Five students took a reading test before and after a summer course… Step by Step

Step by step

The question

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

before: 12, 15, 11, 14, 16; after: 14, 16, 13, 17, 19

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

There are four choices.

First we solve it, the normal way.

Then we pick.

1 The differences

Each row is one person: before and after.

14 minus 12 is 2.

16 minus 15 is 1.

13 minus 11 is 2.

17 minus 14 is 3.

19 minus 16 is 3.

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 11.

Their mean: 11/5 is 2.2.

d̄ = 115 = 2.2

For the SD: each difference minus 2.2, squared.

2 gives 0.04.

1 gives 1.44.

2 gives 0.04.

3 gives 0.64.

They add up to 2.8.

A sample SD divides by n minus 1, not n: a small sample tends to look less spread out than all the numbers, and this makes up for it.

2.8/4 is 0.7.

The square root: about 0.8367.

s_d = √2.85 − 1 = √0.7 ≈ 0.8367

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 more than 0.

H_a : μ_d > 0

More than means only the right end counts.

One tail.

one tail: the right end

4 The test number

The SD of the mean difference: 0.8367 over the square root of 5, about 0.3742.

SD of the mean = s_d√n = 0.8367√5 ≈ 0.3742

t is the mean difference minus 0, over 0.3742: about 5.879.

t = 2.2 − 00.3742 ≈ 5.879

5 The cut-off

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

df = n − 1 = 5 − 1 = 4

One tail: 0.05 in one end.

Row 4, column 0.05.

Row 4, column .05: t* = 2.132.

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

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

6 Decide

Our t, 5.879, is past the cut-off: in the reject part.

So we reject H zero.

t = 5.879 > 2.132 → reject H₀

7 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 scores went up.

the scores went up.

8 Check

Let's check with the mean difference itself.

The cut-off for it is 0 plus 2.132 times 0.3742: 0.7978.

cut-off for the mean difference: 0 + 2.132 × 0.3742 = 0.7978

The mean difference, 2.2, is past it.

The same decision.

So it is right!

2.2: past it, reject

9 Pick the choice

Our answer is reject H₀.

That is choice C.

C) Reject H0

10 Why not the others

A is Do not reject H0.

Wrong: t ≈ 5.88 is more than 2.132.

A) Do not reject H0 ✗ t ≈ 5.88 is more than 2.132

B is Accept H0.

Wrong: a test never shows H0 is true.

B) Accept H0 ✗ a test never shows H0 is true

D is Accept Ha because every score went up.

Wrong: five students alone prove nothing: the test decides.

D) Accept Ha because every score went up ✗ five students alone prove nothing: the test decides

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.

Then: check.

Then: pick the choice.

Last: why not the others.

Answer