14.3 Normal tests for a mean — Practise with help

Attempt each question closed book first. Check that the standard deviation is known before you standardise.

Warm-up: remember

A standard Normal variable has $P(Z \le 1.96) = 0.975$ . Write down $P(Z \ge 1.96)$ .
A Normal distribution has mean 50 and standard deviation 10. Standardise the value $x = 45$ .
A population has standard deviation 3. A sample of size 36 is taken. Find the standard error of the sample mean.

Answers

$P(Z \ge 1.96) = 1 - 0.975 = 0.025$ .
$z = \dfrac{45 - 50}{10} = -0.5$ .
The standard error is $\dfrac{3}{\sqrt{36}} = \dfrac{3}{6} = 0.5$ .

If question 3 was hard, revise the standard error before you start.

Practise with help

1. A machine is set to a mean of 250 g. The standard deviation is known to be 5 g. A sample of 25 packets has a mean of 248 g. Test at the 5% level whether the machine is underfilling the packets.

Hint

The word "underfilling" fixes the direction of the alternative.

Next step

Find the standard error $\dfrac{5}{\sqrt{25}}$ , then standardise the sample mean to get $z$ .

Full working

H_0: \mu = 250, \qquad H_1: \mu < 250, \qquad \alpha = 0.05.

The standard error is

\frac{5}{\sqrt{25}} = 1.

The test statistic is

z = \frac{248 - 250}{1} = -2.

The lower-tail probability is

P(Z \le -2) = 0.0228 \quad (3 \text{ s.f.}).

Since $0.0228 < 0.05$ , the test rejects $H_0$ . There is sufficient evidence at the 5% level that the mean mass is below 250 g.

2. A machine is set to a mean of 60 g. The standard deviation is known to be 12 g. A sample of 36 items has a mean of 62 g. Test at the 5% level whether the mean has changed.

Hint

"Changed" allows a rise or a fall, so the test has two tails.

Next step

Find the standard error, standardise, then double the smaller tail.

Full working

H_0: \mu = 60, \qquad H_1: \mu \ne 60, \qquad \alpha = 0.05.

The standard error is

\frac{12}{\sqrt{36}} = 2.

The test statistic is

z = \frac{62 - 60}{2} = 1.

The smaller tail is

P(Z \ge 1) = 0.1587,

so the two-tailed p-value is

2 \times 0.1587 = 0.3173.

Since $0.3173 > 0.05$ , the test retains $H_0$ . There is not sufficient evidence at the 5% level that the mean has changed.

Notice that the statistic $z = 1$ is small. A sample mean only 2 g above the claim is well within ordinary variation when the standard error is 2 g.

3. A claim sets $\mu = 18$ . The standard deviation is known to be 3. A sample of 36 has mean 19.2. Test at the 5% level whether the mean is higher than the claim.

Hint

"Higher" points to the upper tail. Work out the standard error first.

Next step

The standard error is $\dfrac{3}{\sqrt{36}} = 0.5$ . Standardise 19.2, then find the upper-tail probability.

Full working

H_0: \mu = 18, \qquad H_1: \mu > 18, \qquad \alpha = 0.05.

The standard error is

\frac{3}{\sqrt{36}} = 0.5.

The test statistic is

z = \frac{19.2 - 18}{0.5} = 2.4.

The upper-tail probability is

P(Z \ge 2.4) = 0.0082 \quad (3 \text{ s.f.}).

Since $0.0082 < 0.05$ , the test rejects $H_0$ . There is sufficient evidence at the 5% level that the mean is above 18. The result also rejects at the 1% level, because $0.0082 < 0.01$ .

4. The claim in question 3 is tested with a sample of 9 instead of 36, and the mean is again 19.2. Test at the 5% level whether the mean is higher than the claim.

Hint

Only the sample size has changed. The standard error grows when the sample is smaller.

Next step

The standard error is now $\dfrac{3}{\sqrt{9}} = 1$ . Standardise 19.2 again and find the new upper-tail probability.

Full working

H_0: \mu = 18, \qquad H_1: \mu > 18, \qquad \alpha = 0.05.

The standard error is

\frac{3}{\sqrt{9}} = 1.

The test statistic is

z = \frac{19.2 - 18}{1} = 1.2.

The upper-tail probability is

P(Z \ge 1.2) = 0.1151 \quad (4 \text{ d.p.}).

Since $0.1151 > 0.05$ , the test retains $H_0$ . The same gap of $1.2$ above the claim is not convincing from a sample of 9, because the sample mean varies more when the sample is small.

5. For question 1, the test statistic is $z = -2$ , and the 5% one-tailed critical value for a lower-tail test is $-1.645$ . Decide the test by comparing the two values, and check the decision against the p-value route.

Hint

A lower-tail test rejects when the statistic lies below the critical value.

Next step

Compare $-2$ with $-1.645$ . Decide whether $-2$ lies to the left of $-1.645$ on the number line.

Full working

The critical value is $-1.645$ . Since

-2 < -1.645,

the test statistic lies in the critical region, so the test rejects $H_0$ . The p-value route gave $0.0228 < 0.05$ , which also rejects. The two routes agree, because $z = -1.645$ is exactly the boundary where the lower tail equals $0.05$ .

6. A two-tailed Normal test gives a p-value of $0.0455$ . Decide the test at the 5% level. Then explain what would happen to the decision if the same statistic came from a one-tailed test.

Hint

A two-tailed p-value doubles the smaller tail. A one-tailed p-value uses only one tail.

Next step

Compare $0.0455$ with $0.05$ for the two-tailed decision. Then halve the p-value for the one-tailed case and compare again.

Full working

At the 5% level, $0.0455 < 0.05$ , so the two-tailed test rejects $H_0$ .

If the same statistic came from a one-tailed test, the p-value would be $\dfrac{0.0455}{2} = 0.0228$ . This is also below $0.05$ , so the one-tailed test would also reject. Halving the p-value makes a rejection more likely, so a result close to the level can move from retain to reject when the test becomes one-tailed.