Hypothesis tests for a mean: mixed practice Problem set A Worked solutions

A park ranger records the mass, in kg, of random samples of trout from two lakes. Assume the masses are normally distributed with equal variance. She tests, at the 5% significance level, whether the mean masses in the two lakes are different. Lake P (kg): 1.82, 2.05, 1.67, 2.21, 1.94, 1.78, 2.1 Lake Q (kg): 2.14, 2.32, 1.99, 2.45, 2.27, 2.08

1. (a) State suitable hypotheses.

   • Let μ1 and μ2 be the population mean mass of a trout, in kg, for lake P and lake Q.
   • H0: μ1 = μ2, H1: μ1 ≠ μ2 ('different' is two-tailed)

2. (b) Use your calculator to find the p-value.

   • Two independent samples, variance unknown and assumed equal: two-sample pooled t-test.
   • From the calculator: two-sample t-test, pooled, two-tailed: t = −2.65, p = 0.0224

3. (c) State the conclusion in context.

   • 0.0224 < 0.05, so reject H0.
   • There is sufficient evidence, at the 5% significance level, that the mean masses of trout in the two lakes are different.

A café says that its takeaway coffees contain a mean of 250 ml. A customer measures a random sample of 12 coffees. The mean is 244.8 ml and an unbiased estimate of the population variance is 64.0 ml². He thinks the mean volume is less than 250 ml. Test his belief at the 5% significance level.

1. (a) Explain why a t-test, not a z-test, is used.

   • The population variance is not known, only an estimate from the sample, so use a t-test.
   • Assume the population is normally distributed.

2. (b) Write down hypotheses, defining μ.

   • Let μ ml be the population mean volume of a takeaway coffee.
   • H0: μ = 250, H1: μ < 250

3. (c) Find the p-value.

   • s = √64.0 = 8.00
   • From the calculator: t-test with statistics, μ0 = 250, sample mean 244.8, s = 8.00, n = 12, lower tail: t = −2.25, p = 0.0229

4. (d) State the conclusion in context.

   • 0.0229 < 0.05, so reject H0.
   • There is sufficient evidence, at the 5% significance level, that the mean volume of a takeaway coffee is less than 250 ml.

The mass, in kg, of 8 athletes is recorded at the start and at the end of a training season.

1. (a) Find the difference, end − start, for each athlete.

   • Subtract each start value from the end value in the same column.
   • −1.4, 0.5, −1.5, −1.5, −0.3, −1.2, −0.7, −0.8

2. (b) Use your calculator to find the mean and the standard deviation s of the differences.

   • From the calculator: one-variable statistics of the list of differences.
   • mean = −0.863, s = 0.699 (the sample standard deviation)

The resting heart rate of the members of a gym is normally distributed with mean 72 beats per minute (bpm) and standard deviation 6.4 bpm. A new fitness programme is claimed to lower the mean. A random sample of 25 members follows the programme. Assume the standard deviation is unchanged. Test the claim at the 1% significance level.

1. (a) Write down suitable hypotheses.

   • Let μ bpm be the population mean resting heart rate of members.
   • H0: μ = 72, H1: μ < 72

2. (b) Find the distribution of the sample mean \(\bar X\) when H0 is true.

   • Under H0, \(\bar X \sim N(72, 6.4^2/25)\).
   • The standard deviation is 6.4/√25 = 1.28.

3. (c) Find the critical region.

   • From the calculator: inverse normal, area 0.01, mean 72, standard deviation 1.28: c = 69.0
   • Critical region: \(\bar X < 69.0\)

4. (d) The sample mean is 69.4 bpm. State the conclusion.

   • 69.4 bpm is not in the critical region, so do not reject H0.
   • There is insufficient evidence, at the 1% significance level, that the mean resting heart rate of members has decreased.

A machine fills bottles with a mean of 500 ml and standard deviation 4.2 ml. A supervisor takes a random sample of 36 bottles to test, at the 5% significance level, whether the mean volume has changed. Assume the volumes are normally distributed and the standard deviation is unchanged.

1. (a) Write down the distribution of \(\bar X\) when H0 is true.

   • The population standard deviation is 4.2, so the sample mean has standard deviation 4.2/√36 = 0.700.
   • \(\bar X \sim N(500, 4.2^2/36)\)

2. (b) Find the critical region.

