Question 1206118
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Add up the values and divide by 16 to determine {{{xbar = 287/16 = 17.9375}}} is the exact sample mean (no rounding has been done to it).


Use a <a href="https://www.calculatorsoup.com/calculators/statistics/variance-calculator.php">calculator</a> to determine the sample standard deviation.


If you have a TI84 or similar, then refer to <a href="https://www.statology.org/sample-variance-ti-84/">this page</a>


If you want to know what's going on under the hood, then check out <a href="https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1190757.html">this solution</a> for a similar example. Realistically it's best to use a calculator to avoid such tedious calculations by hand.


You should find that {{{s = 4.1226}}} is the approximate sample standard deviation when rounding to 4 decimal places.


Next we'll need a <a href="https://www.tdistributiontable.com/">T table</a>. 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 <a href="https://www.statology.org/how-to-find-the-t-critical-value-on-a-ti-84-calculator/">TI84</a> (or similar).


Here's an alternative <a href="https://www.statology.org/inverse-t-distribution-calculator/">online calculator</a> 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.
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