Test the claim mean < 100: sample mean 96, sample sd 9, n 18, alpha… Step by Step

Step by step

The question

Question: a sample of 18 has mean 96 and SD 9: at the 5% level, is the true mean less than 100?

μ < 100? x̄ = 96, s = 9, n = 18, α = 0.05

We only know s, the SD of the sample.

So we use t, not z.

σ not known: we use s and t

1 Write the hypotheses

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

Here: the mean is 100.

H₀ : μ = 100

Hₐ is what we want to show.

Here: the mean is less than 100.

H_a : μ < 100

Less than means only the left end counts.

One tail.

one tail: the left end

2 The test number

First the SD of the sample means: s over the square root of n.

9 over the square root of 18 is about 2.1213.

s√n = 9√18 = ≈ 2.1213

t is the sample mean minus 100, over 2.1213.

96 minus 100 is −4.

Divided by 2.1213: about −1.886.

t = x̄ − μ₀SD = 96 − 1002.1213 = −42.1213 ≈ −1.886

3 The cut-off

The t-table needs df: n minus 1, 17.

df = n − 1 = 18 − 1 = 17

Careful!

df is 18 minus 1: 17.

Careful: not df = 18 ✗

One tail: 0.05 in one end.

Row 17, column 0.05.

Row 17, column .05: t* = 1.740.

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

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

4 Decide

Our t, −1.886, is past the cut-off: in the reject part.

So we reject H zero.

= t = −1.886 < −1.740 → reject H₀

5 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 mean is less than 100.

the mean is less than 100.

6 Check

Let's check with the sample mean itself.

The cut-off for it is 100 minus 1.740 times 2.1213: 96.3089.

cut-off for the sample mean: 100 − 1.740 × 2.1213 = 96.3089

The sample mean, 96, is past it.

The same decision.

So it is right!

96: past it, reject

Let's look back.

First: write the hypotheses.

Then: the test number.

Then: the cut-off.

Then: decide.

Then: say it in words.

Last: check.

Answer