Hypothesis testing: exam review Exam review: second paper One question at a time

A tablet maker states that the mean time to charge a tablet fully is 90 minutes. A tester believes that the mean time is greater than this. He will test his belief at the 5% significance level. [5]

1. (a) State the null hypothesis and the alternative hypothesis. Define the symbol you use. [2]

   • A1 for \(H_0:\mu=90\) and \(H_1:\mu>90\)
   • A1 for defining \(\mu\) as the population mean charging time of a tablet, in minutes

2. (b) He tests a random sample of 20 tablets and finds a sample mean of 95 minutes. If the maker's statement is true, the probability of a sample mean of 95 minutes or more is 0.19. State whether the tester has sufficient evidence to support his belief, and give a reason. [3]

   • M1 for comparing 0.19 with 0.05
   • A1 for 0.19 > 0.05, so do not reject \(H_0\)
   • A1 for the conclusion in context: There is insufficient evidence that the mean charging time is greater than 90 minutes.

A plant breeder predicts that the seeds from a cross will be round yellow, round green, wrinkled yellow and wrinkled green in the ratio 9 : 3 : 3 : 1. She grows 320 seeds and records the type of each. The results are in the table. A goodness of fit test is used to test her prediction. [8]

1. (a) State the null hypothesis and the alternative hypothesis. [1]

   • A1 for \(H_0\): the seed types are in the ratio 9 : 3 : 3 : 1; \(H_1\): the proportions are not as stated

2. (b) Find the expected frequency for each type of seed. [2]

   • M1 for expected frequency = total × proportion (320×…)
   • A1 for 180, 60, 60, 20

3. (c) Write down the number of degrees of freedom. [1]

   • A1 for \(4-1=3\)

4. (d) Use your calculator to find the \(\chi^2\) statistic and the \(p\)-value. [2]

   • M1 for using the goodness of fit test with the observed and the expected frequencies
   • A1 for χ² = 1.09, p = 0.780

5. (e) Test at the 5% significance level and state your conclusion in context. [2]

   • R1 for comparing: 0.780 > 0.05
   • A1 for do not reject \(H_0\): there is insufficient evidence, at the 5% significance level, that the seed types are not in the ratio 9 : 3 : 3 : 1

A spinner has eight equal sectors and is spun 160 times. A goodness of fit test, at the 1% significance level, of the null hypothesis that the spinner is fair gives \(\chi^2=10.3\). [6]

1. (a) Write down the number of degrees of freedom. [1]

   • A1 for \(8-1=7\)

2. (b) Find the critical value of the test at the 1% significance level. [2]

   • M1 for using the inverse chi-squared function with area 0.99 to the left and 7 degrees of freedom
   • A1 for critical value \(18.5\)

3. (c) Compare the test statistic with the critical value and state your conclusion in context. [2]

   • R1 for 10.3 < 18.5
   • A1 for do not reject \(H_0\): there is insufficient evidence, at the 1% significance level, that the spinner is not fair

4. (d) Find the \(p\)-value of the test. [1]

   • A1 for \(p=0.172\) (the area to the right of 10.3)

A call centre states that the mean time to answer a call is 40 seconds. A manager believes that the mean time is less than this since new staff were hired. She records the time, in seconds, to answer each of 12 randomly chosen calls: 39.3, 40.5, 37.9, 40.2, 41.8, 40.4, 43.6, 36.7, 39.4, 37.2, 37.0, 38.7. [8]

1. (a) State the null hypothesis and the alternative hypothesis. Define the symbol you use. [2]

   • A1 for \(H_0:\mu=40\) and \(H_1:\mu<40\) (a one-tailed test)
   • A1 for defining \(\mu\) as the population mean time to answer a call, in seconds

2. (b) State an assumption that is needed for the test to be valid. [1]

   • A1 for the times to answer a call are normally distributed

3. (c) Use your calculator to find the mean and the standard deviation \(s_{n-1}\) of the sample. [1]

   • A1 for mean \(=39.4\), \(s_{n-1}=2.06\)

4. (d) Use your calculator to find the \(t\)-statistic and the \(p\)-value for the test. [2]

