Step by step
The question
Question: a sample of 25 has mean 97 and SD 9: at the 5% level, is the true mean different from 100?
We only know s, the SD of the sample.
So we use t, not z.
There are four choices.
First we solve it, the normal way.
Then we pick.
1 Write the hypotheses
H₀ is the "nothing new" idea: it has the equal sign.
Here: the mean is 100.
Hₐ is what we want to show.
Here: the mean is different from 100.
Different from means too big or too small: both ends of the bell count.
Two tails.
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.
9/5 is 1.8.
t is the sample mean minus 100, over 1.8.
97 minus 100 is −3.
Divided by 1.8: about −1.667.
3 The cut-off
The t-table needs df: n minus 1, 24.
Careful!
df is 25 minus 1: 24.
The right way: df = n − 1 = 25 − 1 = 24.
Two tails: 0.05 is split, 0.025 in each end.
Row 24, column 0.025.
Row 24, column .025: t* = 2.064.
The curve is the same on both sides, so the cut-offs are plus and minus 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, −1.667, is not past the cut-off.
So we do not reject H zero.
5 Say it in words
In words: at the 5% level, there is not enough evidence that the mean is different from 100.
This does not prove H zero is true.
We just cannot rule it out.
6 Check
Let's check with the sample mean itself.
The cut-off for it is 100 plus or minus 2.064 times 1.8: 96.2848 and 103.7152.
The sample mean, 97, is not past it.
The same decision.
So it is right!
7 Pick the choice
Our answer is do not reject H₀.
That is choice A.
8 Why not the others
B is Reject H0.
Wrong: |t| ≈ 1.67 is less than 2.064.
C is Accept H0: the mean is 100.
Wrong: a test never shows H0 is true.
D is Accept Ha because 97 is not 100.
Wrong: the test decides.
Let's look back.
First: write the hypotheses.
Then: the test number.
Then: the cut-off.
Then: decide.
Then: say it in words.
Then: check.
Then: pick the choice.
Last: why not the others.