Worked examples
The question
Question: a sample of 16 has mean 28 and SD 4: at the 5% level, is the true mean less than 30?
μ < 30? x̄ = 28, s = 4, n = 16, α = 0.05
We only know s, the SD of the sample.
So we use t, not z.
σ not known: we use s and t
1 Write the hypotheses
H₀ is the "nothing new" idea: it has the equal sign.
Here: the mean is 30.
H0 : μ = 30
Hₐ is what we want to show.
Here: the mean is less than 30.
Ha : μ < 30
Less than means only the left end counts.
One tail.
one tail: the left end
2 The test number
First the SD of the sample means: s over the square root of n.
The square root of 16 is 4.
4/4 is 1.
s√n = 4√16 = 44 = 1
t is the sample mean minus 30, over 1.
28 minus 30 is −2.
Divided by 1: −2.
t = x̄ − μ0SD = 28 − 301 = −21 = −2
3 The cut-off
The t-table needs df: n minus 1, 15.
df = n − 1 = 16 − 1 = 15
Careful!
df is 16 minus 1: 15.
Careful: not df = 16 ✗
One tail: 0.05 in one end.
Row 15, column 0.05.
Row 15, column .05: t* = 1.753.
The t curve, with the cut-off and our t.
Past the cut-off is the reject part: the rare samples.
4 Decide
Our t, −2, is past the cut-off: in the reject part.
So we reject H zero.
= t = −2 < −1.753 → reject H0
5 Say it in words
At the 5% level, there is enough evidence that
In words: at the 5% level, there is enough evidence that the mean is less than 30.
the mean is less than 30.
6 Check
Let's check with the sample mean itself.
The cut-off for it is 30 minus 1.753 times 1: 28.247.
cut-off for the sample mean: 30 − 1.753 × 1 = 28.247
The sample mean, 28, is past it.
The same decision.
So it is right!
28: past it, reject
Let's look back.
First: write the hypotheses.
Then: the test number.
Then: the cut-off.
Then: decide.
Then: say it in words.
Last: check.
Answer = t = −2 < −1.753 → reject H0
▶ Watch it solved
The question
Question: a sample of 25 has mean 43 and SD 5: at the 5% level, is the true mean different from 40?
μ ≠ 40? x̄ = 43, s = 5, n = 25, α = 0.05
We only know s, the SD of the sample.
So we use t, not z.
σ not known: we use s and t
1 Write the hypotheses
H₀ is the "nothing new" idea: it has the equal sign.
Here: the mean is 40.
H0 : μ = 40
Hₐ is what we want to show.
Here: the mean is different from 40.
Ha : μ ≠ 40
Different from means too big or too small: both ends of the bell count.
Two tails.
two tails: far on either side counts
2 The test number
First the SD of the sample means: s over the square root of n.
The square root of 25 is 5.
5/5 is 1.
s√n = 5√25 = 55 = 1
t is the sample mean minus 40, over 1.
43 minus 40 is 3.
Divided by 1: 3.
t = x̄ − μ0SD = 43 − 401 = 31 = 3
3 The cut-off
The t-table needs df: n minus 1, 24.
df = n − 1 = 25 − 1 = 24
Careful!
df is 25 minus 1: 24.
Careful: not df = 25 ✗
Two tails: 0.05 is split, 0.025 in each end.
Row 24, column 0.025.
Row 24, column .025: t* = 2.064.
The t curve, with the cut-offs and our t.
Past the cut-offs is the reject part: the rare samples.
4 Decide
Our t, 3, is past the cut-off: in the reject part.
So we reject H zero.
= |t| = 3 > 2.064 → reject H0
5 Say it in words
At the 5% level, there is enough evidence that
In words: at the 5% level, there is enough evidence that the mean is different from 40.
the mean is different from 40.
6 Check
Let's check with the sample mean itself.
The cut-off for it is 40 plus or minus 2.064 times 1: 37.936 and 42.064.
cut-off for the sample mean: 40 ± 2.064 × 1: 37.936 and 42.064
The sample mean, 43, is past it.
The same decision.
So it is right!
43: past it, reject
Let's look back.
First: write the hypotheses.
Then: the test number.
Then: the cut-off.
Then: decide.
Then: say it in words.
Last: check.
Answer = |t| = 3 > 2.064 → reject H0
▶ Watch it solved
The question
Question: a sample of 12 has mean 67 and SD 4: at the 5% level, is the true mean more than 65?
μ > 65? x̄ = 67, s = 4, n = 12, α = 0.05
We only know s, the SD of the sample.
So we use t, not z.
σ not known: we use s and t
1 Write the hypotheses
H₀ is the "nothing new" idea: it has the equal sign.
Here: the mean is 65.
H0 : μ = 65
Hₐ is what we want to show.
Here: the mean is more than 65.
Ha : μ > 65
More than means only the right end counts.
One tail.
one tail: the right end
2 The test number
First the SD of the sample means: s over the square root of n.
4 over the square root of 12 is about 1.1547.
s√n = 4√12 = ≈ 1.1547
t is the sample mean minus 65, over 1.1547.
67 minus 65 is 2.
Divided by 1.1547: about 1.732.
t = x̄ − μ0SD = 67 − 651.1547 = 21.1547 ≈ 1.732
3 The cut-off
The t-table needs df: n minus 1, 11.
df = n − 1 = 12 − 1 = 11
Careful!
df is 12 minus 1: 11.
Careful: not df = 12 ✗
One tail: 0.05 in one end.
Row 11, column 0.05.
Row 11, column .05: t* = 1.796.
The t curve, with the cut-off and our t.
Past the cut-off is the reject part: the rare samples.
4 Decide
Our t, 1.732, is not past the cut-off.
So we do not reject H zero.
= t = 1.732 < 1.796 → do not reject H0
5 Say it in words
At the 5% level, there is not enough evidence that
In words: at the 5% level, there is not enough evidence that the mean is more than 65.
the mean is more than 65.
This does not prove H zero is true.
We just cannot rule it out.
not rejecting H0 does not prove H0
6 Check
Let's check with the sample mean itself.
The cut-off for it is 65 plus 1.796 times 1.1547: 67.0738.
cut-off for the sample mean: 65 + 1.796 × 1.1547 = 67.0738
The sample mean, 67, is not past it.
The same decision.
So it is right!
67: not past it, do not reject
Let's look back.
First: write the hypotheses.
Then: the test number.
Then: the cut-off.
Then: decide.
Then: say it in words.
Last: check.
Answer = t = 1.732 < 1.796 → do not reject H0
▶ Watch it solved