   • M1 for using the one-sample \(t\)-test on the data list, lower tail
   • A1 for t = −1.02, p = 0.165

5. (e) Conclude at the 5% significance level, in context. [2]

   • R1 for 0.165 > 0.05
   • A1 for do not reject \(H_0\): there is insufficient evidence, at the 5% significance level, that the mean time to answer a call is less than 40 seconds

A teacher compares two revision methods. 10 students use method 1 and have a mean test score of 82.4 and \(s_{n-1}=6.2\). 14 different students use method 2 and have a mean test score of 79.1 and \(s_{n-1}=5.8\). A test is carried out to see whether the mean scores of students using the two methods are different. [9]

1. (a) State two assumptions that are needed for the pooled two-sample \(t\)-test. [2]

   • A1 for both populations are normally distributed
   • A1 for the two populations have equal variance (and the samples are independent)

2. (b) State the null hypothesis and the alternative hypothesis. Define the symbols you use. [2]

   • A1 for \(H_0:\mu_1=\mu_2\) and \(H_1:\mu_1\ne\mu_2\) (two-tailed: the question asks if they differ)
   • A1 for defining \(\mu_1\) and \(\mu_2\): \(\mu_1\) and \(\mu_2\) are the population mean test scores of students using method 1 and method 2

3. (c) Write down the number of degrees of freedom. [1]

   • A1 for \(10+14-2=22\)

4. (d) Use your calculator to find the \(t\)-statistic and the \(p\)-value for the test. [2]

   • M1 for the two-sample \(t\)-test with statistics, pooled, two-tailed
   • A1 for t = 1.34, p = 0.195

5. (e) Conclude at the 5% significance level, in context. [2]

   • R1 for 0.195 > 0.05
   • A1 for do not reject \(H_0\): there is insufficient evidence, at the 5% significance level, that the mean scores using the two methods are different

Two tasters each rank eight cheeses, A to H, from best to worst. Taster 1 orders the cheeses B, G, D, A, E, H, C, F. Taster 2 orders them D, A, H, B, C, E, G, F. A one-tailed test is used, at the 5% significance level, to see whether the tasters tend to agree. The critical value is 0.643. [6]

1. (a) Write down the rank that each taster gave to each cheese, for cheeses A to H. [2]

   • M1 for turning each ordered list into ranks (first in the list is rank 1)
   • A1 for taster 1: 4, 1, 7, 3, 5, 8, 2, 6 and taster 2: 2, 4, 5, 1, 6, 8, 7, 3

2. (b) Use your calculator to find Spearman's rank correlation coefficient, \(r_s\). [1]

   • A1 for \(r_s=0.333\)

3. (c) State the null hypothesis and the alternative hypothesis. [1]

   • A1 for \(H_0\): there is no correlation between the two tasters' ranks; \(H_1\): there is a positive correlation (a one-tailed test)

4. (d) Conclude at the 5% significance level, in context. [2]

   • R1 for 0.333 < 0.643
   • A1 for do not reject \(H_0\): there is insufficient evidence, at the 5% significance level, that the tasters tend to agree

A game uses a fair six-sided die. In each of 400 turns a player rolls the die 4 times and records \(X\), the number of sixes. The results are in the table. A goodness of fit test is used to test whether \(X\sim B(4, \frac{1}{6})\). [13]

1. (a) State the null hypothesis and the alternative hypothesis. [2]

   • A1 for \(H_0\): the number of sixes in a turn follows B(4, 1/6)
   • A1 for \(H_1\): the number of sixes in a turn does not follow B(4, 1/6)

2. (b) Find the expected frequencies for \(X=0\), \(X=1\), \(X=2\), \(X=3\) and \(X=4\). [3]

   • M1 for \(400\times P(X=x)\) with \(X\sim B(4,\frac{1}{6})\)
   • A1 for the first expected frequencies
   • A1 for the rest: 193, 154, 46.3, 6.17, 0.309

3. (c) Explain why \(X=4\) cannot be kept as a separate group. [1]

   • R1 for expected frequencies below 5 (0.309) break the rule that every expected frequency is at least 5

