Hypothesis testing: exam review Exam review Worked solutions

A cereal company states that the mean mass of its boxes is 500 g. An inspector believes that the mean mass is less than this. She will test her 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 \(H_0:\mu=500\) and \(H_1:\mu<500\)
   • A1 for defining \(\mu\) as the population mean mass of a box of cereal, in grams

2. (b) She weighs a random sample of 25 boxes and finds a sample mean of 492 g. If the company's statement is true, the probability of a sample mean of 492 g or less is 0.004. State whether the inspector has sufficient evidence to support her belief, and give a reason. [3]

   • M1 for comparing 0.004 with 0.05
   • A1 for 0.004 < 0.05, so reject \(H_0\)
   • A1 for the conclusion in context: There is sufficient evidence that the mean mass of a box is less than 500 g.

A café states that its customers choose espresso, latte, cappuccino or another drink in the proportions 0.4, 0.3, 0.2 and 0.1. The café records the drink chosen by each of 200 customers. The results are in the table. A goodness of fit test is used to test the café's statement. [8]

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

   • A1 for \(H_0\): the customers choose the drinks in the proportions 0.4, 0.3, 0.2 and 0.1; \(H_1\): the proportions are not as stated

2. (b) Find the expected frequency for each drink. [2]

   • M1 for expected frequency = total × proportion (200×…)
   • A1 for 80, 60, 40, 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 = 12.3, p = 0.00642

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

   • R1 for comparing: 0.00642 < 0.05
   • A1 for reject \(H_0\): there is sufficient evidence, at the 5% significance level, that the proportions of the drinks chosen are not as the café states

A six-sided die is rolled 120 times. A goodness of fit test of the null hypothesis that the die is fair gives \(\chi^2=14.6\). [6]

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

   • A1 for \(6-1=5\)

2. (b) Find the critical value of the test at the 5% significance level. [2]

   • M1 for using the inverse chi-squared function with area 0.95 to the left and 5 degrees of freedom
   • A1 for critical value \(11.1\)

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

   • R1 for 14.6 > 11.1
   • A1 for reject \(H_0\): there is sufficient evidence, at the 5% significance level, that the die is not fair

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

   • A1 for \(p=0.0122\) (the area to the right of 14.6)

A manufacturer states that the mean mass of its bags of sugar is 1000 g. A trading standards officer believes that the mean mass is less than this. She weighs 10 randomly chosen bags. The masses, in grams, are 1002, 1003, 996, 992, 991, 996, 991, 989, 997, 997. [8]

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

   • A1 for \(H_0:\mu=1000\) and \(H_1:\mu<1000\) (a one-tailed test)
   • A1 for defining \(\mu\) as the population mean mass of a bag of sugar, in grams

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

   • A1 for the masses of the bags 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 \(=995\), \(s_{n-1}=4.70\)

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 = −3.10, p = 0.00638

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

   • R1 for 0.00638 < 0.05
   • A1 for reject \(H_0\): there is sufficient evidence, at the 5% significance level, that the mean mass of a bag of sugar is less than 1000 g

Two machines fill tubs of margarine. A random sample of 12 tubs from machine 1 has a mean mass of 251.8 g and \(s_{n-1}=3.1\) g. A random sample of 15 tubs from machine 2 has a mean mass of 248.2 g and \(s_{n-1}=3.4\) g. A test is carried out to see whether the mean masses of the tubs from the two machines 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 masses of tubs from machine 1 and machine 2, in grams

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

   • A1 for \(12+15-2=25\)

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 = 2.84, p = 0.00881

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

   • R1 for 0.00881 < 0.05
   • A1 for reject \(H_0\): there is sufficient evidence, at the 5% significance level, that the mean masses of tubs from the two machines are different

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

1. (a) Write down the rank that each judge gave to each dancer, for dancers A to H. [2]

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

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

   • A1 for \(r_s=0.905\)

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

   • A1 for \(H_0\): there is no correlation between the two judges' 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 0.905 > 0.643
   • A1 for reject \(H_0\): there is sufficient evidence, at the 5% significance level, that the judges tend to agree

A battery maker states that each battery is faulty with probability 0.15, independently of the others. An inspector tests 160 packs of 4 batteries and records \(X\), the number of faulty batteries in each pack. The results are in the table. A goodness of fit test is used to test whether \(X\sim B(4, 0.15)\). [13]

