1. P($82,864.02 le U ge $84,902.64)=.90
2. P($78,916.24 le Xnew ge $88,850.42)=.90
What did you get for problems 3 and 4 (skip 5)? Please post your answer by commenting to this post.
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6 comments:
#3: P(83,474 le M ge 88,125.35)=90%
#4: P(74,634.93 le M ge 91,965.08)=90%
For #2 I am getting a different answer. The k value i get is 3.16228. The answer I get is:
P(82,052.91 le M ge 85,713.75)=90%.
Dr. Hansz, am I doing something wrong when calculating K? Does anyone else get the same answer I came up with?
Homework Problem #1:
GIVEN:
- Sample Mean(xbar)=83,883.33
- Sample Size(n)=15
- Degrees of freedom=n–1=14
- Sample standard deviation(S)=2,241.78
- Confidence Interval≈90%
A confidence interval of 90% means that α = 10% → α⁄2 = 0.05. Because the population standard deviation is not known (sigma=?), we must use the t-distribution table and use the sample standard deviation(S). A sample size of 15 means that our degrees of freedom is 14. At df = 14, t0.05 = 1.761. From here on out, it’s plug and chug…
μ=xbar±{t(α⁄2)}S/√n
=83,883.33±(1.761)(2,241.78)/√15
=83,883.33±1,019.31
=[82,864.02,84,902.64]
=P(82,864..02≤μ≤84,902.64)=0.90
Homework Problem #2:
GIVEN:
- Sample Mean(xbar)=83,883.33
- Sample Size(n)=15
- Sample standard deviation(S)=2,241.78
- Confidence Interval≈90%
- Unknown population functional form
Because the population standard deviation (sigma) and the population functional form are not known, we must use Chebychev’s Theorem to establish bounds on the possible values of the random variable with certain probability. A confidence interval of 90% means that Pm=90%.
Pm=1-1/k^2
∴ k=√(1/(1-Pm ))
=√(1/(1-0.90))=3.162
μ=xbar±k{S/√n}
=83,883.33±(3.162)(2,241.78)/√15
=83,883.33±1,830.24
=[82,053.09,85,713.57]
=P(82,053.09≤μ≤85,713.57)≥0.90
Homework Problem #3:
GIVEN:
- Sample Mean(xbar)=83,300
- Sample Size(n)=15
- Degrees of freedom=n–1=14
- Sample standard deviation(S)=2,177.81
- Confidence Interval≈90%
A confidence interval of 90% means that α = 10% → α⁄2 = 0.05. Because the population standard deviation is not known (sigma=?), we must use the t-distribution table and use the sample standard deviation(S). A sample size of 15 means that our degrees of freedom is 14. At df = 14, t0.05 = 1.761. From here on out, it’s plug and chug…
xnew= xbar±{t(α⁄2)}(S+S/√n)
=83,300±1.761{2,177.81+(2,177.81/√15)}
=83,300±4,825.35
=[78,474.65,88,125.35]
=P(78,474.65≤x_new≤88,125.35)=0.90
Homework Problem #4:
GIVEN:
- Sample Mean(xbar)=83,300
- Sample Size(n)=15
- Sample standard deviation(S)=2,177.78
- Confidence Interval≈90%
- Unknown population functional form
Because the population standard deviation (sigma) and the population functional form are not known, we must use Chebychev’s Theorem to establish bounds on the possible values of the random variable with certain probability. A confidence interval of 90% means that Pm = 90%.
Pm=1-1/k^2
∴k=√(1/(1-Pm))
=√(1/(1-0.90))=3.162
xnew=xbar±k(S+S/√n)
=83,300±3.162(2177.78+2,177.78/√15)
=83,300±8,664.13
=[74,635.87,91,964.13]
=P(74,635.87≤xnew≤91,964.13)≥0.90
I would be grateful if someone would critique my procedures and answers and let me know if I did something wrong or if this agrees with your own procedures and answers.
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