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

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