Paired test: before 40, 44, 38, 42, 45, 41; after 42, 43, 41, 46… Step by Step

Step by step

The question

Question: the same 6 people are measured before and after: at the 5% level, did the scores change?

before: 40, 44, 38, 42, 45, 41; after: 42, 43, 41, 46, 44, 46

The same people are measured twice.

So we look at each person's change: one list of differences.

the same people twice: work with each one's difference

1 The differences

Each row is one person: before and after.

42 minus 40 is 2.

43 minus 44 is −1.

41 minus 38 is 3.

46 minus 42 is 4.

44 minus 45 is −1.

46 minus 41 is 5.

Careful!

The two lists are the same people.

So we test the differences, not two separate groups.

Careful: one list of differences, not two groups

2 Mean and SD of d

The differences add up to 12.

Their mean: 12/6 is 2.

d̄ = 126 = 2

For the SD: each difference minus 2, squared.

2 gives 0.

−1 gives 9.

3 gives 1.

4 gives 4.

−1 gives 9.

5 gives 9.

They add up to 32.

A sample SD divides by n minus 1: 32/5 is 6.4.

The square root: about 2.5298.

s_d = √326 − 1 = √6.4 ≈ 2.5298

3 Write the hypotheses

H₀ is the "nothing new" idea: it has the equal sign.

Here: the mean change is 0.

H₀ : μ_d = 0

Hₐ is what we want to show.

Here: the mean change is different from 0.

H_a : μ_d ≠ 0

Different from means too big or too small: both ends of the bell count.

Two tails.

two tails: far on either side counts

4 The test number

The SD of the mean difference: 2.5298 over the square root of 6, about 1.0328.

SD of the mean = s_d√n = 2.5298√6 ≈ 1.0328

t is the mean difference minus 0, over 1.0328: about 1.936.

t = 2 − 01.0328 ≈ 1.936

5 The cut-off

The t-table needs df: n minus 1, 5.

df = n − 1 = 6 − 1 = 5

Two tails: 0.05 is split, 0.025 in each end.

Row 5, column 0.025.

Row 5, column .025: t* = 2.571.

The t curve, with the cut-offs and our t.

Past the cut-offs is the reject part: the rare samples.

6 Decide

Our t, 1.936, is not past the cut-off.

So we do not reject H zero.

= |t| = 1.936 < 2.571 → do not reject H₀

7 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 scores changed.

the scores changed.

This does not prove H zero is true.

We just cannot rule it out.

not rejecting H₀ does not prove H₀

8 Check

Let's check with the mean difference itself.

The cut-off for it is 0 plus or minus 2.571 times 1.0328: −2.6553 and 2.6553.

cut-off for the mean difference: 0 ± 2.571 × 1.0328: −2.6553 and 2.6553

The mean difference, 2, is not past it.

The same decision.

So it is right!

2: not past it, do not reject

Let's look back.

First: the differences.

Then: mean and SD of d.

Then: write the hypotheses.

Then: the test number.

Then: the cut-off.

Then: decide.

Then: say it in words.

Last: check.

Answer