4. (d) The groups are combined so that every expected frequency is at least 5. Write down the number of degrees of freedom. [1]

   • A1 for 4 groups, so \(4-1=3\) (the probability \frac{1}{6} was given, so no parameter is estimated)

5. (e) Use your calculator to find the \(\chi^2\) statistic and the \(p\)-value. [2]

   • M1 for the goodness of fit test with the grouped observed and expected frequencies
   • A1 for χ² = 0.990, p = 0.804

6. (f) Test at the 5% significance level and state your conclusion in context. [2]

   • R1 for 0.804 > 0.05
   • A1 for do not reject \(H_0\): there is insufficient evidence, at the 5% significance level, that the number of sixes in a turn does not follow B(4, 1/6)

7. (g) Find which group contributes most to the \(\chi^2\) statistic, and say whether it has more or fewer observations than expected. [2]

   • M1 for finding \(\dfrac{(O-E)^2}{E}\) for each group: 0.261, 0.259, 0.114, 0.356
   • A1 for X = 3 or more, with more observations than expected

A school asks 154 students how they travel to school and whether they arrived late on a given day. The results are in the table. A \(\chi^2\) test for independence is used to test whether arriving late is independent of the way a student travels. [13]

1. (a) State the null hypothesis and the alternative hypothesis. [2]

   • A1 for \(H_0\): arriving late and the way a student travels are independent
   • A1 for \(H_1\): arriving late and the way a student travels are not independent

2. (b) Write down the number of degrees of freedom. [1]

   • A1 for \((3-1)(2-1)=2\)

3. (c) Show how to find the expected frequency for Walking and Late. [2]

   • M1 for row total × column total ÷ grand total: \(48\times62\div154\)
   • A1 for \(19.3\)

4. (d) Use your calculator to find the \(\chi^2\) statistic and the \(p\)-value. [2]

   • M1 for entering the observed table (without totals) and using the \(\chi^2\) test for independence
   • A1 for χ² = 1.45, p = 0.485

5. (e) Test at the 5% significance level and state your conclusion in context. [2]

   • R1 for 0.485 > 0.05
   • A1 for do not reject \(H_0\): there is insufficient evidence, at the 5% significance level, that arriving late is not independent of the way a student travels

6. (f) Find the cell with the greatest contribution to the \(\chi^2\) statistic, and compare its observed and expected frequencies. [2]

   • M1 for the expected table and the values \(\dfrac{(O-E)^2}{E}\) for each cell (or comparing observed with expected)
   • A1 for Walking, Late: observed 16, expected 19.3, so fewer than expected

7. (g) State the conclusion if the significance level is 1%. [2]

   • R1 for 0.485 > 0.01
   • A1 for do not reject \(H_0\) at the 1% level: there is insufficient evidence, at the 5% significance level, that arriving late is not independent of the way a student travels

Nine flats are at different distances from a city centre. The distance from the centre, \(x\) km, and the monthly rent, \(y\) euros, of each flat are recorded. The values of \(x\) are 1.2, 2.5, 3.1, 4.8, 6.0, 6.0, 9.5, 11.0, 18.2 and the values of \(y\) are 640, 720, 900, 640, 800, 560, 700, 780, 610. A test is carried out, at the 5% significance level, to see whether rent tends to be lower further from the centre. The critical value for this one-tailed test is 0.6. [14]

1. (a) Write down the ranks of the values of \(x\) and of the values of \(y\), giving tied values their average rank. [3]

   • M1 for ranking each variable from smallest to largest
   • A1 for ranks of the values of \(x\): 1, 2, 3, 4, 5.5, 5.5, 7, 8, 9 (the tied values share the average rank)
   • A1 for ranks of the values of \(y\): 3.5, 6, 9, 3.5, 8, 1, 5, 7, 2

2. (b) Use your calculator to find Spearman's rank correlation coefficient, \(r_s\). [1]

   • A1 for \(r_s=-0.168\)

3. (c) State the null hypothesis and the alternative hypothesis. [2]