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

   • A1 for \(H_0\): the number of faulty batteries in a pack follows B(4, 0.15)
   • A1 for \(H_1\): the number of faulty batteries in a pack does not follow B(4, 0.15)

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

   • M1 for \(160\times P(X=x)\) with \(X\sim B(4,0.15)\)
   • A1 for the first expected frequencies
   • A1 for the rest: 83.5, 59.0, 15.6, 1.84, 0.081

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

   • R1 for expected frequencies below 5 (1.84, 0.081) 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 3 groups, so \(3-1=2\) (the probability 0.15 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 = 10.8, p = 0.00456

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

   • R1 for 0.00456 < 0.05
   • A1 for reject \(H_0\): there is sufficient evidence, at the 5% significance level, that the number of faulty batteries in a pack does not follow B(4, 0.15)

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: 3.68, 0.842, 6.26
   • A1 for X = 2 or more, with more observations than expected

A school asks 166 students in Year 10 and Year 11 which type of music they prefer. The results are in the table. A \(\chi^2\) test for independence is used to test whether music preference is independent of year group. [13]

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

   • A1 for \(H_0\): music preference and year group are independent
   • A1 for \(H_1\): music preference and year group are not independent

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

   • A1 for \((2-1)(3-1)=2\)

3. (c) Show how to find the expected frequency for Year 11 and Rock. [2]

   • M1 for row total × column total ÷ grand total: \(89\times63\div166\)
   • A1 for \(33.8\)

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 = 14.2, p = 8.29 × 10⁻⁴

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

   • R1 for 8.29 × 10⁻⁴ < 0.05
   • A1 for reject \(H_0\): there is sufficient evidence, at the 5% significance level, that music preference is not independent of year group

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 Year 10, Hip-hop: observed 13, expected 24.1, so fewer than expected

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

   • R1 for 8.29 × 10⁻⁴ < 0.01
   • A1 for reject \(H_0\) at the 1% level: there is sufficient evidence, at the 5% significance level, that music preference is not independent of year group

Ten students record the time they spend on social media each day, \(x\) hours, and the time they sleep that night, \(y\) hours. The values of \(x\) are 1.5, 2.0, 2.5, 3.0, 3.0, 3.5, 4.0, 4.5, 5.0, 7.5 and the values of \(y\) are 8.5, 8.0, 8.2, 7.5, 7.0, 7.6, 6.5, 6.8, 6.0, 5.5. A test is carried out, at the 5% significance level, to see whether more time on social media goes with less sleep. The critical value for this one-tailed test is 0.564. [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, 4.5, 6, 7, 8, 9, 10 (the tied values share the average rank)
   • A1 for ranks of the values of \(y\): 10, 8, 9, 6, 5, 7, 3, 4, 2, 1

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

   • A1 for \(r_s=-0.936\)

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

   • A1 for \(H_0\): there is no correlation between the time on social media and the time asleep
   • 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 −0.936 with the critical value \(-0.564\)
   • R1 for −0.936 < \(-0.564\)
   • A1 for reject \(H_0\): there is sufficient evidence, at the 5% significance level, that more time on social media goes with less sleep

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

   • A1 for \(r_s=-0.936\), a strong negative correlation
   • R1 for in context: more time on social media tends to go with less sleep

6. (f) The largest value of \(x\), 7.5, was recorded wrongly. The correct value is 6.0. 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 hours to minutes. State how this changes \(r_s\). [1]

   • A1 for no change, because multiplying every value by 60 keeps the order, so the ranks are the same

A shop asks 194 customers how they paid and which age group they are in. A \(\chi^2\) test for independence is to be used to test whether the way a customer pays is independent of age group. [10]

1. (a) Show how to find the expected frequency for Card and 65 or over. [2]

   • M1 for row total × column total ÷ grand total: \(97\times9\div194\)
   • A1 for \(4.50\)

2. (b) Write down one expected frequency that is less than 5, and explain why 45 to 64 and 65 or over 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 \(4.50\) for Card and 65 or over, as a reason for combining (the test is unreliable when an expected frequency is 5 or less)
   • M1 for combining the neighbouring columns 45 to 64 and 65 or over (whole columns)
   • A1 for the combined table: Card 31, 40, 26; Cash 9, 18, 28; Phone app 14, 19, 9

