SOLUTION: A random sample of 16 pharmacy customers showed the waiting times below (in minutes). 11 25 19 17 24 16 18 20 19 23 21 17 1

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Question 1206118: A random sample of 16 pharmacy customers showed the waiting times below (in minutes).
11 25 19 17 24 16 18 20
19 23 21 17 17 14 15 11
Waiting TImes for Prescription (n = 16)

Minutes
11
25
19
17
24
16
18
20
19
23
21
17
17
14
15
11
Find a 90 percent confidence interval for μ, assuming that the sample is from a normal population. (Round your standard deviation answer to 4 decimal places and t-value to 3 decimal places.)
The 90% confidence interval ____ to ____

Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!

Add up the values and divide by 16 to determine is the exact sample mean (no rounding has been done to it).

Use a calculator to determine the sample standard deviation.

If you have a TI84 or similar, then refer to this page

If you want to know what's going on under the hood, then check out this solution for a similar example. Realistically it's best to use a calculator to avoid such tedious calculations by hand.

You should find that is the approximate sample standard deviation when rounding to 4 decimal places.

Next we'll need a T table. Such tables are found in the back of your stats textbook.
For exam purposes, your teacher should hand out the table if s/he expects you to use it.

The degrees of freedom are df = n-1 = 16-1 = 15
Look at the row that starts with df = 15.
Also, look at the column that has "two tails = 0.10" at the top.
The 0.10 refers to the fact 1 - 0.90 = 0.10 is the area of the combined tails when the main body is 0.90 to represent the 90% confidence level.

The value at the row df = 15 and column "two tails = 0.10" is roughly 1.753
What this means is P(-1.753 < t < 1.753) = 0.90 approximately when df = 15.

Another way to determine this value is to use a TI84 (or similar).

Here's an alternative online calculator that will reach the same goal. The only downside is that there's no option to adjust rounding precision.
Refer to the "Two-sided t-Score".
Feel free to explore other calculators.

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Summary so far:
xbar = 17.9375(exact; no rounding)
s = 4.1226 (approximate)
t = 1.753 (approximate)
n = 16

Let's calculate the margin of error for the mean.
E = t*s/sqrt(n)
E = 1.753*4.1226/sqrt(16)
E = 1.806729 approximately

Then we can compute lower and upper bounds (L and U) of this confidence interval.
L = lower bound
L = xbar - E
L = 17.9375 - 1.806729
L = 16.130771
L = 16.13
Your teacher doesn't mention rounding instructions for lower or upper bounds.
I'll assume the standard convention of 2 decimal places. I recommend asking your teacher for clarification.

U = upper bound
U = xbar + E
U = 17.9375 + 1.806729
U = 19.744229
U = 19.74

The 90% confidence interval of the form L < mu < U is roughly 16.13 < mu < 19.74

We can condense this down to the shorthand notation (16.13, 19.74)
Some textbooks will use the notation [16.13, 19.74] instead.