Question 1060693: 1. You are interested in the average annual salary of American public health professionals. So, you took a random sample and found that the mean annual salary is $53,145. Suppose you knew from previous studies that the standard deviation of the annual salary is $16, 451. (Hint: given the population standard deviation is given, use the Z Table).
1). Please find a 95% confidence interval for the mean salary, given the following sample size: (a) n = 36, (b) n = 64, (c) n = 144.
(2) How does the range of values (i.e. Margin of Error) change when the sample size (n) changes?
(3) When the confidence level changes from 95% to 99%, what will be the confidence interval for a sample of n=36?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! 1) 95% confidence interval(C.I.)
alpha(a) = 1 - (95/100) = 0.05
critical probability(p*) = (1 - a/2) = 1 - 0.025 = 0.975
critical value(cv) = 1.96
1.a) standard error(SE) = 16451 / square root(36) is approximately 2741.83
margin of error(M.E.) = 1.96 * 2741.83 = 5373.9868 approximately 5373.99
95% C.I. = $53145 + or - $5373.99
1.b) SE = 16451 / square root(64) is approximately 2056.38
M.E. = 1.96 * 2056.38 = 4030.5048 is approximately 4030.51
95% C.I. = $53145 + or - $4030.51
1.c) SE = 16451 / square root(144) is approximately 1370.92
M.E. = 1.96 * 1370.92 = 2687.0032 is approximately 2687.00
95% C.I. = $53145 + or - $2687.00
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2) M.E. decreases with increased sample sizes
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3) 99% C.I.
a = 1 - (99/100) = 0.01
p* = 1 - 0.01/2 = 0.995
cv = 2.57
M.E. = 2.57 * 2741.83 = 7046.5031 is approximately 7046.50
99% C.I. = $53145 + or - $7046.50
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