SOLUTION: Another Question , Help Please
A survey of 20 randomly sampled judges employed by the state of Florida found that they earned an average wage (including benefits) of $57.00 per
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A survey of 20 randomly sampled judges employed by the state of Florida found that they earned an average wage (including benefits) of $57.00 per
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Question 1112685: Another Question , Help Please
A survey of 20 randomly sampled judges employed by the state of Florida found that they earned an average wage (including benefits) of $57.00 per hour. The sample standard deviation was $5.60 per hour. (Use t Distribution Table.)
Q1) What is the best estimate of the population mean?
Q2) Develop a 90% confidence interval for the population mean wage (including benefits) for these employees. (Round your answers to 2 decimal places.)
Confidence interval for the population mean wage is between ...?... and ...?...
Q3)How large a sample is needed to assess the population mean with an allowable error of $1.00 at 90% confidence? (Round up your answer to the next whole number.) Sample size ?
You can put this solution on YOUR website! A survey of 20 randomly sampled judges employed by the state of Florida found that they earned an average wage (including benefits) of $57.00 per hour. The sample standard deviation was $5.60 per hour. (Use t Distribution Table.)
Q1) What is the best estimate of the population mean wage ?
Ans:: $57.00
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Q2) Develop a 90% confidence interval for the population mean wage (including benefits) for these employees. (Round your answers to 2 decimal places.)
Margin of Error = t*s/sqrt(20) = |invT(0.05,19)|*s/sqrt(20) = 1.7291*5.6/sqrt(20)
= 2.`65
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Confidence interval for the population mean wage is between
57-2.165 and 57+2.165
-------------------------------.
Q3)How large a sample is needed to assess the population mean with an allowable error of $1.00 at 90% confidence? (Round up your answer to the next whole
number.)
n = [t*s/E]^2 = [1.7291*5.6/1]^2 = 94 when rounded up
Sample size ? = 94
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Note:: I only used the t-Distribution because your post called for it.
I would normally use the z-distribution.
Cheers,
Stan H.
You can put this solution on YOUR website! Q1. Use the sample mean of $57.00
:
Q2. Standard Error(SE) = $5.60/square root(2) = 1.2522
:
alpha(a) = 1 - (90/100) = 0.10
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critical probability(p*) = 1 - (a/2) = 0.95
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degrees of freedom(DF) = 20 - 1 = 19
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the critical value(CV) is the t-statistic having 19 DF and a cumulative probability of 0.95. From the t-distribution tables, the CV is 1.729
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Margin of Error(ME) is CV * SE = 1.729 * 1.2522 = 2.1650 approximately 2.17
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90% confidence interval is $57.00 + or - 2.17 = ($54.83, $59.17)
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Q3. We have no standard deviation for the population, so we must estimate it.
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Since our sample size is 20 which is < 30, we must use the t-statistic and the standard deviation of the sample($5.60)
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The problem states that the allowable error is $1.00, which I assume describes the margin of error(ME)
:
ME = 1
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estimated population standard deviation = 5.60
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CV is 1.729
:
ME = CV * standard deviation/square root(n)
:
n = ((1.729 * 5.60)/1)^2 = 93.7488
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sample size(n) is 94
:
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