Worked examples
The question
Question: the SD of all the numbers is 15: how big a sample do we need to be 95% confident with an error bound of 2?
n? 95% confidence, σ = 15, EBM = 2
The error bound is how far the sample mean may be from the true mean.
EBM: how far the sample mean may be from the true mean
1 Find z*
95 percent of the bell is in the middle.
That leaves 5 percent: 2.5 percent in each end.
95% in the middle: 5% left over, 2.5% in each end
z* marks the cut between the middle and the ends.
Left of z* are the middle and one end: 0.95 plus 0.025 is 0.9750.
left of z* = 0.95 + 0.025 = 0.9750
Now we look for 0.9750 inside the table.
0.9750 is in the table: row 1.9, column .06.
So z = 1.96.
z* = 1.96
2 The formula
The error bound is z* times σ over the square root of n.
Solved for n, it gives this formula.
from EBM = z* × σ ÷ √n, solved for n
n is z* squared times σ squared, over the error bound squared.
n = z*2 × σ2EBM2
3 Put the numbers in
1.96 squared is 3.8416.
15 squared is 225.
2 squared is 4.
3.8416 times 225 is 864.36.
Divided by 4: 216.09.
n = 1.962 × 15222 = 3.8416 × 2254 = 864.364 = 216.09
4 Round up
A sample size must be a whole number.
We always round up: a smaller sample would miss the goal.
always round UP: a bigger sample is safe
216.09 rounds up to 217.
n = 217
Careful!
Rounding down to 216 gives an error bound a bit too big.
Always round up.
Careful: not n = 216 ✗
5 Check
Let's check: with 217 people the error bound is about 1.9958.
That is not more than 2.
Good.
n = 217: EBM ≈ 1.9958 ≤ 2
With one person less, 216, it is about 2.0004: too big.
So 217 is the smallest that works.
So it is right!
n = 216: EBM ≈ 2.0004 > 2
Let's look back.
First: find z*.
Then: the formula.
Then: put the numbers in.
Then: round up.
Last: check.
Answer Careful: not n = 216 ✗
▶ Watch it solved
The question
Question: the SD of all the numbers is 8: how big a sample do we need to be 90% confident with an error bound of 1?
n? 90% confidence, σ = 8, EBM = 1
The error bound is how far the sample mean may be from the true mean.
EBM: how far the sample mean may be from the true mean
1 Find z*
90 percent of the bell is in the middle.
That leaves 10 percent: 5 percent in each end.
90% in the middle: 10% left over, 5% in each end
z* marks the cut between the middle and the ends.
Left of z* are the middle and one end: 0.9 plus 0.05 is 0.9500.
left of z* = 0.9 + 0.05 = 0.9500
Now we look for 0.9500 inside the table.
0.9500 is halfway between 0.9495 and 0.9505.
So z is halfway between 1.64 and 1.65: 1.645.
z* = 1.645
2 The formula
The error bound is z* times σ over the square root of n.
Solved for n, it gives this formula.
from EBM = z* × σ ÷ √n, solved for n
n is z* squared times σ squared, over the error bound squared.
n = z*2 × σ2EBM2
3 Put the numbers in
1.645 squared is about 2.706.
8 squared is 64.
1 squared is 1.
2.706 times 64 is 173.184.
Divided by 1: about 173.18.
n = 1.6452 × 8212 = 2.706 × 641 = 173.1841 ≈ 173.18
4 Round up
A sample size must be a whole number.
We always round up: a smaller sample would miss the goal.
always round UP: a bigger sample is safe
173.18 rounds up to 174.
n = 174
Careful!
Rounding down to 173 gives an error bound a bit too big.
Always round up.
Careful: not n = 173 ✗
5 Check
Let's check: with 174 people the error bound is about 0.9977.
That is not more than 1.
Good.
n = 174: EBM ≈ 0.9977 ≤ 1
With one person less, 173, it is about 1.0005: too big.
So 174 is the smallest that works.
So it is right!
n = 173: EBM ≈ 1.0005 > 1
Let's look back.
First: find z*.
Then: the formula.
Then: put the numbers in.
Then: round up.
Last: check.
Answer Careful: not n = 173 ✗
▶ Watch it solved
The question
Question: the SD of all the numbers is 2.5: how big a sample do we need to be 95% confident with an error bound of 0.5?
n? 95% confidence, σ = 2.5, EBM = 0.5
The error bound is how far the sample mean may be from the true mean.
EBM: how far the sample mean may be from the true mean
1 Find z*
95 percent of the bell is in the middle.
That leaves 5 percent: 2.5 percent in each end.
95% in the middle: 5% left over, 2.5% in each end
z* marks the cut between the middle and the ends.
Left of z* are the middle and one end: 0.95 plus 0.025 is 0.9750.
left of z* = 0.95 + 0.025 = 0.9750
Now we look for 0.9750 inside the table.
0.9750 is in the table: row 1.9, column .06.
So z = 1.96.
z* = 1.96
2 The formula
The error bound is z* times σ over the square root of n.
Solved for n, it gives this formula.
from EBM = z* × σ ÷ √n, solved for n
n is z* squared times σ squared, over the error bound squared.
n = z*2 × σ2EBM2
3 Put the numbers in
1.96 squared is 3.8416.
2.5 squared is 6.25.
0.5 squared is 0.25.
3.8416 times 6.25 is 24.01.
Divided by 0.25: 96.04.
n = 1.962 × 2.520.52 = 3.8416 × 6.250.25 = 24.010.25 = 96.04
4 Round up
A sample size must be a whole number.
We always round up: a smaller sample would miss the goal.
always round UP: a bigger sample is safe
96.04 rounds up to 97.
n = 97
Careful!
Rounding down to 96 gives an error bound a bit too big.
Always round up.
Careful: not n = 96 ✗
5 Check
Let's check: with 97 people the error bound is about 0.4975.
That is not more than 0.5.
Good.
n = 97: EBM ≈ 0.4975 ≤ 0.5
With one person less, 96, it is about 0.5001: too big.
So 97 is the smallest that works.
So it is right!
n = 96: EBM ≈ 0.5001 > 0.5
Let's look back.
First: find z*.
Then: the formula.
Then: put the numbers in.
Then: round up.
Last: check.
Answer Careful: not n = 96 ✗
▶ Watch it solved