   • A1 for \(H_0\): there is no correlation between the distance from the centre and the rent
   • A1 for \(H_1\): there is a negative correlation (a one-tailed test)

4. (d) Use the critical value to conclude, at the 5% significance level, in context. [3]

   • M1 for comparing −0.168 with the critical value \(-0.6\)
   • R1 for −0.168 > \(-0.6\)
   • A1 for do not reject \(H_0\): there is insufficient evidence, at the 5% significance level, that rent is lower further from the centre

5. (e) Interpret the value of \(r_s\) in the context of the question. [2]

   • A1 for \(r_s=-0.168\), a weak negative correlation
   • R1 for in context: there is only a weak tendency for rent to be lower further from the centre

6. (f) The largest value of \(x\), 18.2, was recorded wrongly. The correct value is 16.0. Without calculating, state how this changes \(r_s\), and give a reason. [2]

   • A1 for \(r_s\) does not change
   • R1 for the correct value is still the largest value of \(x\), so no rank changes

7. (g) The values of \(y\) are converted from euros to cents. State how this changes \(r_s\). [1]

   • A1 for no change, because multiplying every value by 100 keeps the order, so the ranks are the same

A gym asks 209 members which time of day they usually train and which type of membership they hold. A \(\chi^2\) test for independence is to be used to test whether the time of day is independent of membership type. [10]

1. (a) Show how to find the expected frequency for Morning and Corporate. [2]

   • M1 for row total × column total ÷ grand total: \(86\times9\div209\)
   • A1 for \(3.70\)

2. (b) Write down one expected frequency that is less than 5, and explain why Family and Corporate should be combined. Write down the table of observed frequencies after the columns are combined. [3]

   • A1 for an expected frequency of 5 or less, for example \(3.70\) for Morning and Corporate, as a reason for combining (the test is unreliable when an expected frequency is 5 or less)
   • M1 for combining the neighbouring columns Family and Corporate (whole columns)
   • A1 for the combined table: Morning 28, 35, 23; Afternoon 19, 26, 26; Evening 17, 15, 20

3. (c) Write down the number of degrees of freedom for the test on the combined table. [1]

   • A1 for \((3-1)(3-1)=4\)

4. (d) Use your calculator to find the \(\chi^2\) statistic and the \(p\)-value for the combined table, and state the conclusion at the 5% significance level, in context. [4]

   • M1 for entering the combined observed table and using the \(\chi^2\) test for independence
   • A1 for \(\chi^2=3.55\), \(p=0.471\)
   • R1 for 0.471 > 0.05
   • A1 for Do not reject \(H_0\): there is insufficient evidence, at the 5% significance level, that time of day and membership type are not independent

A website records the number of server errors in each of 100 hours. The results are in the table. The owner thinks that the number of errors in an hour follows a Poisson distribution, and uses a \(\chi^2\) goodness of fit test with the mean estimated from the data. [10]

1. (a) State the null hypothesis, and find the mean number of errors in an hour. [2]

   • A1 for \(H_0\): the number of errors in an hour follows a Poisson distribution
   • A1 for mean \(=140\div100=1.40\)

2. (b) Using this mean, show that the expected frequency of 2 errors is 24.2, correct to 3 significant figures. [2]

   • M1 for \(100\times P(X=2)\) with \(X\sim\text{Po}(1.40)\)
   • A1 for \(24.2\)

3. (c) The expected frequency of exactly 4 errors is 3.95. Explain why the groups 4, 5 and 6 are combined with any larger value into one group, and state the number of degrees of freedom. [2]

   • A1 for the expected frequency \(3.95\) being less than 5 (the tail groups would also be small), so the groups are combined as '4 or more'
   • A1 for 5 groups, 1 parameter estimated: \(5-1-1=3\)

4. (d) Use your calculator to find the \(\chi^2\) statistic and the \(p\)-value, and state the conclusion at the 5% significance level, in context. [4]

   • M1 for the observed frequencies 0, 1, 2, 3, 4 or more with the expected frequencies, using the goodness of fit test with \(\nu=3\)
   • A1 for \(\chi^2=8.80\), \(p=0.0321\)
   • R1 for 0.0321 < 0.05
   • A1 for Reject \(H_0\): there is sufficient evidence, at the 5% significance level, that the number of errors does not follow a Poisson distribution