   • Two-tailed, so 2.5% in each tail.
   • From the calculator: inverse normal, areas 0.025 and 0.975, mean 500, standard deviation 0.700: 498.6 and 501.4
   • Critical region: \(\bar X < 498.6\) or \(\bar X > 501.4\)

3. (c) The sample mean is 498.3 ml. State the conclusion.

   • 498.3 ml is in the critical region, so reject H0.
   • There is sufficient evidence, at the 5% significance level, that the mean volume of a bottle has changed.

A psychologist tests whether listening to music reduces the reaction time of students. Nine students' reaction times, in milliseconds, are measured without and with music. Test at the 5% significance level.

1. (a) Explain why a paired test is suitable.

   • The same students appear in both rows, so each pair of values belongs together.
   • The samples are not independent, so use a paired test on the differences.

2. (b) Write down hypotheses in terms of μD, where D = with music − without music.

   • Let μD milliseconds be the population mean of with music − without music.
   • H0: μD = 0, H1: μD < 0

3. (c) Find the p-value.

   • Differences: −7, −2, −14, 2, −12, −7, −9, −7, −7
   • From the calculator: t-test on the differences, μ0 = 0, lower tail: t = −4.38, p = 0.00118

4. (d) State the conclusion in context.

   • 0.00118 < 0.05, so reject H0.
   • There is sufficient evidence, at the 5% significance level, that listening to music reduces the mean reaction time.

A teacher wants to compare the achievement of her class in two tests.

1. (a) She compares the marks of the same 15 students in an algebra test and a geometry test. Should she use a paired test or a two-sample test? Give a reason.

   • The same 15 students sit both tests, so each algebra mark pairs with a geometry mark.
   • Use a paired test on the differences.

2. (b) She also compares the mean heights of 12 boys and 14 girls from the class. Should she use a paired test or a two-sample test? Give a reason.

   • The boys and the girls are different people, with no natural pairing, and the samples have different sizes.
   • Use a two-sample test.

3. (c) State the assumption needed for a paired t-test.

   • A paired t-test works with the differences.
   • Assume the population of differences is normally distributed.

4. (d) State two assumptions needed for a pooled two-sample t-test.

   • Assume both populations are normally distributed.
   • Assume the two populations have equal variance, so the sample variances can be pooled.

5. (e) State one advantage of a paired design over two independent samples.

   • Each person is compared with themselves.
   • The variation between individuals drops out, so a smaller real difference can show up.

A coach splits a squad into two groups for sprint training. After a season, 9 swimmers in group 1 have a mean 50 m time of 52.6 s (sample variance 4.4) and 11 swimmers in group 2 have a mean time of 49.1 s (sample variance 3.9). The coach thinks the mean time of group 1 is more than 2 seconds greater than that of group 2. Assume the times are normally distributed with equal variance.

1. (a) Find unbiased estimates of the two population variances.

   • s1² = 9/8 × 4.4 = 4.95
   • s2² = 11/10 × 3.9 = 4.29

2. (b) Test the claim at the 5% significance level. Give the hypotheses, the pooled estimate of the variance, the test statistic and the p-value.

   • H0: μ1 − μ2 = 2, H1: μ1 − μ2 > 2, where μ1 and μ2 are the population means for group 1 and group 2.
   • sp² = (8 × 4.95 + 10 × 4.29) / 18 = 4.58
   • t = (52.6 − 49.1 − 2) / √(4.58 × (1/9 + 1/11)) = 1.56
   • From the calculator: t cdf, 18 degrees of freedom, upper tail from 1.56: p = 0.0682

3. (c) State the conclusion in context.

   • 0.0682 > 0.05, so do not reject H0.
   • There is insufficient evidence, at the 5% significance level, that the mean time of group 1 is more than 2 seconds greater than that of group 2.

Marks in a trial exam are normally distributed. For exam board X the standard deviation is known to be 9 and for exam board Y it is known to be 11. A random sample of 40 students sits the X paper and has a mean mark of 71.8. A different random sample of 50 students sits the Y paper and has a mean mark of 65.2. A tutor believes that the mean mark with board X is more than 4 marks greater than with board Y. Test her belief at the 5% significance level.