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 \(\chi^2=13.0\), \(p=0.0113\)
   • R1 for 0.0113 < 0.05
   • A1 for Reject \(H_0\): there is sufficient evidence, at the 5% significance level, that payment method and age group are not independent

A city bus company records the number of breakdowns in each of 100 weeks. The results are in the table. The company thinks that the number of breakdowns in a week 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 breakdowns in a week. [2]

   • A1 for \(H_0\): the number of breakdowns in a week follows a Poisson distribution
   • A1 for mean \(=151\div100=1.51\)

2. (b) Using this mean, show that the expected frequency of 2 breakdowns is 25.2, correct to 3 significant figures. [2]

   • M1 for \(100\times P(X=2)\) with \(X\sim\text{Po}(1.51)\)
   • A1 for \(25.2\)

3. (c) The expected frequency of exactly 4 breakdowns is 4.79. 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 \(4.79\) 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 \(\chi^2=2.10\), \(p=0.551\)
   • R1 for 0.551 > 0.05
   • A1 for Do not reject \(H_0\): there is insufficient evidence, at the 5% significance level, that the number of breakdowns does not follow a Poisson distribution

A bottling plant states that the mean volume of a bottle of juice is 330 ml. An inspector believes that the mean volume is less than this. The volume of juice in a bottle is normally distributed with standard deviation 2.5 ml, which is known. A random sample of 40 bottles has a mean volume of 329.2 ml. [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 \(H_0:\mu=330\) and \(H_1:\mu<330\) (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 \(\sigma=2.5\), \(n=40\), \(\bar x=329.2\), lower tail
   • A1 for \(z=-2.02\), \(p=0.0215\)

3. (c) The inspector then tests a second production line, where the standard deviation is not known. The volumes, in ml, of a random sample of 10 bottles from this line are: 328.1, 330.4, 327.9, 329.5, 328.8, 330.1, 327.6, 329.0, 328.4, 329.7. Assuming that these volumes are normally distributed, test the same belief for this line. 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 (lower tail)
   • A1 for \(p=0.0036\)
   • R1 for 0.0036 < 0.05
   • A1 for Reject \(H_0\): there is sufficient evidence, at the 5% significance level, that the mean volume of a bottle from the second line is less than 330 ml

A gardener compares two fertilisers, X and Y. She grows 30 plants with fertiliser X and 35 different plants with fertiliser Y. The heights of the plants are normally distributed. The population standard deviations are known to be 5.1 cm for X and 6.0 cm for Y. The mean height of the plants with fertiliser X is 41.2 cm and the mean height of the plants with fertiliser Y is 44.0 cm. She believes that fertiliser Y gives taller plants. [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 height of a plant for fertiliser X and for fertiliser Y

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 \(\sigma_1=5.1\), \(\sigma_2=6.0\), lower tail
   • A1 for \(z=−2.03\), \(p=0.0210\)

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

   • R1 for 0.0210 < 0.05
   • A1 for Reject \(H_0\): there is sufficient evidence, at the 5% significance level, that the mean height of a plant is greater with fertiliser Y than with fertiliser X

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)

Eight athletes take part in a six-week training programme. Each athlete's resting heart rate, in beats per minute, is measured before and after the programme. A coach believes that the programme reduces the mean resting heart rate. [8]

1. (a) State the null hypothesis and the alternative hypothesis, taking \(d\) as the after value minus the before 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 \(d\) = after − before)
   • A1 for defining \(\mu_d\) as the population mean difference for athletes

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 \(8\) differences (after − before) and using a one-sample \(t\)-test on them, lower tail
   • A1 for \(t=−3.07\), \(p=0.00902\)

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

   • R1 for 0.00902 < 0.05
   • A1 for Reject \(H_0\): there is sufficient evidence, at the 5% significance level, that the programme reduces the mean resting heart rate

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 company compares two methods, A and B, of assembling a unit. A random sample of 40 units assembled by method A has a mean assembly time of 31.2 minutes. An independent random sample of 45 units assembled by method B has a mean assembly time of 27.9 minutes. The population standard deviations are known to be 4.8 minutes for method A and 5.1 minutes for method B. The distributions of the assembly times are not known. The company claims that method A takes more than 2 minutes longer, on average, than method B. Let \(\mu_A\) and \(\mu_B\) be the population mean assembly times for methods 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 \(\dfrac{4.8^2}{40}+\dfrac{5.1^2}{45}=1.154\)

