Testing a mean: z-test or t-test Problem set B One question at a time

For each situation, write down the null and alternative hypotheses, defining \(\mu\), and say whether a z-test or a t-test is needed. The population is normally distributed in each case.

1. (a) A tile factory says its tiles have a mean thickness of 8.0 mm. Calibration of the machine gives a population standard deviation of 0.3 mm. A sample of 25 tiles has a mean thickness of 7.9 mm. Has the mean thickness decreased?

   • Let \(\mu\) be the mean thickness of a tile, in millimetres. \(H_0: \mu = 8.0\), \(H_1: \mu < 8.0\)
   • The population standard deviation is known (0.3 mm), so use z. 'Decreased' means one-tailed.
   • Use a z-test

2. (b) A school canteen says pupils wait a mean of 6 minutes to be served. A sample of 18 pupils has a mean wait of 6.9 minutes and a sample standard deviation of 1.8 minutes. Is the mean waiting time different from 6 minutes?

   • Let \(\mu\) be the mean waiting time, in minutes. \(H_0: \mu = 6\), \(H_1: \mu \neq 6\)
   • 1.8 minutes is a sample standard deviation, so use t. 'Different from' means two-tailed.
   • Use a t-test

3. (c) A bus company says its journeys take a mean of 35 minutes. The population variance is known to be 16 min². A sample of 24 journeys has a mean of 36.4 minutes. Is the mean journey time more than 35 minutes?

   • Let \(\mu\) be the mean journey time, in minutes. \(H_0: \mu = 35\), \(H_1: \mu > 35\)
   • The population variance is known, so \(\sigma = 4\) and use z. 'More than' means one-tailed.
   • Use a z-test

A rope maker cuts lengths with a mean of 2.50 m. A supervisor suspects the cutting machine has moved and checks a sample. The population standard deviation is 0.06 m. A random sample of 25 ropes has a mean of 2.53 m. Assume the population is normally distributed. Test, at the 5% significance level, whether the mean length has changed. Give the p-value and your conclusion.

• Let \(\mu\) be the mean length of a rope, in metres. \(H_0: \mu = 2.50\), \(H_1: \mu \neq 2.50\). Two-tailed test, because H1 uses ≠.
• The population standard deviation is known and the population is normal, so use a z-test.
• From the calculator: \(\sigma = 0.06\), \(\bar{x} = 2.53\), \(n = 25\): z = 2.50, p = 0.0124
• 0.0124 < 0.05, so reject \(H_0\). There is sufficient evidence, at the 5% significance level, that the mean length has changed.

A railway says the mean journey between two cities is 60 minutes. Passengers think it is quicker. The sample standard deviation is 7.4 minutes. A random sample of 20 journeys has a mean of 56.9 minutes. Assume the population is normally distributed. Test, at the 5% significance level, whether the mean journey time is less than 60 minutes. Give the p-value and your conclusion.

• Let \(\mu\) be the mean journey time, in minutes. \(H_0: \mu = 60\), \(H_1: \mu < 60\). One-tailed test, because H1 uses <.
• The standard deviation comes from the sample, so it is an estimate. The population is normal, so use a t-test.
• From the calculator: \(s = 7.4\), \(\bar{x} = 56.9\), \(n = 20\): t = −1.87, df = 19, p = 0.0382
• 0.0382 < 0.05, so reject \(H_0\). There is sufficient evidence, at the 5% significance level, that the mean journey time is less than 60 minutes.

A farmer says her tomatoes have a mean mass of 320 g. A customer weighs 7 tomatoes, in grams. The data are 310, 324, 305, 318, 330, 312, 321 Assume the population is normally distributed. Test, at the 5% significance level, whether the mean mass of a tomato is less than 320 g. Give the p-value and your conclusion.

• Let \(\mu\) be the mean mass of a tomato, in grams. \(H_0: \mu = 320\), \(H_1: \mu < 320\). One-tailed test, because H1 uses <.
• The standard deviation comes from the sample, so it is an estimate. The population is normal, so use a t-test.
• Enter the data in a list. From the calculator: \(\bar{x} = 317\), \(s = 8.69\), t = −0.870, df = 6, p = 0.209
• 0.209 > 0.05, so do not reject \(H_0\). There is insufficient evidence, at the 5% significance level, that the mean mass of a tomato is less than 320 g.

A candle maker's new wax is said to burn for a mean of 11.9 hours. A sample of 8 candles is tested, with burn times in hours. The population standard deviation is 0.5 hours. The data are 12.3, 11.8, 12.9, 12.4, 11.6, 12.7, 12.1, 12.5 Assume the population is normally distributed. Test, at the 5% significance level, whether the mean burn time is more than 11.9 hours. Give the p-value and your conclusion.

