Step by step
The question
Question: in a sample of 2000, 1100 vote yes: find the 99% confidence interval for the true proportion.
1 Name the numbers
p prime is the part of the sample that says yes: x divided by n.
1100 divided by 2000 is 0.55.
q prime is the rest: 1 minus 0.55 is 0.45.
Careful!
p prime is the fraction 0.55, not the count 1100.
Both counts, 1100 and 900, are at least 5.
So the normal bell works here.
2 Find z*
99 percent of the bell is in the middle.
That leaves 1 percent: 0.5 percent in each end.
z* marks the cut between the middle and the ends.
Left of z* are the middle and one end: 0.99 plus 0.005 is 0.9950.
Now we look for 0.9950 inside the table.
0.9950 is halfway between 0.9949 and 0.9951.
So z is halfway between 2.57 and 2.58: 2.575.
With more digits it is 2.576.
That is the number books use.
3 Error bound
The error bound for a proportion: z* times the square root of p prime times q prime over n.
0.55 times 0.45 is 0.2475.
Divided by 2000: 0.00012375.
The square root of that is about 0.0111.
Times 2.576: about 0.0287.
4 The interval
Take the error bound away from the middle: 0.55 minus 0.0287 is 0.5213.
And add it: 0.55 plus 0.0287 is 0.5787.
The 99 percent confidence interval is (0.5213, 0.5787).
5 Say it in words
In words: we are 99 percent confident that the true proportion is between 52.13% and 57.87%.
6 Check
Let's check: the middle of the interval should be 0.55.
0.5213 plus 0.5787, divided by 2, is 0.55.
Half the width should be the error bound: 0.5787 minus 0.5213, divided by 2, is 0.0287.
So it is right!
Let's look back.
First: name the numbers.
Then: find z*.
Then: error bound.
Then: the interval.
Then: say it in words.
Last: check.