Given Data Set = {15,19,16,18,10,20,19,13,22,9,19,22,18,20,24,19}
n = 16
because n > 30 is not true, and because we don't know sigma (population standard deviation), we use a T distribution
df = n-1 = 16-1 = 15 degrees of freedom
Use a calculator or spreadsheet program to find these two values
xbar = sample mean = 17.6875
s = sample standard deviation = 4.1748253456 which is approximate
In the back of your stats textbook is a t table that looks similar to this shown here
http://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
Using that table specifically, highlight the entire row that starts with df = 15. Highlight the column that represents the 90% confidence level (see the bottom portion of the table).
The value 1.753 is in this row and column. Therefore, t = 1.753 is the approximate critical value.
Note: P(|T| < 1.753) = 0.90 which is another way of saying P(-1.753 < T < 1.753) = 0.90
Now onto computing the lower (L) and upper (U) bounds of the confidence interval.
L = lower bound of confidence interval
L = xbar - t*(s/sqrt(n))
L = 17.6875 - 1.753*(4.1748253456/sqrt(16))
L = 17.6875 - 1.753*(4.1748253456/4)
L = 17.6875 - 1.753*(1.0437063364)
L = 17.6875 - 1.8296172077092
L = 15.8578827922908
L = 15.86
U = upper bound of confidence interval
U = xbar + t*(s/sqrt(n))
U = 17.6875 + 1.753*(4.1748253456/sqrt(16))
U = 17.6875 + 1.753*(4.1748253456/4)
U = 17.6875 + 1.753*(1.0437063364)
U = 17.6875 + 1.8296172077092
U = 19.5171172077092
U = 19.52
(L,U) = (15.86,19.52)
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Answer:
The 90% confidence interval is approximately (15.86,19.52)
We can write the confidence interval in the form L < mu < U which means we would have 15.86 < mu < 19.52
Yet another alternative is to write the confidence interval in the form