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

A teacher times how long, in seconds, students take to solve a puzzle. One random group had hints and a different random group had none. Assume the times are normally distributed with equal variance. He tests, at the 10% significance level, whether the mean times of the two groups are different. With hints (s): 48, 55, 62, 41, 57, 50, 66, 53 No hints (s): 61, 72, 58, 69, 75, 64, 70

1. (a) State suitable hypotheses.

   • Let μ1 and μ2 be the population mean time to solve the puzzle, in seconds, for the group with hints and the group without hints.
   • 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 = −3.50, p = 0.00388

3. (c) State the conclusion in context.

   • 0.00388 < 0.1, so reject H0.
   • There is sufficient evidence, at the 10% significance level, that the mean times of the two groups are different.

A bus company says that the journey to school takes a mean of 35 minutes. A parent times a random sample of 15 journeys. The mean is 37.9 minutes and an unbiased estimate of the population variance is 28.8 min². She thinks the mean journey time is more than 35 minutes. Test her 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 minutes be the population mean journey time of the school bus.
   • H0: μ = 35, H1: μ > 35

3. (c) Find the p-value.

   • s = √28.8 = 5.37
   • From the calculator: t-test with statistics, μ0 = 35, sample mean 37.9, s = 5.37, n = 15, upper tail: t = 2.09, p = 0.0275

4. (d) State the conclusion in context.

   • 0.0275 < 0.05, so reject H0.
   • There is sufficient evidence, at the 5% significance level, that the mean journey time of the school bus is more than 35 minutes.

The number of sit-ups that 8 pupils can do in one minute is counted before and after a month of practice.

1. (a) Find the difference, after − before, for each pupil.

   • Subtract each before value from the after value in the same column.
   • 5, 3, 7, 3, 1, 6, 4, 6

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 = 4.38, s = 2.00 (the sample standard deviation)

The drying time of a load of washing in a tumble dryer is normally distributed with mean 55 minutes and standard deviation 7 minutes. After a repair the owner thinks the mean drying time has increased. She times a random sample of 30 loads and assumes the standard deviation is unchanged. Test her belief at the 5% significance level.

1. (a) Write down suitable hypotheses.

   • Let μ minutes be the population mean drying time of a load.
   • H0: μ = 55, H1: μ > 55

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

   • Under H0, \(\bar X \sim N(55, 7^2/30)\).
   • The standard deviation is 7/√30 = 1.28.

3. (c) Find the critical region.

   • From the calculator: inverse normal, area 0.95, mean 55, standard deviation 1.28: c = 57.1
   • Critical region: \(\bar X > 57.1\)

4. (d) The sample mean is 57.6 minutes. State the conclusion.

   • 57.6 minutes is in the critical region, so reject H0.
   • There is sufficient evidence, at the 5% significance level, that the mean drying time has increased.

A saw cuts rods with a mean length of 120 cm and standard deviation 0.9 cm. An engineer takes a random sample of 25 rods to test, at the 1% significance level, whether the mean length has changed. Assume the lengths 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 0.9, so the sample mean has standard deviation 0.9/√25 = 0.180.
   • \(\bar X \sim N(120, 0.9^2/25)\)

2. (b) Find the critical region.

   • Two-tailed, so 0.5% in each tail.
   • From the calculator: inverse normal, areas 0.005 and 0.995, mean 120, standard deviation 0.180: 119.5 and 120.5
   • Critical region: \(\bar X < 119.5\) or \(\bar X > 120.5\)

3. (c) The sample mean is 120.5 cm. State the conclusion.

   • 120.5 cm is in the critical region, so reject H0.
   • There is sufficient evidence, at the 1% significance level, that the mean length of a rod has changed.

A coach tests whether a ten-minute walk raises the resting pulse rate of pupils. Eight pupils' pulse rates, in beats per minute, are measured before and after the walk. Test at the 5% significance level.

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

   • The same pupils 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 = after − before.

   • Let μD beats per minute be the population mean of after − before.
   • H0: μD = 0, H1: μD > 0

3. (c) Find the p-value.

   • Differences: 6, 3, −1, 7, 5, 3, 6, 4
   • From the calculator: t-test on the differences, μ0 = 0, upper tail: t = 4.61, p = 0.00123

4. (d) State the conclusion in context.

   • 0.00123 < 0.05, so reject H0.
   • There is sufficient evidence, at the 5% significance level, that the walk raises the mean pulse rate.

A sports scientist is comparing training methods.

1. (a) She measures the jump height of the same 10 athletes before and after a month of training. Should she use a paired test or a two-sample test? Give a reason.

   • The same 10 athletes are measured before and after, so each pair of heights belongs together.
   • Use a paired test on the differences.

2. (b) She also compares the mean jump heights of 9 athletes who trained with weights and 11 different athletes who trained with bands. Should she use a paired test or a two-sample test? Give a reason.

