Two-sample tests for means Problem set A Worked solutions

State the null and alternative hypotheses for each test. Define μ1 and μ2.

1. (a) Does the mean fuel use differ between petrol cars (group 1) and hybrid cars (group 2)?

   • Let μ1 and μ2 be the population mean fuel use of petrol cars (group 1) and fuel use of hybrid cars (group 2).
   • H0: μ1 = μ2
   • H1: μ1 ≠ μ2 (different, so two-tailed)

2. (b) A gardener thinks plants given fertiliser X (group 1) grow taller on average than plants given fertiliser Y (group 2).

   • Let μ1 and μ2 be the population mean height of plants given fertiliser X (group 1) and height of plants given fertiliser Y (group 2).
   • H0: μ1 = μ2
   • H1: μ1 > μ2 (group 1 is greater)

Use a two-sample z-test with H0: μ1 = μ2 and H1: μ1 < μ2 to find the p-value, from these summary statistics for two normal populations with known standard deviations.

1. (a) σ1 = 2.4, n1 = 25, x̄1 = 14.8; σ2 = 3.1, n2 = 30, x̄2 = 16.1.

   • H0: μ1 = μ2, H1: μ1 < μ2
   • Two-sample z-test, because both σ are known.
   • From the calculator: z = −1.75, p = 0.0399

2. (b) σ1 = 6.5, n1 = 20, x̄1 = 84.2; σ2 = 8.0, n2 = 24, x̄2 = 87.9.

   • H0: μ1 = μ2, H1: μ1 < μ2
   • Two-sample z-test, because both σ are known.
   • From the calculator: z = −1.69, p = 0.0453

Two normal populations have known standard deviations σ1 = 0.7 and σ2 = 0.7. Sample 1: 8.4, 9.1, 7.9, 9.6, 8.8, 9.3, 8.5. Sample 2: 7.8, 8.2, 8.9, 7.5, 8.1, 8.6. Use a two-sample z-test with H0: μ1 = μ2 and H1: μ1 > μ2 to find the p-value.

• H0: μ1 = μ2, H1: μ1 > μ2
• From the calculator, two-sample z-test with lists, σ1 = 0.7, σ2 = 0.7: z = 1.58
• p = 0.0567

Use a two-sample pooled t-test with H0: μ1 = μ2 and H1: μ1 ≠ μ2 to find the p-value. Assume that the populations are normal with equal variances. Each s is a sample standard deviation.

1. (a) x̄1 = 72.5, s1 = 8.1, n1 = 12; x̄2 = 66.8, s2 = 9.4, n2 = 15.

   • H0: μ1 = μ2, H1: μ1 ≠ μ2
   • Two-sample t-test, pooled, two-tailed.
   • From the calculator: t = 1.66, p = 0.109

2. (b) x̄1 = 5.6, s1 = 1.2, n1 = 9; x̄2 = 6.9, s2 = 1.5, n2 = 11.

   • H0: μ1 = μ2, H1: μ1 ≠ μ2
   • Two-sample t-test, pooled, two-tailed.
   • From the calculator: t = −2.10, p = 0.0497

Sample 1: 23.1, 25.4, 22.8, 24.0, 21.9, 23.5. Sample 2: 25.9, 27.2, 24.8, 26.5, 25.1, 27.8, 26.0. Use a two-sample pooled t-test with H0: μ1 = μ2 and H1: μ1 < μ2 to find the p-value. Assume that the populations are normal with equal variances.

• H0: μ1 = μ2, H1: μ1 < μ2
• Two-sample t-test from lists, pooled, lower tail.
• From the calculator: t = −4.36, p = 0.000570

Two machines fill bags of rice. The masses, in grams, are normally distributed with known standard deviations of 4.2 g for machine 1 and 5.1 g for machine 2. A random sample of 30 bags from machine 1 has mean mass 502.8 g. A random sample of 25 bags from machine 2 has mean mass 499.6 g. Test, at the 5% significance level, whether the machines fill bags to different mean masses.

1. (a) State the hypotheses, defining μ1 and μ2.

   • μ1 and μ2 are the population mean mass of a bag from machine 1 and mass of a bag from machine 2.
   • H0: μ1 = μ2, H1: μ1 ≠ μ2

2. (b) Find the p-value.

   • Both σ are known: two-sample z-test, two-tailed.
   • From the calculator: z = 2.51, p = 0.0122

3. (c) State the conclusion in context.

   • 0.0122 < 0.05, so reject H0. There is sufficient evidence, at the 5% significance level, that the machines fill bags to different mean masses.

A teacher tests whether a memory technique helps. Eight students used the technique (group 1) and seven different students did not (group 2). The numbers of words recalled were: technique: 14, 17, 15, 18, 16, 13, 17, 19; no technique: 12, 15, 13, 16, 14, 12, 15. Assume that the numbers are normally distributed with equal variances. Test, at the 5% significance level, whether the technique increases the mean number of words recalled.

1. (a) State the hypotheses, defining μ1 and μ2.

   • μ1 and μ2 are the population mean number of words recalled with the technique and number recalled without it.
   • H0: μ1 = μ2, H1: μ1 > μ2

2. (b) Find the p-value.

   • Unknown σ, assumed equal: two-sample pooled t-test, upper tail.
   • From the calculator, with lists: t = 2.39, p = 0.0164

3. (c) State the conclusion in context.

   • 0.0164 < 0.05, so reject H0. There is sufficient evidence, at the 5% significance level, that the technique increases the mean number of words recalled.

Two classes sit the same test. The marks in each class are normally distributed with standard deviation 6.5. Class 1 marks: 68, 74, 71, 66, 79, 72, 70, 77. Class 2 marks: 64, 70, 66, 69, 62, 71, 67. Test, at the 10% significance level, whether the mean marks of the two classes differ.

1. (a) Choose the test and state the hypotheses, defining μ1 and μ2.

   • The standard deviations are known, so use a two-sample z-test. μ1 and μ2 are the population mean mark of class 1 and mark of class 2.
   • H0: μ1 = μ2, H1: μ1 ≠ μ2

2. (b) Find the p-value.

   • From the calculator, with lists: z = 1.52, p = 0.128

3. (c) State the conclusion in context.

   • 0.128 > 0.1, so do not reject H0. There is insufficient evidence, at the 10% significance level, that the mean marks of the two classes differ.

Two suppliers sell steel cables. The breaking loads, in kN, are normally distributed with known standard deviations of 12 kN (supplier 1) and 15 kN (supplier 2). A sample of 40 cables from supplier 1 has mean breaking load 95.5 kN. A sample of 35 cables from supplier 2 has mean breaking load 88.0 kN. A test is carried out of whether the mean breaking loads differ.

1. (a) Find the p-value.

   • H0: μ1 = μ2, H1: μ1 ≠ μ2. Both σ are known, and they are different, which does not matter for the z-test.
   • From the calculator: z = 2.37, p = 0.0179

2. (b) State the conclusion at the 5% significance level.

   • 0.0179 < 0.05, so reject H0. There is sufficient evidence, at the 5% significance level, that the mean breaking loads of the two suppliers differ.

3. (c) State the conclusion at the 1% significance level.

   • 0.0179 > 0.01, so do not reject H0. There is insufficient evidence, at the 1% significance level, that the mean breaking loads of the two suppliers differ.