Question 1170607
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Part (a)


The population mean is unknown. We'll let mu be the population mean. 


The goal of statistics is to estimate population parameters based on sample statistics. This is due to the high cost (of money and time) dealing with taking a census, so resorting to a representative sample is the next best thing. Therefore, it is quite common to not know the population mean. The same goes for the population standard deviation; however, this was given to be sigma = 3000. 


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Part (b)


The variable xbar is the best estimate of the population mean. The statistic xbar is an unbiased estimate of the parameter mu. 


In this problem, xbar = 20000


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Part (c)


n = 40 is the sample size
sigma = 3000 is the population standard deviation


We'll use the z distribution because sigma is known. 


At 99% confidence, the z critical value is about 2.576 which is determined through use of a calculator or table.


The margin of error is
E = z*sigma/sqrt(n)
E = 2.576*3000/sqrt(40)
E = 1221.90408788907


The confidence interval for mu is
xbar-E < mu < xbar+E
20000-1221.90408788907 < mu < 20000+1221.90408788907
18778.095912111 < mu < 21221.904087889
18778.096 < mu < 21221.904


The 99% confidence interval for mu is 18778.096 < mu < 21221.904


This is equivalent to saying (18778.096, 21221.904)


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Part (d)


We are 99% confident that mu is between 18778.096 and 21221.904


In the context of the problem, it means we're 99% confident that the population mean of daily sales is between $18,778.096 and $21,221.904
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