1. (a) Explain why a two-sample z-test can be used.

   • The two samples are independent, both populations are normal and both population standard deviations are known.
   • That makes a z-test suitable.

2. (b) Find the distribution of the difference between the two sample means when H0 is true.

   • H0: μ1 − μ2 = 4, H1: μ1 − μ2 > 4
   • Variance of the difference = 9²/40 + 11²/50 = 4.45
   • \(\bar X_1 - \bar X_2 \sim N(4, 4.45)\), standard deviation 2.11

3. (c) Find the p-value.

   • Observed difference = 71.8 − 65.2 = 6.60
   • From the calculator: normal cdf from 6.60 to ∞, mean 4, standard deviation 2.11: p = 0.109

4. (d) State the conclusion in context.

   • 0.109 > 0.05, so do not reject H0.
   • There is insufficient evidence, at the 5% significance level, that the mean mark with exam board X is more than 4 marks greater than with exam board Y.

A farmer thinks a new feed raises the daily milk yield of cows. The yield of one cow, in litres, is normally distributed with standard deviation 1.2, both before and after the change, and the yields are independent. The yields of 10 cows are measured before and after the change. Let D = yield after − yield before. Test at the 5% significance level.

1. (a) Write down suitable hypotheses in terms of μD.

   • Let μD litres be the population mean of D = milk yield after − milk yield before.
   • H0: μD = 0, H1: μD > 0

2. (b) Find Var(D).

   • Var(D) = 1.2² + 1.2² = 2.88
   • The two measurements are independent, so the variances add.

3. (c) Find the distribution of \(\bar D\) when H0 is true.

   • Under H0, \(\bar D \sim N(0, 2.88/10)\)
   • = N(0, 0.2880), standard deviation 0.537

4. (d) Find the critical region.

   • From the calculator: inverse normal, area 0.95, mean 0, standard deviation 0.537: c = 0.883
   • Critical region: \(\bar D > 0.883\)

5. (e) The sample mean difference is 0.9 litres. State the conclusion.

   • 0.9 litres is in the critical region, so reject H0.
   • There is sufficient evidence, at the 5% significance level, that the new feed raises the mean milk yield.

Customers' waiting times at a service desk are skewed, with mean 8.0 minutes and standard deviation 5.2 minutes. A new booking system is introduced, and the manager believes it reduces the mean waiting time. A random sample of 64 customers has a mean waiting time of 6.9 minutes. Assume the standard deviation is unchanged. Test at the 5% significance level.

1. (a) Explain why a test based on the normal distribution can be used although the distribution of waiting times is not normal.

   • The sample size, 64, is large.
   • By the central limit theorem, \(\bar X\) is approximately normally distributed, whatever the distribution of one value.
   • σ is known, so use a z-test.

2. (b) Find the p-value.

   • H0: μ = 8.0, H1: μ < 8.0
   • Under H0, \(\bar X \approx N(8.0, 5.2^2/64)\), standard deviation 0.650
   • From the calculator: normal cdf, lower tail from 6.9: p = 0.0453

3. (c) State the conclusion in context.

   • 0.0453 < 0.05, so reject H0.
   • There is sufficient evidence, at the 5% significance level, that the new system has reduced the mean waiting time.

A farm shop says its jars of jam contain a mean of 115 g. A customer thinks they contain more. The masses of a random sample of 9 jars are recorded. Assume the masses are normally distributed. Masses (g): 118, 124, 109, 131, 122, 127, 115, 129, 120

1. (a) Use your calculator to find the mean and the sample standard deviation s.

   • From the calculator: one-variable statistics of the data.
   • mean = 122, s = 7.04

2. (b) Find the p-value of a one-sample t-test of the claim.

   • H0: μ = 115, H1: μ > 115
   • From the calculator: t-test on the data, μ0 = 115, upper tail: t = 2.84, p = 0.0109

3. (c) The customer says: 'The p-value is small, so the jars are definitely heavier.' Comment on this statement.

   • A small p-value is evidence, not proof: if the mean really were 115 g, a sample like this would be unlikely, but it could still happen.
   • The test shows evidence at the chosen significance level. It does not show the mean is greater with certainty.