Question 1041239: Twelve different video games showing substance use were observed and the duration times of game play (in seconds) are listed below. The design of the study justifies the assumption that the sample can be treated as a simple random sample. Use the data to construct a 90% confidence interval estimate the mean duration of game play.
4042,3883,3862,4034,3310,4804,4667, 4033, 5010, 4819, 3332, 3315.
What is the confidence interval of population mean?
_< mean < _
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Add up the values to get 49111.
Then divide that sum by 12 to get the sample mean to be 4092.58333333333.
So xbar = 4092.58333333333
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Use a calculator to compute the standard deviation to be 611.079292330506.
Doing the standard deviation by hand would take a very long time.
So s = 611.079292330506
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The degrees of freedom is df = n-1 = 12-1 = 11
Use this table to highlight the "90% confidence level" column and the row that has df = 11. The value at this row and column combo is 1.796. This is the t-critical value
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Summary
sample mean = xbar = 4092.58333333333
t critical value = t = 1.796
standard deviation = s = 611.079292330506
sample size = n = 12
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We'll use those values in the summary (above) to compute the lower and upper boundaries of the confidence interval
Lower boundary L
L = xbar - t*s/sqrt(n)
L = 4092.58333333333 - 1.796*611.079292330506/sqrt(12)
L = 3775.76283239028
L = 3775.763 ... rounding to three decimal places
Upper boundary U
U = xbar + t*s/sqrt(n)
U = 4092.58333333333 + 1.796*611.079292330506/sqrt(12)
U = 4409.40383427639
U = 4409.404 ... rounding to three decimal places
So the confidence interval, accurate to three decimal places, is 3775.763 < mean < 4409.404
If your book requires different decimal precision, then be sure to round in a different way.
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