Step by step
The question
Question: a sample of 9 has mean 55 and SD 9: find the 90% confidence interval for the true mean.
x̄, the sample mean, is 55: the middle of our interval.
s is the SD of the sample: 9.
And n is 9.
1 Why t
We do not know σ, the SD of all the numbers.
We only have s, the SD of the sample.
So we use t instead of z.
The t curve is a bit wider than the bell.
It makes the interval a bit wider, to be safe.
The t-table has a row for each df, the degrees of freedom: n minus 1.
9 minus 1 is 8.
2 Find t*
90 percent of the bell is in the middle.
That leaves 10 percent: 5 percent in each end.
t* marks the cut between the middle and the ends.
In the t-table we take row 8 and the 90 percent column.
Row 8, column 90%: t* = 1.860.
Careful!
df is n minus 1: 8, not 9.
3 Error bound
The error bound: t* times s over the square root of n.
The square root of 9 is 3.
9/3 is 3.
Times 1.86: 5.58.
4 The interval
Take the error bound away from the middle: 55 minus 5.58 is 49.42.
And add it: 55 plus 5.58 is 60.58.
The 90 percent confidence interval is (49.42, 60.58).
5 Say it in words
In words: we are 90 percent confident that the true mean is between 49.42 and 60.58.
6 Check
Let's check: the middle of the interval should be 55.
49.42 plus 60.58, divided by 2, is 55.
Half the width should be the error bound: 60.58 minus 49.42, divided by 2, is 5.58.
So it is right!
Let's look back.
First: why t.
Then: find t*.
Then: error bound.
Then: the interval.
Then: say it in words.
Last: check.