Paired test: before 64, 70, 58, 66, 72; after 66, 69, 61, 70, 71… Step by Step

Step by step

The question

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

before: 64, 70, 58, 66, 72; after: 66, 69, 61, 70, 71

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.

66 minus 64 is 2.

69 minus 70 is −1.

61 minus 58 is 3.

70 minus 66 is 4.

71 minus 72 is −1.

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 7.

Their mean: 7/5 is 1.4.

d̄ = 75 = 1.4

For the SD: each difference minus 1.4, squared.

2 gives 0.36.

−1 gives 5.76.

3 gives 2.56.

4 gives 6.76.

−1 gives 5.76.

They add up to 21.2.

A sample SD divides by n minus 1: 21.2/4 is 5.3.

The square root: about 2.3022.

s_d = √21.25 − 1 = √5.3 ≈ 2.3022

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: 2.3022 over the square root of 5, about 1.0296.

SD of the mean = s_d√n = 2.3022√5 ≈ 1.0296

t is the mean difference minus 0, over 1.0296: about 1.36.

t = 1.4 − 01.0296 ≈ 1.36

5 The cut-off

The t-table needs df: n minus 1, 4.

df = n − 1 = 5 − 1 = 4

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

Row 4, column 0.025.

Row 4, column .025: t* = 2.776.

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, 1.36, is not past the cut-off.

So we do not reject H zero.

= |t| = 1.36 < 2.776 → 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 weights changed.

the weights 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 2.776 times 1.0296: −2.8582 and 2.8582.

cut-off for the mean difference: 0 ± 2.776 × 1.0296: −2.8582 and 2.8582

The mean difference, 1.4, is not past it.

The same decision.

So it is right!

1.4: 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