   • The two groups are different athletes, with no natural pairing, and the groups 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 athlete is compared with themselves.
   • The variation between individuals drops out, so a smaller real difference can show up.

A school compares two teaching methods. Students taught with the old method (12 students) have a mean mark of 71.5 (sample variance 16.8) and students taught with the new method (10 students) have a mean mark of 66.2 (sample variance 14.4). A teacher thinks the mean mark with the old method is more than 3 marks greater. Assume the marks are normally distributed with equal variance.

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

   • s1² = 12/11 × 16.8 = 18.3
   • s2² = 10/9 × 14.4 = 16.0

2. (b) Test the claim at the 10% significance level. Give the hypotheses, the pooled estimate of the variance, the test statistic and the p-value.

   • H0: μ1 − μ2 = 3, H1: μ1 − μ2 > 3, where μ1 and μ2 are the population means for the old method and the new method.
   • sp² = (11 × 18.3 + 9 × 16.0) / 20 = 17.3
   • t = (71.5 − 66.2 − 3) / √(17.3 × (1/12 + 1/10)) = 1.29
   • From the calculator: t cdf, 20 degrees of freedom, upper tail from 1.29: p = 0.106

3. (c) State the conclusion in context.

   • 0.106 > 0.1, so do not reject H0.
   • There is insufficient evidence, at the 10% significance level, that the mean mark with the old method is more than 3 marks greater than with the new method.

Daily sales, in items, are normally distributed. At shop M the standard deviation is known to be 14 and at shop N it is known to be 10. Over a random sample of 35 days, shop M sells a mean of 253 items. Over a different random sample of 45 days, shop N sells a mean of 232 items. The owner believes that the mean daily sales at shop M are more than 15 greater than at shop N. Test his belief at the 10% 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 = 15, H1: μ1 − μ2 > 15
   • Variance of the difference = 14²/35 + 10²/45 = 7.82
   • \(\bar X_1 - \bar X_2 \sim N(15, 7.82)\), standard deviation 2.80

3. (c) Find the p-value.

   • Observed difference = 253 − 232 = 21.0
   • From the calculator: normal cdf from 21.0 to ∞, mean 15, standard deviation 2.80: p = 0.0160

4. (d) State the conclusion in context.

   • 0.0160 < 0.1, so reject H0.
   • There is sufficient evidence, at the 10% significance level, that the mean daily sales at shop M are more than 15 greater than at shop N.

A canteen manager thinks a new layout shortens the time students queue for lunch. The queue time of one student, in minutes, is normally distributed with standard deviation 0.8, both before and after the change, and the times are independent. The queue times of 12 students are measured before and after the change. Let D = queue time before − queue time after. Test at the 10% significance level.

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

   • Let μD minutes be the population mean of D = queue time before − queue time after.
   • H0: μD = 0, H1: μD > 0

2. (b) Find Var(D).

   • Var(D) = 0.8² + 0.8² = 1.28
   • The two measurements are independent, so the variances add.

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

   • Under H0, \(\bar D \sim N(0, 1.28/12)\)
   • = N(0, 0.1067), standard deviation 0.327

4. (d) Find the critical region.

   • From the calculator: inverse normal, area 0.9, mean 0, standard deviation 0.327: c = 0.419
   • Critical region: \(\bar D > 0.419\)

5. (e) The sample mean difference is 0.6 minutes. State the conclusion.

   • 0.6 minutes is in the critical region, so reject H0.
   • There is sufficient evidence, at the 10% significance level, that the new layout shortens the mean queue time.

The daily rainfall in a valley is skewed, with mean 3.4 mm and standard deviation 2.9 mm. A scientist thinks the mean daily rainfall has increased over the last 50 days, which have a mean of 4.2 mm. Assume the standard deviation is unchanged and treat the 50 days as a random sample. Test at the 5% significance level.

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

   • The sample size, 50, 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 = 3.4, H1: μ > 3.4
   • Under H0, \(\bar X \approx N(3.4, 2.9^2/50)\), standard deviation 0.410
   • From the calculator: normal cdf, upper tail from 4.2: p = 0.0255

3. (c) State the conclusion in context.

   • 0.0255 < 0.05, so reject H0.
   • There is sufficient evidence, at the 5% significance level, that the mean daily rainfall has increased.

A gym says its fitness classes last a mean of 50 minutes. A member thinks they are shorter. The lengths of a random sample of 9 classes are recorded. Assume the lengths are normally distributed. Times (min): 46, 52, 44, 49, 55, 47, 51, 43, 50

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 = 48.6, s = 3.91

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

   • H0: μ = 50, H1: μ < 50
   • From the calculator: t-test on the data, μ0 = 50, lower tail: t = −1.11, p = 0.150

3. (c) The member says: 'The p-value is large, so the classes last exactly 50 minutes.' Comment on this statement.

   • A large p-value means there is insufficient evidence that the mean is below 50 minutes. It does not prove the mean is 50.
   • Not rejecting H0 is not the same as showing H0 is true. A larger sample might find a difference.