Question 1163876: The manager of the Super Mall, wants to estimate the mean amount spent per shopping visit
by customers. A sample of 20 customers reveals the following amounts spent.
$48.16 $42.22 $46.82 $51.45 $23.78 $41.86 $54.86 $37.92 $52.64 $48.59 $50.82
$46.94 $61.83 $61.69 $49.17 $61.46 $51.35 $52.68 $58.84 $43.88
What is the best estimate of the population mean? Determine a 99 percent confidence interval.
Interpret the result. Would it be reasonable to conclude that the population mean is $50? What
about $60?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
The best estimate of the population mean (mu) is the sample mean (xbar). To find the population mean, we would have to find all the values in the population, add them up, and divide by the population size. To find the sample mean, we do the same thing but with the small sample instead of the entire population.
If you were to add up all of those dollar amounts given, you would get 986.96; I recommend using a spreadsheet program to get the sum. Divide this over 20 (since the sample size is 20) to get 986.96/20 = 49.348
So xbar = 49.348 is the best estimate of the mean. I'll leave it unrounded, though if you want to round to the nearest cent, then you'd have xbar = 49.35
In addition to the sample mean, we'll also need the sample standard deviation. This is where a calculator or spreadsheet will need to be used. Doing so by hand takes quite a bit of time especially for n = 20 values. You should get the approximate standard deviation of s = 9.01211
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At 99% confidence, the t critical value is approximately t = 2.8609 when the degrees of freedom is df = n-1 = 20-1 = 19
You'll need a calculator or a table for this step. I recommend a calculator. This is the calculator I used
https://www.statology.org/inverse-t-distribution-calculator/
Though you could use a calculator such as a TI83/TI84 just as well.
The degrees of freedom is df = 19 and you'll have 0.99 as the confidence level. The two-sided t score is roughly 2.8609
This means that
P(-2.8609 < T < 2.8609) = 0.99
approximately
So because the t critical value is t = 2.8609, we can say
L = xbar - t*s/sqrt(n)
L = 49.348 - 2.8609*9.01211/sqrt(20)
L = 43.582802841766
L = 43.58
is the lower bound of the confidence interval
and
U = xbar + t*s/sqrt(n)
U = 49.348 + 2.8609*9.01211/sqrt(20)
U = 55.113197158234
U = 55.11
is the upper bound of the confidence interval
The 99% confidence interval is therefore
(L,U) = (43.58, 55.11)
This is the same as writing
43.58 < mu < 55.11
saying that mu is somewhere between 43.58 and 55.11, in which we are 99% confident of this statement.
Since mu = 50 is between the lower and upper bounds mentioned, this means it is reasonable that the population mean could be $50. It's not reasonable to have mu = 60 since this is larger than the upper bound value.
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