Step by step
The question
Question: a sample of 20 has mean 48.6 and SD 2.5: find the 95% confidence interval for the true mean.
x̄, the sample mean, is 48.6: the middle of our interval.
s is the SD of the sample: 2.5.
And n is 20.
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, more so for small samples: a small sample makes us less sure.
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.
20 minus 1 is 19.
2 Find t*
95 percent of the bell is in the middle.
That leaves 5 percent: 2.5 percent in each end.
t* marks the cut between the middle and the ends.
In the t-table we take row 19.
Its columns are named by the middle part: the 95 percent column, which leaves 2.5 percent in each end.
Row 19, column 95%: t* = 2.093.
Careful!
df is n minus 1: 19, not 20.
3 Error bound
The error bound: t* times s over the square root of n.
s over the square root of n is the SD of sample means, from the sample means lesson.
t* of those SDs on each side holds the middle part we want.
2.5 over the square root of 20 is about 0.559.
Times 2.093: about 1.17.
4 The interval
Take the error bound away from the middle: 48.6 minus 1.17 is 47.43.
And add it: 48.6 plus 1.17 is 49.77.
The 95 percent confidence interval is (47.43, 49.77).
5 Say it in words
In words: we are 95 percent confident that the true mean is between 47.43 and 49.77.
6 Check
Let's check: the middle of the interval should be 48.6.
47.43 plus 49.77, divided by 2, is 48.6.
Half the width should be the error bound: 49.77 minus 47.43, divided by 2, is 1.17.
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.