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 \(H_0:\mu_A-\mu_B=2\) and \(H_1:\mu_A-\mu_B>2\)

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

   • M1 for a two-sample \(z\)-test statistic \(\dfrac{(31.2-27.9)-2}{\sqrt{1.154}}\) and an upper-tail probability
   • A1 for \(z=1.21\), \(p=0.1131\)

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

   • R1 for 0.1131 > 0.05
   • A1 for Do not reject \(H_0\): there is insufficient evidence, at the 5% significance level, that the mean assembly time for method A is more than 2 minutes greater than that for method B

A machine fills tubs with soup. The mass of soup in a tub, \(X\) g, is normally distributed with standard deviation 6 g, and the mass is claimed to have mean 400 g. An inspector takes a random sample of 36 tubs and tests, at the 5% significance level, whether the mean mass differs from 400 g. Let \(\mu\) be the population mean mass. [9]

1. (a) Write down the distribution of \(\bar X\), the mean of the sample of 36, when \(H_0\) is true. [1]

   • A1 for \(\bar X\sim N\left(400,\dfrac{6^2}{36}\right)\), that is standard deviation 1

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 \(\bar X<398.04\)
   • A1 for \(\bar X>401.96\)

3. (c) The sample mean is 402.3 g. State the conclusion of the test, in context. [1]

   • A1 for Reject \(H_0\): there is sufficient evidence, at the 5% significance level, that the mean mass of soup in a tub differs from 400 g (402.3 is inside the critical region)

4. (d) A second inspector rejects \(H_0\) if \(\bar X\) differs from 400 by more than 1.5 g. Find the probability that this inspector makes a Type I error. [2]

   • M1 for \(2P\left(\bar X>400+1.5\right)=2P\left(Z>1.500\right)\)
   • A1 for 0.1336

5. (e) The true value of \(\mu\) is in fact 402. Using the critical region in part (b), find the probability of a Type II error. [2]

   • M1 for \(P(398.04<\bar X<401.96)\) with \(\bar X\sim N(402,1)\)
   • A1 for 0.4840

A seed supplier states that 70% of its seeds germinate. A gardener believes that the proportion is lower. She sows 25 seeds, chosen at random, and counts the number \(X\) that germinate. 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 \(H_0:p=0.7\) and \(H_1:p<0.7\) (a one-tailed test)
   • A1 for defining \(p\) as the probability that a seed germinates

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

   • M1 for using \(X\sim B(25,0.7)\) and cumulative probabilities at the boundary
   • A1 for \(P(X \le 13)=0.0442<0.05\) and \(P(X \le 14)=0.0978>0.05\)
   • A1 for the critical region \(X\le 13\)

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 \(0.0442\)

4. (d) The observed value is 12. State the conclusion of the test, in context. [2]

   • R1 for 12 being in the critical region (or the \(p\)-value 0.0175 < 0.05)
   • A1 for Reject \(H_0\): there is sufficient evidence, at the 5% significance level, that the proportion of seeds that germinate is less than 70%

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

   • A1 for concluding that the proportion of seeds that germinate is not less than 70%, when it is less than 70% (not rejecting H0 when H0 is false)

6. (f) In fact the probability is 0.55. Find the probability of a Type II error. [3]

   • M1 for using \(X\sim B(25,0.55)\)
   • M1 for \(P(X\ge 14)\), the probability of the complement of the critical region, which is \(1-P(X\le 13)\)
   • A1 for \(0.543\)

A weaving mill has averaged 3.5 flaws per week in the cloth it produces. After new looms are fitted, the manager records 7 flaws in the next 4 weeks. She believes that the mean number of flaws per week has decreased, and tests her belief at the 5% significance level. Assume that flaws occur at random. [10]

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

   • A1 for \(H_0:\lambda=3.5\) and \(H_1:\lambda<3.5\)
   • A1 for defining \(\lambda\) as the mean number of flaws per week

2. (b) Write down the distribution of \(X\), the number of flaws in 4 weeks, when \(H_0\) is true. [1]

   • A1 for \(X\sim\text{Po}(14)\) (the rate scaled to 4 weeks: \(3.5\times4\))

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

   • M1 for using \(X\sim\text{Po}(14)\) with the observed value 7
   • A1 for \(P(X\le 7)=0.0316\)
   • R1 for 0.0316 < 0.05
   • A1 for Reject \(H_0\): there is sufficient evidence, at the 5% significance level, that the mean number of flaws per week has decreased