A flour mill states that the mean mass of a sack of flour is 25.0 kg. A trading standards officer believes that the mean mass is greater than this. The mass of a sack is normally distributed with standard deviation 0.40 kg, which is known. A random sample of 50 sacks has a mean mass of 25.11 kg. [7]

1. (a) Write down the null and alternative hypotheses, and state, with a reason, whether a \(z\)-test or a \(t\)-test should be used. [2]

   • A1 for \(H_0:\mu=25.0\) and \(H_1:\mu>25.0\) (one-tailed)
   • A1 for a \(z\)-test because the population standard deviation \(\sigma\) is known

2. (b) Find the \(p\)-value for the test. [2]

   • M1 for a one-sample \(z\)-test from summary statistics with \(\sigma=0.4\), \(n=50\), \(\bar x=25.11\), upper tail
   • A1 for \(z=1.94\), \(p=0.0259\)

3. (c) The officer then tests a second mill, where the standard deviation is not known. The masses, in kg, of a random sample of 10 sacks from this mill are: 25.2, 24.9, 25.3, 25.1, 25.4, 24.8, 25.2, 25.3, 25.0, 25.35. Assuming that these masses are normally distributed, test the same belief for this mill. Find the \(p\)-value and state the conclusion at the 5% significance level, in context. [3]

   • M1 for a one-sample \(t\)-test on the data, because the standard deviation is estimated from the sample (upper tail)
   • A1 for \(p=0.0186\)
   • R1 for 0.0186 < 0.05
   • A1 for Reject \(H_0\): there is sufficient evidence, at the 5% significance level, that the mean mass of a sack from the second mill is greater than 25.0 kg

A courier compares two routes, 1 and 2, between the same two depots. She times 45 journeys on route 1 and 40 different journeys on route 2. The journey times are normally distributed. The population standard deviations are known to be 4.2 minutes for route 1 and 3.9 minutes for route 2. The mean time on route 1 is 36.4 minutes and the mean time on route 2 is 35.2 minutes. She believes that route 1 takes longer. [8]

1. (a) State the null hypothesis and the alternative hypothesis, defining the symbols you use. [2]

   • A1 for \(H_0:\mu_1=\mu_2\) and \(H_1:\mu_1>\mu_2\) (a one-tailed test)
   • A1 for defining \(\mu_1\) and \(\mu_2\) as the population mean journey time for route 1 and for route 2

2. (b) Use your calculator to find the \(p\)-value for a two-sample \(z\)-test. [2]

   • M1 for using the two-sample \(z\)-test from summary statistics with \(\sigma_1=4.2\), \(\sigma_2=3.9\), upper tail
   • A1 for \(z=1.37\), \(p=0.0860\)

3. (c) Conclude at the 5% significance level, in context. [2]

   • R1 for 0.0860 > 0.05
   • A1 for Do not reject \(H_0\): there is insufficient evidence, at the 5% significance level, that the mean journey time is greater on route 1 than on route 2

4. (d) Suppose that the population standard deviations were not known, but were assumed to be equal. State the test that would be used, and two assumptions that are needed for it. [2]

   • A1 for the two-sample pooled \(t\)-test
   • A1 for any two of: normal populations; equal variances (independent random samples)

Nine students sit a practice test and, after a revision course, a second test of the same difficulty. Their scores are in the table. A teacher believes that the revision course increases the mean score. [8]

1. (a) State the null hypothesis and the alternative hypothesis, taking \(d\) as the second value minus the first value and defining \(\mu_d\). [2]

   • A1 for \(H_0:\mu_d=0\) and \(H_1:\mu_d>0\) (the sign of \(\mu_d\) matches \(d\) = second − first)
   • A1 for defining \(\mu_d\) as the population mean difference for students

2. (b) State one advantage of using a paired test rather than a two-sample test here. [1]

   • R1 for an advantage in context: each person acts as their own control, so the differences between people are removed (accept: the pairs are matched)

