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IB Mathematics: Applications and Interpretation HL · 13.4: Hypothesis Testing · 13.4.22

Hypothesis testing: exam review Exam review Answers

Section A: short questions

    1. (a) \(H_0:\mu=500\), \(H_1:\mu<500\), where \(\mu\) is the population mean mass of a box of cereal, in grams
    2. (b) Unusual; reject H0. There is sufficient evidence that the mean mass of a box is less than 500 g.
    1. (a) H0: the proportions are 0.4, 0.3, 0.2, 0.1; H1: they are not
    2. (b) 80, 60, 40, 20
    3. (c) 3
    4. (d) χ² = 12.3, p = 0.00642
    5. (e) Reject H0: there is sufficient evidence, at the 5% significance level, that the proportions of the drinks chosen are not as the café states
    1. (a) 5
    2. (b) 11.1
    3. (c) Reject H0: there is sufficient evidence, at the 5% significance level, that the die is not fair
    4. (d) 0.0122
    1. (a) \(H_0:\mu=1000\), \(H_1:\mu<1000\), where \(\mu\) is the population mean mass of a bag of sugar, in grams
    2. (b) the masses of the bags are normally distributed
    3. (c) mean \(=995\), \(s_{n-1}=4.70\)
    4. (d) t = −3.10, p = 0.00638
    5. (e) Reject H0: there is sufficient evidence, at the 5% significance level, that the mean mass of a bag of sugar is less than 1000 g
    1. (a) Both populations are normally distributed; the population variances are equal
    2. (b) \(H_0:\mu_1=\mu_2\), \(H_1:\mu_1\ne\mu_2\), where \(\mu_1\) and \(\mu_2\) are the population mean masses of tubs from machine 1 and machine 2, in grams
    3. (c) 25
    4. (d) t = 2.84, p = 0.00881
    5. (e) Reject H0: there is sufficient evidence, at the 5% significance level, that the mean masses of tubs from the two machines are different
    1. (a) Judge 1: 2, 4, 1, 6, 7, 3, 8, 5; judge 2: 1, 6, 2, 5, 7, 3, 8, 4 (for A to H)
    2. (b) 0.905
    3. (c) \(H_0\): no correlation between the judges' ranks; \(H_1\): positive correlation
    4. (d) Reject H0: there is sufficient evidence, at the 5% significance level, that the judges tend to agree

Section B: long questions

    1. (a) \(H_0\): \(X\sim B(4,0.15)\)
    2. (b) 83.5, 59.0, 15.6, 1.84, 0.081
    3. (c) The expected frequencies for X = 3 and X = 4 are below 5
    4. (d) 3 groups, so 2
    5. (e) χ² = 10.8, p = 0.00456
    6. (f) Reject H0: 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) X = 2 or more: more than expected (observed 28, expected 17.5)
    1. (a) H0: music preference is independent of year group; H1: it is not
    2. (b) (2−1)(3−1) = 2
    3. (c) 89 × 63 ÷ 166 = 33.8
    4. (d) χ² = 14.2, p = 8.29 × 10⁻⁴
    5. (e) Reject H0: there is sufficient evidence, at the 5% significance level, that music preference is not independent of year group
    6. (f) Year 10, Hip-hop: observed 13, expected 24.1 (fewer than expected)
    7. (g) Reject H0 at 1%
    1. (a) ranks of the values of \(x\): 1, 2, 3, 4.5, 4.5, 6, 7, 8, 9, 10; ranks of the values of \(y\): 10, 8, 9, 6, 5, 7, 3, 4, 2, 1
    2. (b) −0.936
    3. (c) \(H_0\): no correlation; \(H_1\): negative correlation
    4. (d) Reject H0: there is sufficient evidence, at the 5% significance level, that more time on social media goes with less sleep
    5. (e) r_s = −0.936: a strong negative association; students who spend more time on social media tend to sleep less
    6. (f) No change: 6.0 is still the largest value of x, so the ranks do not change
    7. (g) No change: the order of the values does not change

