Step by step
The question
Question: a sample of 36 has mean 7.7, and the SD of all the numbers is 1.2: at the 5% level, is the true mean less than 8?
Alpha, 0.05, is the level: a sample rarer than 5% counts as too rare.
1 Write the hypotheses
H₀ is the "nothing new" idea: it has the equal sign.
Here: the mean is 8.
Hₐ is what we want to show.
Here: the mean is less than 8.
Less than means only the left end counts.
One tail.
2 The test number
We act as if H zero is true: the mean is 8.
How far is our sample mean from it?
First the SD of the sample means: σ over the square root of n.
The square root of 36 is 6.
1.2/6 is 0.2.
z is the sample mean minus 8, over 0.2.
7.7 minus 8 is −0.3.
Divided by 0.2: −1.5.
3 The p-value
The p-value is the chance of getting a sample this far out, if H zero is true.
On the bell of H zero: 8 in the middle, our sample at z = −1.5.
We shade the left end, from our z down.
The z-table gives the area to the left of z.
Row −1.5, column .00.
For z = −1.5: the row −1.5 and the column .00 meet at 0.0668.
Careful!
Here the end we want is on the left, so the table number is the p-value.
The left end is the table number itself: 0.0668.
4 Decide
The p-value, 0.0668, is more than alpha, 0.05.
Our sample is not rare enough.
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 less than 8.
This does not prove H zero is true.
We just cannot rule it out.
6 Check
Let's check the other way, with the cut-off.
0.05 in the left end gives the cut-off −1.645.
z is −1.5: not past the cut-off.
The same decision.
So it is right!
Let's look back.
First: write the hypotheses.
Then: the test number.
Then: the p-value.
Then: decide.
Then: say it in words.
Last: check.