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

   • M1 for cumulative probabilities at the boundary: \(P(X\le 7)=0.0316<0.05\), \(P(X\le 8)=0.0621>0.05\)
   • A1 for the critical region \(X \le 7\)

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

   • A1 for \(0.0316\)

Customers arrive at a help desk at random, at a mean rate of 0.5 per hour. After a change to the opening hours, the manager believes that the rate has increased. She counts the number of customers who arrive in a 6-hour period, \(Y\), and will test her belief at the 5% significance level. [10]

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

   • A1 for \(H_0:\lambda=0.5\) and \(H_1:\lambda>0.5\)
   • A1 for defining \(\lambda\) as the mean number of customers arriving per hour

2. (b) Write down the distribution of \(Y\), the number of customers arriving in 6 hours, when \(H_0\) is true. [1]

   • A1 for \(Y\sim\text{Po}(3)\) (the rate scaled to 6 hours: 0.5\times6)

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

   • M1 for cumulative probabilities at the boundary: \(P(Y\ge 7)=0.0335\le 0.05\), \(P(Y\ge 6)=0.0839>0.05\)
   • A1 for the critical region \(Y\ge 7\)

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

   • A1 for 0.0335 (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 1 per hour. [2]

   • M1 for \(P(Y\le 6)\) with \(Y\sim\text{Po}(6)\)
   • A1 for 0.6063

6. (f) Another manager counts the customers arriving in a 4-hour period, \(Z\), to test whether the rate has decreased below 0.5 per hour 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 \(Z\sim\text{Po}(2)\) and \(P(Z=0)=0.1353>0.05\)
   • A1 for no critical region exists at the 5% level; the smallest level for which one exists is 0.1353

A teacher records the number of hours, \(x\), that nine students spent revising for a test and their score, \(y\), out of 80. She believes that there is positive correlation between the time spent revising and the score. [7]

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

   • A1 for \(H_0:\rho=0\) and \(H_1:\rho>0\) (one-tailed)
   • A1 for defining \(\rho\) as the population correlation coefficient between revision time and test score

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 \(9\) pairs and using the linear correlation test with \(\rho>0\)
   • A1 for \(r=0.933\), \(p=1.17 × 10⁻⁴\)

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

   • R1 for 1.17 × 10⁻⁴ < 0.05
   • A1 for Reject \(H_0\): there is sufficient evidence, at the 5% significance level, that there is positive correlation between revision time and test score

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

   • A1 for both variables being normally distributed

The mass, \(X\) grams, of a bag of rice packed by a machine is normally distributed with mean 500 and standard deviation 6. [13]

1. (a) Find the probability that one bag has a mass greater than 508 g. [1]

   • A1 for \(0.0912\)

2. (b) Two bags are chosen at random. Find the probability that their total mass is greater than 1012 g. [3]

   • M1 for the mean \(1000\) and variance \(2\times6^2=72\) of the total
   • A1 for \(X_1+X_2\sim N(1000,72)\)
   • A1 for \(0.0786\)

3. (c) The mass is thought to have increased. A random sample of 16 bags is used to test this at the 5% significance level, using the sample mean \(\bar X\) and the known standard deviation \(6\). Write down the null hypothesis and the alternative hypothesis. [1]

   • A1 for \(H_0:\mu=500\) and \(H_1:\mu>500\)

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

   • M1 for \(\bar X\sim N\left(500,\dfrac{6^2}{16}\right)\), so the standard deviation of \(\bar X\) is \(1.50\)
   • M1 for using the inverse normal with probability 0.95 (or \(z=1.645\))
   • A1 for the critical region \(\bar x>502.5\)

5. (e) The test is carried out after a change in the process. Every bag now has its mass increased by 0.8%. Find the new mean and the new standard deviation of the mass. [2]

   • M1 for multiplying the mean and the standard deviation by \(1.008\)
   • A1 for mean \(504\) and standard deviation \(6.048\)

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

   • M1 for \(\bar X\sim N\left(504,\dfrac{6.048^2}{16}\right)\)
   • M1 for finding the probability of the complement of the critical region, \(P(\bar X<502.5)\)
   • A1 for \(0.155\) (using the unrounded critical value; accept \(0.161\) from the rounded value \(502.5\))