14.3 Normal tests for a mean — Practise

Work through these on your own. For every test, check that the standard deviation is known, find the standard error, and name the significance level in the conclusion.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A machine fills bottles with a mean of 330 ml. The standard deviation is known to be 6 ml. A sample of 36 bottles has a mean of 328 ml. Test at the 5% level whether the machine is underfilling the bottles.
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.
A bottle is filled with a mean of 20 ml. The standard deviation is known to be 2 ml. A sample of 16 bottles has a mean of 20.8 ml. Test at the 5% level whether the mean has changed.
The same bottle in question 3 is checked with a sample of 64 bottles instead, and the mean is again 20.8 ml. Test at the 5% level whether the mean has changed.
A claim sets $\mu = 250$ and the standard deviation is known to be 5. A sample of 25 has mean 248. Find the value of the test statistic $z$ .
A population has standard deviation 10. A sample of size 25 is taken. Find the standard error of the sample mean.
A claim sets $\mu = 100$ . The standard deviation is known to be 10. A sample of 25 has mean 104. Test at the 5% level whether the mean is higher than the claim.
State, with a symbol, how the two-tailed 5% critical value $1.96$ compares with the one-tailed 5% critical value $1.645$ .
A Normal test gives a p-value of $0.0228$ . State, with a symbol, how it compares with the 5% significance level.
The same test gives a p-value of $0.0228$ . State, with a symbol, how it compares with the 1% significance level.
A claim sets $\mu = 100$ and the standard deviation is known to be 15. A sample of 25 has mean 104. Find the test statistic $z$ , as an exact fraction.
A Normal test has an upper-tail probability of $0.0038$ . Find the two-tailed p-value.

Answers and concise solutions

1. The p-value and the decision

Answer: $H_0: \mu = 330$ and $H_1: \mu < 330$ . The standard error is $\dfrac{6}{\sqrt{36}} = 1$ , so $z = \dfrac{328 - 330}{1} = -2$ . The p-value is $0.0228 < 0.05$ , so reject $H_0$ . There is sufficient evidence at the 5% level that the mean is below 330 ml.

2. The p-value and the decision

Answer: $H_0: \mu = 60$ and $H_1: \mu \ne 60$ . The standard error is $\dfrac{12}{\sqrt{36}} = 2$ , so $z = \dfrac{62 - 60}{2} = 1$ . The two-tailed p-value is $0.3173 > 0.05$ , so retain $H_0$ . There is not sufficient evidence at the 5% level that the mean has changed.

3. The p-value and the decision

Answer: $H_0: \mu = 20$ and $H_1: \mu \ne 20$ . The standard error is $\dfrac{2}{\sqrt{16}} = 0.5$ , so $z = \dfrac{20.8 - 20}{0.5} = 1.6$ . The two-tailed p-value is $0.1096 > 0.05$ , so retain $H_0$ . There is not sufficient evidence at the 5% level that the mean has changed.

4. The p-value and the decision

Answer: The standard error is $\dfrac{2}{\sqrt{64}} = 0.25$ , so $z = \dfrac{20.8 - 20}{0.25} = 3.2$ . The two-tailed p-value is $0.00137 < 0.05$ , so reject $H_0$ . The same sample mean now gives a rejection, because the larger sample makes the gap more convincing.

5. The test statistic

Answer: The standard error is $\dfrac{5}{\sqrt{25}} = 1$ , so $z = \dfrac{248 - 250}{1} = -2$ .

6. The standard error

Answer: The standard error is $\dfrac{10}{\sqrt{25}} = \dfrac{10}{5} = 2$ .

7. The p-value and the decision

Answer: $H_0: \mu = 100$ and $H_1: \mu > 100$ . The standard error is $\dfrac{10}{\sqrt{25}} = 2$ , so $z = \dfrac{104 - 100}{2} = 2$ . The p-value is $0.0228 < 0.05$ , so reject $H_0$ . There is sufficient evidence at the 5% level that the mean is above 100.

8. The comparison

Answer: $1.96 > 1.645$ . A two-tailed test splits the 5% between two tails, so each boundary sits closer to the centre and more of the statistic is accepted.

9. The comparison

Answer: $0.0228 < 0.05$ , so the test rejects $H_0$ at the 5% level.

10. The comparison

Answer: $0.0228 > 0.01$ , so the test retains $H_0$ at the 1% level. The same evidence gives two decisions at the two levels.

11. The test statistic

Answer: The standard error is $\dfrac{15}{\sqrt{25}} = 3$ , so $z = \dfrac{4}{3}$ . The value $1.96$ lies above $1.33$ , so a two-tailed 5% test retains $H_0$ .

12. The two-tailed p-value

Answer: Double the smaller tail: $2 \times 0.0038 = 0.0076$ .