3. (c) Use your calculator to find the \(t\)-statistic and the \(p\)-value for the test. [2]

   • M1 for finding the \(9\) differences (second − first) and using a one-sample \(t\)-test on them, upper tail
   • A1 for \(t=1.58\), \(p=0.0768\)

4. (d) Conclude at the 5% significance level, in context. [2]

   • R1 for 0.0768 > 0.05
   • A1 for Do not reject \(H_0\): there is insufficient evidence, at the 5% significance level, that the revision course increases the mean score

5. (e) State an assumption that is needed for the test to be valid. [1]

   • A1 for the population of differences is normally distributed

A farm compares two feeds, A and B, for chickens. A random sample of 50 chickens given feed A has a mean mass gain of 620 g over a month. An independent random sample of 60 chickens given feed B has a mean mass gain of 566 g. The population standard deviations are known to be 70 g for feed A and 64 g for feed B. The distributions of the mass gains are not known. The supplier claims that feed A gives a mean mass gain more than 30 g greater than feed B. Let \(\mu_A\) and \(\mu_B\) be the population mean mass gains for feeds A and B. [8]

1. (a) Write down the distribution of \(\bar X_A-\bar X_B\) in terms of \(\mu_A\) and \(\mu_B\). [2]

   • A1 for mean \(\mu_A-\mu_B\)
   • A1 for variance \(\dfrac{70^2}{50}+\dfrac{64^2}{60}=166.3\)

2. (b) Explain why this distribution can be used even though the distributions of the individual times are not known. [1]

   • R1 for the central limit theorem: the samples are large, so the sample means are approximately normal

3. (c) Write down the null and alternative hypotheses. [1]

   • A1 for \(H_0:\mu_A-\mu_B=30\) and \(H_1:\mu_A-\mu_B>30\)

4. (d) Find the \(p\)-value for the test. [2]

   • M1 for a two-sample \(z\)-test statistic \(\dfrac{(620-566)-30}{\sqrt{166.3}}\) and an upper-tail probability
   • A1 for \(z=1.86\), \(p=0.0314\)

5. (e) State the conclusion at the 5% significance level, in context. [2]

   • R1 for 0.0314 < 0.05
   • A1 for Reject \(H_0\): there is sufficient evidence, at the 5% significance level, that the mean mass gain with feed A is more than 30 g greater than with feed B

A machine cuts lengths of cable. The length of a cable, \(X\) cm, is normally distributed with standard deviation 0.8 cm, and the length is claimed to have mean 150 cm. An inspector takes a random sample of 25 cables and tests, at the 5% significance level, whether the mean length differs from 150 cm. Let \(\mu\) be the population mean length. [9]

1. (a) Write down the distribution of \(\bar X\), the mean of the sample of 25, when \(H_0\) is true. [1]

   • A1 for \(\bar X\sim N\left(150,\dfrac{0.8^2}{25}\right)\), that is standard deviation 0.16

2. (b) Find the critical region for \(\bar X\) at the 5% significance level. [3]

   • M1 for a two-tailed test with \(z=\pm1.96\) (2.5% in each tail)
   • A1 for \(\bar X<149.69\)
   • A1 for \(\bar X>150.31\)

3. (c) The sample mean is 150.18 cm. State the conclusion of the test, in context. [1]

   • A1 for Do not reject \(H_0\): there is insufficient evidence, at the 5% significance level, that the mean length of a cable differs from 150 cm (150.18 is outside the critical region)

4. (d) A second inspector rejects \(H_0\) if \(\bar X\) differs from 150 by more than 0.25 cm. Find the probability that this inspector makes a Type I error. [2]

   • M1 for \(2P\left(\bar X>150+0.25\right)=2P\left(Z>1.562\right)\)
   • A1 for 0.1182

5. (e) The true value of \(\mu\) is in fact 150.4. Using the critical region in part (b), find the probability of a Type II error. [2]

   • M1 for \(P(149.69<\bar X<150.31)\) with \(\bar X\sim N(150.4,0.0256)\)
   • A1 for 0.2946

