}[/tex])
are the lower and upper limits, respectively.
Substituing values:
([tex]\frac{11.59.29}{21.92} , \frac{11.59.29}{3.816}[/tex])
(29.7532, 170.9093)
The interval for variance is: 29.7532 ≤ [tex]\sigma^{2}[/tex] ≤ 170.9093
