Hypothesis testing: exam review Exam review: second paper Worked solutions

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