A cafe manager says that 15% of customers order a decaffeinated drink. After a health campaign the owner believes that the proportion has increased. She asks 40 customers, chosen at random, and counts the number \(X\) who order a decaffeinated drink. She tests her belief at the 5% significance level. [13]

1. (a) Write down suitable null and alternative hypotheses, defining the parameter you use. [2]

   • A1 for \(H_0:p=0.15\) and \(H_1:p>0.15\) (a one-tailed test)
   • A1 for defining \(p\) as the probability that a customer orders a decaffeinated drink

2. (b) Find the critical region for the test at the 5% significance level. [3]

   • M1 for using \(X\sim B(40,0.15)\) and cumulative probabilities at the boundary
   • A1 for \(P(X \ge 11)=0.0299<0.05\) and \(P(X \ge 10)=0.0672>0.05\)
   • A1 for the critical region \(X\ge 11\)

3. (c) Hence find the probability of a Type I error for this test. [2]

   • M1 for the probability of the critical region under \(H_0\)
   • A1 for \(0.0299\)

4. (d) The observed value is 10. State the conclusion of the test, in context. [2]

   • R1 for 10 not being in the critical region (or the \(p\)-value 0.0672 > 0.05)
   • A1 for Do not reject \(H_0\): there is insufficient evidence, at the 5% significance level, that the proportion of customers who order a decaffeinated drink has increased

5. (e) Describe a Type II error in this context. [1]

   • A1 for concluding that the proportion of customers who order a decaffeinated drink has not increased, when it has increased (not rejecting H0 when H0 is false)

6. (f) In fact the probability is 0.3. Find the probability of a Type II error. [3]

   • M1 for using \(X\sim B(40,0.3)\)
   • M1 for \(P(X\le 10)\), the probability of the complement of the critical region, which is \(P(X\le 10)\)
   • A1 for \(0.309\)

A harbour has averaged 0.8 boats arriving late per day. After the timetable is changed, 13 boats arrive late in the next 10 days. The harbour master believes that the mean number of late boats per day has increased, and tests his belief at the 5% significance level. Assume that late arrivals occur at random. [10]

1. (a) Write down suitable null and alternative hypotheses, defining the parameter you use. [2]

   • A1 for \(H_0:\lambda=0.8\) and \(H_1:\lambda>0.8\)
   • A1 for defining \(\lambda\) as the mean number of late boats per day

2. (b) Write down the distribution of \(X\), the number of late boats in 10 days, when \(H_0\) is true. [1]

   • A1 for \(X\sim\text{Po}(8)\) (the rate scaled to 10 days: \(0.8\times10\))

3. (c) Find the \(p\)-value for the test and state the conclusion at the 5% significance level, in context. [4]

   • M1 for using \(X\sim\text{Po}(8)\) with the observed value 13
   • A1 for \(P(X\ge 13)=1-P(X\le 12)=0.0638\)
   • R1 for 0.0638 > 0.05
   • A1 for Do not reject \(H_0\): there is insufficient evidence, at the 5% significance level, that the mean number of late boats per day has increased

4. (d) Find the critical region for the test at the 5% significance level, for a count over 10 days. [2]

   • M1 for cumulative probabilities at the boundary: \(P(X\ge 14)=0.0342<0.05\), \(P(X\ge 13)=0.0638>0.05\)
   • A1 for the critical region \(X \ge 14\)

5. (e) Find the probability of a Type I error for this test. [1]

   • A1 for \(0.0342\)

Faults occur in a pipeline at random, at a mean rate of 0.4 per kilometre. After a repair programme, the engineer believes that the rate has increased. He counts the number of faults in a 5 km section, \(Y\), and will test his belief at the 5% significance level. [10]

1. (a) Write down suitable null and alternative hypotheses, defining the parameter you use. [2]

   • A1 for \(H_0:\lambda=0.4\) and \(H_1:\lambda>0.4\)
   • A1 for defining \(\lambda\) as the mean number of faults per kilometre

2. (b) Write down the distribution of \(Y\), the number of faults in 5 kilometres, when \(H_0\) is true. [1]

   • A1 for \(Y\sim\text{Po}(2)\) (the rate scaled to 5 kilometres: 0.4\times5)