Section C: HL questions

    1. (a) 97 × 9 ÷ 194 = 4.50
    2. (b) e.g. expected frequency for Card and 65 or over is 4.50 (the expected frequencies in the 65 or over column are all 5 or less); combined table: Card: 31, 40, 26; Cash: 9, 18, 28; Phone app: 14, 19, 9
    3. (c) (3 − 1)(3 − 1) = 4
    4. (d) χ² = 13.0, p = 0.0113; Reject H0: there is sufficient evidence, at the 5% significance level, that payment method and age group are not independent
    1. (a) H0: the number of breakdowns in a week follows a Poisson distribution; mean = 151 ÷ 100 = 1.51
    2. (b) 100 × P(X = 2) = 25.2
    3. (c) expected frequency 4.79 < 5 for 4, so combine to '4 or more'; 5 groups − 1 − 1 (mean estimated) = 3
    4. (d) χ² = 2.10, p = 0.551; Do not reject H0: there is insufficient evidence, at the 5% significance level, that the number of breakdowns does not follow a Poisson distribution
    1. (a) H0: μ = 330, H1: μ < 330; z-test, because the population standard deviation is known
    2. (b) z = -2.02, p = 0.0215
    3. (c) p = 0.0036 (t = -3.46, df = 9); 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
    1. (a) H0: μ1 = μ2, H1: μ1 < μ2, where μ1 is the population mean height of a plant for fertiliser X and μ2 is the population mean height of a plant for fertiliser Y
    2. (b) z = −2.03, p = 0.0210
    3. (c) Reject H0: 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) two-sample pooled t-test; both populations are normally distributed; the population variances are equal
    1. (a) H0: μd = 0, H1: μd < 0, where μd is the population mean of (the after value − the before value) for athletes
    2. (b) it removes the differences between individuals, so only the change is tested (the samples are not independent)
    3. (c) d: −2, −3, −1, −4, 0, −3, −3, 1; t = −3.07, p = 0.00902
    4. (d) Reject H0: there is sufficient evidence, at the 5% significance level, that the programme reduces the mean resting heart rate
    5. (e) the differences are normally distributed
    1. (a) \(\bar X_A-\bar X_B\sim N(\mu_A-\mu_B,\ 1.154)\)
    2. (b) by the central limit theorem, because both samples are large, the sample means are approximately normally distributed
    3. (c) H0: μA − μB = 2, H1: μA − μB > 2
    4. (d) z = 1.21, p = 0.1131
    5. (e) 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
    1. (a) \(\bar X\sim N(400,\ 1)\)
    2. (b) \(\bar X<398.04\) or \(\bar X>401.96\)
    3. (c) 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
    4. (d) 0.1336
    5. (e) 0.4840
    1. (a) H0: p = 0.7, H1: p < 0.7, where p is the probability that a seed germinates
    2. (b) X ≤ 13
    3. (c) 0.0442
    4. (d) Reject H0: there is sufficient evidence, at the 5% significance level, that the proportion of seeds that germinate is less than 70%
    5. (e) 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) 0.543
    1. (a) H0: λ = 3.5, H1: λ < 3.5, where λ is the mean number of flaws per week
    2. (b) X ~ Po(14)
    3. (c) p = 0.0316; Reject H0: there is sufficient evidence, at the 5% significance level, that the mean number of flaws per week has decreased
    4. (d) X ≤ 7
    5. (e) 0.0316
    1. (a) H0: λ = 0.5, H1: λ > 0.5, where λ is the mean number of customers arriving per hour
    2. (b) Y ~ Po(3)
    3. (c) Y ≥ 7
    4. (d) 0.0335
    5. (e) 0.6063
    6. (f) P(Z = 0) = 0.1353 > 0.05, so no value of Z can have a probability of at most 0.05; the smallest significance level for which a critical region exists is 0.1353
    1. (a) H0: ρ = 0, H1: ρ > 0, where ρ is the population correlation coefficient between revision time and test score
    2. (b) r = 0.933, p = 1.17 × 10⁻⁴
    3. (c) Reject H0: there is sufficient evidence, at the 5% significance level, that there is positive correlation between revision time and test score
    4. (d) both variables are normally distributed (the data come from a bivariate normal population)
    1. (a) 0.0912
    2. (b) X1 + X2 ~ N(1000, 72); 0.0786
    3. (c) H0: μ = 500, H1: μ > 500
    4. (d) x̄ > 502.5
    5. (e) mean 504, standard deviation 6.048
    6. (f) 0.155

jmaths.xyz · AI HL · 13.4.22 Unit review