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

Hypothesis testing: exam review Exam review: second paper Answers

Section A: short questions

    1. (a) \(H_0:\mu=90\), \(H_1:\mu>90\), where \(\mu\) is the population mean charging time of a tablet, in minutes
    2. (b) Not unusual; do not reject H0. There is insufficient evidence that the mean charging time is greater than 90 minutes.
    1. (a) H0: the types are in the ratio 9 : 3 : 3 : 1; H1: they are not
    2. (b) 180, 60, 60, 20
    3. (c) 3
    4. (d) χ² = 1.09, p = 0.780
    5. (e) Do not reject H0: there is insufficient evidence, at the 5% significance level, that the seed types are not in the ratio 9 : 3 : 3 : 1
    1. (a) 7
    2. (b) 18.5
    3. (c) Do not reject H0: there is insufficient evidence, at the 1% significance level, that the spinner is not fair
    4. (d) 0.172
    1. (a) \(H_0:\mu=40\), \(H_1:\mu<40\), where \(\mu\) is the population mean time to answer a call, in seconds
    2. (b) the times to answer a call are normally distributed
    3. (c) mean \(=39.4\), \(s_{n-1}=2.06\)
    4. (d) t = −1.02, p = 0.165
    5. (e) Do not reject H0: there is insufficient evidence, at the 5% significance level, that the mean time to answer a call is less than 40 seconds
    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 test scores of students using method 1 and method 2
    3. (c) 22
    4. (d) t = 1.34, p = 0.195
    5. (e) Do not reject H0: there is insufficient evidence, at the 5% significance level, that the mean scores using the two methods are different
    1. (a) Taster 1: 4, 1, 7, 3, 5, 8, 2, 6; taster 2: 2, 4, 5, 1, 6, 8, 7, 3 (for A to H)
    2. (b) 0.333
    3. (c) \(H_0\): no correlation between the tasters' ranks; \(H_1\): positive correlation
    4. (d) Do not reject H0: there is insufficient evidence, at the 5% significance level, that the tasters tend to agree