• Let \(\mu\) be the mean burn time of a candle, in hours. \(H_0: \mu = 11.9\), \(H_1: \mu > 11.9\). One-tailed test, because H1 uses >.
• The population standard deviation is known and the population is normal, so use a z-test.
• Enter the data in a list. From the calculator, with \(\sigma = 0.5\): \(\bar{x} = 12.3\), z = 2.19, p = 0.0142
• 0.0142 < 0.05, so reject \(H_0\). There is sufficient evidence, at the 5% significance level, that the mean burn time is more than 11.9 hours.

A gym says its members attend a mean of 120 minutes of exercise a week. The sample variance is 7.84 minutes². A random sample of 15 members has a mean of 122.1 minutes. Assume the population is normally distributed. Test, at the 5% significance level, whether the mean weekly exercise time is different from 120 minutes. Give the p-value and your conclusion.

• Let \(\mu\) be the mean weekly exercise time, in minutes. \(H_0: \mu = 120\), \(H_1: \mu \neq 120\). Two-tailed test, because H1 uses ≠.
• The standard deviation comes from the sample, so it is an estimate. The population is normal, so use a t-test.
• The sample variance is given, so \(s = \sqrt{7.84} = 2.8\)
• From the calculator: \(s = 2.8\), \(\bar{x} = 122.1\), \(n = 15\): t = 2.90, df = 14, p = 0.0115
• 0.0115 < 0.05, so reject \(H_0\). There is sufficient evidence, at the 5% significance level, that the mean weekly exercise time is different from 120 minutes.

A mill makes thread with a mean thickness of 1.20 mm. The population variance is 0.16 mm². A random sample of 30 lengths of thread has a mean of 1.35 mm. Assume the population is normally distributed. Test, at the 5% significance level, whether the mean thickness is more than 1.20 mm. Give the p-value and your conclusion.

• Let \(\mu\) be the mean thickness of the thread, in millimetres. \(H_0: \mu = 1.20\), \(H_1: \mu > 1.20\). One-tailed test, because H1 uses >.
• The population standard deviation is known and the population is normal, so use a z-test.
• The variance is given, so \(\sigma = \sqrt{0.16} = 0.4\)
• From the calculator: \(\sigma = 0.4\), \(\bar{x} = 1.35\), \(n = 30\): z = 2.05, p = 0.0200
• 0.0200 < 0.05, so reject \(H_0\). There is sufficient evidence, at the 5% significance level, that the mean thickness is more than 1.20 mm.

A water tank fills with a mean volume of 1500 litres. From years of data the population standard deviation is 30 litres. A sample of 16 fills has a mean of 1485 litres and a sample standard deviation of 34 litres. The population is normally distributed. Test, at the 5% significance level, whether the mean volume of a fill has changed.

1. (a) State the hypotheses and say which test you will use.

   • Let \(\mu\) be the mean volume of a fill, in litres. \(H_0: \mu = 1500\), \(H_1: \mu \neq 1500\). Two-tailed test, because H1 uses ≠.
   • The population standard deviation is given, so use a z-test with \(\sigma = 30\). Ignore the sample standard deviation.

2. (b) Find the p-value.

   • From the calculator: \(\bar{x} = 1485\), \(n = 16\): z = −2.00, p = 0.0455

3. (c) Write your conclusion at the 5% significance level, and at the 1% significance level.

   • 0.0455 < 0.05, so reject \(H_0\). There is sufficient evidence, at the 5% significance level, that the mean volume of a fill has changed.
   • 0.0455 > 0.01, so do not reject \(H_0\) at the 1% level: there is insufficient evidence, at the 1% significance level, that the mean volume of a fill has changed.

A plum grower says the mean mass of a plum is 6.9 g. After a new treatment she weighs 9 plums, in grams: 7.2, 6.6, 7.9, 7.4, 6.9, 7.7, 7.1, 6.5, 7.5 Assume the population is normally distributed.

1. (a) Find the sample mean and the sample standard deviation.

   • Enter the data in a list. From the calculator: \(\bar{x} = 7.20\), \(s = 0.477\)

2. (b) Test, at the 5% significance level, whether the treatment has increased the mean mass.

   • Let \(\mu\) be the mean mass of a plum after the treatment, in grams. \(H_0: \mu = 6.9\), \(H_1: \mu > 6.9\). One-tailed test, because H1 uses >.
   • From the calculator: t = 1.89, df = 8, p = 0.0479
   • 0.0479 < 0.05, so reject \(H_0\). There is sufficient evidence, at the 5% significance level, that the treatment has increased the mean mass.

3. (c) Say why a t-test and not a z-test is used here.

   • No population standard deviation is given.
   • The standard deviation comes from a small sample, so it is an estimate, and the t-distribution allows for that.