Question 128115: Problem One:
According to police records, from 1987 to 1995, the city of Davis, California experienced the following burglary rate (offenses per 1,000 residents for a given year):
(1) What is the mean burglary rate?
(2) What is the standard deviation of the burglary rate?
(3) What is the degree of freedom?
(4) What is the t-statistic @ 95% confidence?
(5) What is the standard error?
(6) What is the amount of error?
(7) What is the upper confidence bound per thousand offenses @ 95% confidence the mean annual burglary rate in Davis, California?
(8) What is the lower confidence bound per thousand offences @ 95% confidence the mean annual burglary rate in Davis, California?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website!
Problem One:
According to police records, from 1987 to 1995, the city of Davis, California experienced the following burglary rate (offenses per 1,000 residents for a given year):
year: 1987 data 15.8
year: 1988 data 17.2
year: 1989 data 16.5
year: 1990 data 15.5
year 1991 data 17.1
year 1992 data 18.6
year 1993 data 21.4
year 1994 data 29.0
year 1995 data 19.3
(1) What is the mean burglary rate?
Add the numbers and divide by nine to get 18.9333
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(2) What is the standard deviation of the burglary rate?
I used a calculator to get: s = 4.2065
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(3) What is the degree of freedom?
9-1 = 8
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(4) What is the t-statistic @ 95% confidence?
t(18.9333) = (18.9333 - population mean)/[4.2065/(sqrt(9)]
Comment: You did not include the population mean in your posting.
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(5) What is the standard error?
E = 1.96= 2.7482
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(6) What is the amount of error?
Not sure how that differs from (5)
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(7) What is the upper confidence bound per thousand offenses @ 95% confidence the mean annual burglary rate in Davis, California?
18.9333+2.7482
(8) What is the lower confidence bound per thousand offences @ 95% confidence the mean annual burglary rate in Davis, California?
18.9333-2.7482
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Cheers,
Stan H.
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