Section B: long questions

    1. (a) \(H_0\): \(X\sim B(4,\frac{1}{6})\)
    2. (b) 193, 154, 46.3, 6.17, 0.309
    3. (c) The expected frequency for X = 4 is below 5
    4. (d) 4 groups, so 3
    5. (e) χ² = 0.990, p = 0.804
    6. (f) Do not reject H0: 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) X = 3 or more: more than expected (observed 8, expected 6.48)
    1. (a) H0: arriving late is independent of the way a student travels; H1: it is not
    2. (b) (3−1)(2−1) = 2
    3. (c) 48 × 62 ÷ 154 = 19.3
    4. (d) χ² = 1.45, p = 0.485
    5. (e) Do not reject H0: there is insufficient evidence, at the 5% significance level, that arriving late is not independent of the way a student travels
    6. (f) Walking, Late: observed 16, expected 19.3 (fewer than expected)
    7. (g) Do not reject H0 at 1%
    1. (a) ranks of the values of \(x\): 1, 2, 3, 4, 5.5, 5.5, 7, 8, 9; ranks of the values of \(y\): 3.5, 6, 9, 3.5, 8, 1, 5, 7, 2
    2. (b) −0.168
    3. (c) \(H_0\): no correlation; \(H_1\): negative correlation
    4. (d) Do not reject H0: there is insufficient evidence, at the 5% significance level, that rent is lower further from the centre
    5. (e) r_s = −0.168: a weak negative association; rent tends to fall slightly with distance, but the link is weak
    6. (f) No change: 16.0 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) 86 × 9 ÷ 209 = 3.70
    2. (b) e.g. expected frequency for Morning and Corporate is 3.70 (the expected frequencies in the Corporate column are all 5 or less); combined table: Morning: 28, 35, 23; Afternoon: 19, 26, 26; Evening: 17, 15, 20
    3. (c) (3 − 1)(3 − 1) = 4
    4. (d) χ² = 3.55, p = 0.471; Do not reject H0: there is insufficient evidence, at the 5% significance level, that time of day and membership type are not independent
    1. (a) H0: the number of errors in an hour follows a Poisson distribution; mean = 140 ÷ 100 = 1.40
    2. (b) 100 × P(X = 2) = 24.2
    3. (c) expected frequency 3.95 < 5 for 4, so combine to '4 or more'; 5 groups − 1 − 1 (mean estimated) = 3
    4. (d) χ² = 8.80, p = 0.0321; Reject H0: there is sufficient evidence, at the 5% significance level, that the number of errors does not follow a Poisson distribution
    1. (a) H0: μ = 25.0, H1: μ > 25.0; z-test, because the population standard deviation is known
    2. (b) z = 1.94, p = 0.0259
    3. (c) p = 0.0186 (t = 2.44, df = 9); 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
    1. (a) H0: μ1 = μ2, H1: μ1 > μ2, where μ1 is the population mean journey time for route 1 and μ2 is the population mean journey time for route 2
    2. (b) z = 1.37, p = 0.0860
    3. (c) Do not reject H0: 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) 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 second value − the first value) for students
    2. (b) it removes the differences between individuals, so only the change is tested (the samples are not independent)
    3. (c) d: −1, 3, 0, 2, −1, 2, 3, −1, 1; t = 1.58, p = 0.0768
    4. (d) Do not reject H0: there is insufficient evidence, at the 5% significance level, that the revision course increases the mean score
    5. (e) the differences are normally distributed
    1. (a) \(\bar X_A-\bar X_B\sim N(\mu_A-\mu_B,\ 166.3)\)
    2. (b) by the central limit theorem, because both samples are large, the sample means are approximately normally distributed
    3. (c) H0: μA − μB = 30, H1: μA − μB > 30
    4. (d) z = 1.86, p = 0.0314
    5. (e) 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
    1. (a) \(\bar X\sim N(150,\ 0.0256)\)
    2. (b) \(\bar X<149.69\) or \(\bar X>150.31\)
    3. (c) 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
    4. (d) 0.1182
    5. (e) 0.2946
    1. (a) H0: p = 0.15, H1: p > 0.15, where p is the probability that a customer orders a decaffeinated drink
    2. (b) X ≥ 11
    3. (c) 0.0299
    4. (d) Do not reject H0: there is insufficient evidence, at the 5% significance level, that the proportion of customers who order a decaffeinated drink has increased
    5. (e) 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) 0.309
    1. (a) H0: λ = 0.8, H1: λ > 0.8, where λ is the mean number of late boats per day
    2. (b) X ~ Po(8)
    3. (c) p = 0.0638; Do not reject H0: there is insufficient evidence, at the 5% significance level, that the mean number of late boats per day has increased
    4. (d) X ≥ 14
    5. (e) 0.0342
    1. (a) H0: λ = 0.4, H1: λ > 0.4, where λ is the mean number of faults per kilometre
    2. (b) Y ~ Po(2)
    3. (c) Y ≥ 6
    4. (d) 0.0166
    5. (e) 0.7029
    6. (f) P(Z = 0) = 0.3012 > 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 0.3012
    1. (a) H0: ρ = 0, H1: ρ ≠ 0, where ρ is the population correlation coefficient between height and reaction time
    2. (b) r = −0.0404, p = 0.924
    3. (c) Do not reject H0: there is insufficient evidence, at the 5% significance level, that there is correlation between height and reaction time
    4. (d) both variables are normally distributed (the data come from a bivariate normal population)
    1. (a) 0.106
    2. (b) X1 + X2 ~ N(240, 128); 0.188
    3. (c) H0: μ = 120, H1: μ < 120
    4. (d) x̄ < 117.8
    5. (e) mean 116.4, standard deviation 7.76
    6. (f) 0.138

jmaths.xyz · AI HL · 13.4.22 Unit review