SOLUTION: In this problem, how do they get the 2.682? Forty-nine items are randomly selected from a population of 500 items. The sample mean is 40 and the sample standard deviation 9.

Question 1030262: In this problem, how do they get the 2.682?
Forty-nine items are randomly selected from a population of 500 items. The sample mean is 40 and the sample standard deviation 9.

Found 2 solutions by rothauserc, addingup:
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
what is the problem asking for?
:
student states that a 99% confidence interval is needed
:
Margin of Error(M.E.) = critical value * standard deviation of statistic
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Confidence interval(C.I.) = critical value + or - M.E.
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the sample mean is 40 and sample standard deviation is 9
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since sample size is > 30, we can assume normal distribution
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M.E. = 9 / sqrt(49) = 1
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critical value(CV) is calculated using these steps
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1) alpha = 1 - (99/100) = 0.01
critical probability = 1 - (0.01/2) = 0.995
consult the z-tables for the associated z-value
z-value = 2.576
:
*********************
Normal Distribution
C.I. = 2.576 + or - 1
*********************
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Now use the student t-distribution to calculate the critical value
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the degrees of freedom(D.F.) = sample size - 1
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D.F. = 49 - 1 = 48
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the C.V. is the t-score having 48 degrees of freedom and a cumulative probability equal to 0.995
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consult the table of t-scores
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t-score = 2.682
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***********************
Student t-distribution
C.I. = 2.682 + or - 1
***********************
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Answer by addingup(3677) (Show Source): You can put this solution on YOUR website!
look at your t-value chart. Also, if you are looking for help you need to provide the entire problem, you're missing important information

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