Test the claim mean > 300: sample mean 318, sample sd 30, n 10… Step by Step

Step by step

The question

Question: a sample of 10 has mean 318 and SD 30: at the 5% level, is the true mean more than 300?

μ > 300? x̄ = 318, s = 30, n = 10, 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 300.

H₀ : μ = 300

Hₐ is what we want to show.

Here: the mean is more than 300.

H_a : μ > 300

More than means only the right end counts.

One tail.

one tail: the right end

2 The test number

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

30 over the square root of 10 is about 9.4868.

s√n = 30√10 = ≈ 9.4868

t is the sample mean minus 300, over 9.4868.

318 minus 300 is 18.

Divided by 9.4868: about 1.897.

t = x̄ − μ₀SD = 318 − 3009.4868 = 189.4868 ≈ 1.897

3 The cut-off

The t-table needs df: n minus 1, 9.

df = n − 1 = 10 − 1 = 9

Careful!

df is 10 minus 1: 9.

Careful: not df = 10 ✗

One tail: 0.05 in one end.

Row 9, column 0.05.

Row 9, column .05: t* = 1.833.

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, 1.897, is past the cut-off: in the reject part.

So we reject H zero.

= t = 1.897 > 1.833 → 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 more than 300.

the mean is more than 300.

6 Check

Let's check with the sample mean itself.

The cut-off for it is 300 plus 1.833 times 9.4868: 317.3893.

cut-off for the sample mean: 300 + 1.833 × 9.4868 = 317.3893

The sample mean, 318, is past it.

The same decision.

So it is right!

318: 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