Question 1175141
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Problem 1
Use a spreadsheet or calculator to compute the following
xbar = 67.8551020408163
s = 15.8895151649404


xbar is the sample mean, while s is the sample standard deviation
Because we have n = 49 items here, this means we have enough to use the Z distribution. We can use this if we know sigma, or if n > 30. When n > 30, the student T distribution is approximately the same as the standard normal distribution.


At 90% confidence, the z critical value is roughly z = 1.645; use a table or calculator to compute this


L = lower bound of confidence interval
L = xbar - z*s/sqrt(n)
L = 67.8551020408163 - 1.645*15.8895151649404/sqrt(49)
L = 67.8551020408163 - 3.734036063761
L = 64.1210659770552
L = 64.12


U = upper bound of confidence interval
U = xbar + z*s/sqrt(n)
U = 67.8551020408163 + 1.645*15.8895151649404/sqrt(49)
U = 67.8551020408163 + 3.734036063761
U = 71.5891381045772
U = 71.59


The 90% confidence interval of the form L < mu < U is therefore <font color=red>64.12 < mu < 71.59</font>


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Problem 2


z = 1.645 from earlier (critical value for 90% confidence interval) 
sigma = 36.7 = given population standard deviation
E = 0.1 = desired error


n = minimum sample size needed
n = (z*sigma/E)^2
n = (1.645*36.7/0.1)^2
n = 364,471.801225
n = <font color=red>364,472</font>
Always round up to the nearest whole number for this type of problem.


The min sample size needed is <font color=red>364,472</font>


Side note: you may need to ignore the comma if you are typing this answer into a computer system
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