Respuesta :
Answer:
1.- CI = 94% ( μ₀ - 1,04 < x < μ₀ + 1,04 )
2.- CI = 98 % ( μ₀ - 2,05 < x < μ₀ + 2,05 )
3.- CI = 96 % ( μ₀ - 1,75 < x < μ₀ + 1,75 )
Step-by-step explanation:
Sample size n = 3000000
Sample mean x = 200
Standard deviation s = 30
From z-table values of z(c):
CI 94 % Confidencial level α = 6 % α = 0,06 z(c) = 1,55
CI 98 % Confidencial level α = 2 % α = 0,02 z(c) = 2,05
CI 96 % Confidencial level α = 4 % α = 0,04 z(c) = 1,75
MOE = z(c) * σ/√n
1.-MOE = 1,55* 30 / √2000 MOE = 1,04
2.-MOE = 2,05*30/√2000 MOE = 1,38
3.-MOE = 1,75*30/√2000 MOE = 1,17
Then CI
1.- CI = 94 % ( μ₀ - MOE < x < μ₀ - MOE )
CI = ( μ₀ - 1,04 < x < μ₀ + 1,04 )
2.-CI = 98 %
CI = ( μ₀ - 2,05 < x < μ₀ + 2,05 )
3.- CI = 96 %
CI = ( μ₀ - 1,75 < x < μ₀ + 1,75 )
Using the t-distribution, it is found that:
- The 94% confidence interval is (198.73, 201.27).
- The 96% confidence interval is (198.61, 201.39).
- The 98% confidence interval is (198.43, 201.57).
We are given the standard deviation for the sample, which is why the t-distribution is used to solve this question.
The information given is:
- Sample mean of [tex]\overline{x} = 200[/tex].
- Sample standard deviation of [tex]s = 30[/tex].
- Sample size of [tex]n = 2000[/tex].
The interval is:
[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]
- In which t is the critical value for the two-tailed confidence interval.
Considering a 94% confidence level, using a calculator, with 200 - 1 = 199 df, the critical value is t = 1.8916, hence:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 200 - 1.8916\frac{30}{\sqrt{2000}} = 198.73[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 200 + 1.8916\frac{30}{\sqrt{2000}} = 201.27[/tex]
The 94% confidence interval is (198.73, 201.27).
Considering a 96% confidence level, using a calculator, with 200 - 1 = 199 df, the critical value is t = 2.0673, hence:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 200 - 2.0673\frac{30}{\sqrt{2000}} = 198.61[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 200 + 2.0673\frac{30}{\sqrt{2000}} = 201.39[/tex]
The 96% confidence interval is (198.61, 201.39).
Considering a 98% confidence level, using a calculator, with 200 - 1 = 199 df, the critical value is t = 2.3452, hence:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 200 - 2.3452\frac{30}{\sqrt{2000}} = 198.43[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 200 + 2.3452\frac{30}{\sqrt{2000}} = 201.57[/tex]
The 98% confidence interval is (198.43, 201.57).
You can learn more about the use of the t-distribution to build a confidence interval at https://brainly.com/question/16162795
