Question 1193902
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There are n = 15 values in the list, which is the sample size.


To get the sample mean, first we add up the values
12+8+11+10+11+6+6+9+6+11+11+11+7+7+9 = 135
Then divide this over the sample size (n = 15) to get
135/15 = 9
The sample mean is 9
We call this xbar because x has a horizontal bar over top.
xbar = 9


You could calculate the sample standard deviation by hand, but it's preferable to use technology instead. There are many free calculators online to help with that if you don't have a TI83/84.
You should get a sample standard deviation of roughly s = 2.17
The sample standard deviation s estimates the population standard deviation sigma. 
s = sample standard deviation = 2.17 approximately
sigma = population standard deviation = unknown


Because we don't know sigma and because n > 30 is not true, we must use a T distribution. 
The degrees of freedom (df) is equal to the sample size minus 1
df = n-1
df = 15-1
df = 14


Use the T table in the back of your textbook, or use a free online resource like this
<a href = "https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf">https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf</a>
Start by looking at the row labeled "two tails". The value 0.05 corresponds to a confidence interval of 95% since 1 - 0.05 = 0.95
The 0.05 refers to the combined area of both tails.
Highlight this entire column.
Then look in the row that has df = 14 at the very left.
The intersection of the row and column leads to the approximate t critical value of 2.145
What is this value saying? It says that P(-2.145 < T < 2.145) = 0.95 approximately when df = 14.
95% of the area under the T curve (df = 14) is between roughly t = -2.145 and t = 2.145



Let's compute the margin of error
{{{E = t[c]*expr(s/sqrt(n))}}}


{{{E =  2.145*expr(2.17/sqrt(15))}}}


{{{E =  1.20182546216162}}}


{{{E =  1.20}}}


And lastly, let's compute the confidence interval for the mean mu (symbol 𝜇)
{{{xbar - E < mu < xbar + E}}}


{{{9 - 1.20 < mu < 9 + 1.20}}}


{{{7.80 < mu < 10.20}}}
We could shorten this to (7.80, 10.20) which is common notation for confidence intervals. 
Some books and research papers use the notation [7.80, 10.20] to mean the same thing. 
We are 95% confident the true population mean (mu) is somewhere between 7.80 and 10.20


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Answers:
Sample mean = 9
Sample standard deviation = 2.17
Degrees of Freedom = 14
{{{t[c]}}} (t critical value) is roughly  2.145
Margin of error: E = 1.20
The 95% confidence interval is roughly  {{{7.80 < mu < 10.20}}}
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