3. (c) Find the critical region for the test at the 5% significance level. [2]

   • M1 for cumulative probabilities at the boundary: \(P(Y\ge 6)=0.0166\le 0.05\), \(P(Y\ge 5)=0.0527>0.05\)
   • A1 for the critical region \(Y\ge 6\)

4. (d) Write down the actual significance level of the test. [1]

   • A1 for 0.0166 (the probability of falling in the critical region when \(H_0\) is true)

5. (e) Find the probability of a Type II error when the true rate is 0.9 per kilometre. [2]

   • M1 for \(P(Y\le 5)\) with \(Y\sim\text{Po}(4.5)\)
   • A1 for 0.7029

6. (f) Another engineer counts the faults in a 3 km section, \(Z\), to test whether the rate has decreased below 0.4 per kilometre at the 5% significance level. Show that this test has no critical region, and find the smallest significance level for which a critical region exists. [2]

   • M1 for \(Z\sim\text{Po}(1.2)\) and \(P(Z=0)=0.3012>0.05\)
   • A1 for no critical region exists at the 5% level; the smallest level for which one exists is 0.3012

A researcher records the height, \(x\) cm, and the reaction time, \(y\) milliseconds, of eight adults. She wishes to test whether there is any correlation between height and reaction time. [7]

1. (a) State the null hypothesis and the alternative hypothesis, defining the symbol you use. [2]

   • A1 for \(H_0:\rho=0\) and \(H_1:\rho\ne0\) (two-tailed)
   • A1 for defining \(\rho\) as the population correlation coefficient between height and reaction time

2. (b) Use your calculator to find the product moment correlation coefficient \(r\) and the \(p\)-value for the test. [2]

   • M1 for entering the \(8\) pairs and using the linear correlation test with \(\rho\ne0\)
   • A1 for \(r=−0.0404\), \(p=0.924\)

3. (c) Conclude at the 5% significance level, in context. [2]

   • R1 for 0.924 > 0.05
   • A1 for Do not reject \(H_0\): there is insufficient evidence, at the 5% significance level, that there is correlation between height and reaction time

4. (d) State an assumption that is needed for the test to be valid. [1]

   • A1 for both variables being normally distributed

The time, \(X\) seconds, that a robot takes to complete one task is normally distributed with mean 120 and standard deviation 8. [13]

1. (a) Find the probability that one task has a time less than 110 s. [1]

   • A1 for \(0.106\)

2. (b) Two tasks are chosen at random. Find the probability that their total time is less than 230 s. [3]

   • M1 for the mean \(240\) and variance \(2\times8^2=128\) of the total
   • A1 for \(X_1+X_2\sim N(240,128)\)
   • A1 for \(0.188\)

3. (c) The time is thought to have decreased. A random sample of 36 tasks is used to test this at the 5% significance level, using the sample mean \(\bar X\) and the known standard deviation \(8\). Write down the null hypothesis and the alternative hypothesis. [1]

   • A1 for \(H_0:\mu=120\) and \(H_1:\mu<120\)

4. (d) Find the critical region for \(\bar X\). [3]

   • M1 for \(\bar X\sim N\left(120,\dfrac{8^2}{36}\right)\), so the standard deviation of \(\bar X\) is \(1.33\)
   • M1 for using the inverse normal with probability 0.05 (or \(z=-1.645\))
   • A1 for the critical region \(\bar x<117.8\)

5. (e) The test is carried out after a change in the process. Every task now has its time decreased by 3%. Find the new mean and the new standard deviation of the time. [2]

   • M1 for multiplying the mean and the standard deviation by \(0.97\)
   • A1 for mean \(116.4\) and standard deviation \(7.76\)

6. (f) Use these values to find the probability of a Type II error. [3]

   • M1 for \(\bar X\sim N\left(116.4,\dfrac{7.76^2}{36}\right)\)
   • M1 for finding the probability of the complement of the critical region, \(P(\bar X>117.8)\)
   • A1 for \(0.138\) (using the unrounded critical value; accept \(0.140\) from the rounded value \(117.8\))