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IB Mathematics: Applications and Interpretation HL · 13.4: Hypothesis Testing · 13.4.20

Type II errors Problem set B One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Skills

  1. Find the probability of a Type II error for each test of a normal mean. Give answers to 3 significant figures.

    1. (a) \(X \sim N(\mu, 1.8^2)\), sample size 12. H0: μ = 40, H1: μ < 40. Critical region \(\bar X < 38.9\). True mean μ = 38.0.

      • True distribution: \(\bar X \sim N(38.0, 1.8^2/12)\), standard deviation \(1.8/\sqrt{12} = 0.520\).
      • Reject H0 if \(\bar X < 38.9\). A Type II error is \(\bar X > 38.9\).
      • From the calculator: normal cdf, lower bound 38.9, upper bound 1E99, mean 38.0, standard deviation 0.520: 0.0416
      • P(Type II error) = 0.0416

    2. (b) \(X \sim N(\mu, 0.35^2)\), sample size 10. H0: μ = 8, H1: μ ≠ 8. Critical region \(\bar X < 7.8\) or \(\bar X > 8.25\). True mean μ = 7.7.

      • True distribution: \(\bar X \sim N(7.7, 0.35^2/10)\), standard deviation \(0.35/\sqrt{10} = 0.111\).
      • Reject H0 if \(\bar X < 7.80\) or \(\bar X > 8.25\). A Type II error is \(7.80 < \bar X < 8.25\).
      • From the calculator: normal cdf, lower bound 7.80, upper bound 8.25, mean 7.7, standard deviation 0.111: 0.183
      • P(Type II error) = 0.183

  2. Find the probability of a Type II error for each test. Give answers to 3 significant figures.

    1. (a) \(X \sim B(30, p)\). H0: p = 0.5, H1: p > 0.5. Critical region \(X \ge 20\). True p = 0.7.

      • True distribution: X ~ B(30, 0.7).
      • Reject H0 if X ≥ 20. A Type II error is X ≤ 19.
      • From the calculator: binomial cdf, n = 30, p = 0.7, upper bound 19: 0.270
      • P(Type II error) = 0.270

    2. (b) \(X \sim Po(\lambda)\). H0: λ = 5, H1: λ < 5. Critical region \(X \le 1\). True λ = 2.6.

      • True distribution: X ~ Po(2.6).
      • Reject H0 if X ≤ 1. A Type II error is X ≥ 2.
      • P(X ≥ 2) = 1 − P(X ≤ 1)
      • From the calculator: Poisson cdf, mean 2.6, upper bound 1: 0.267, so 1 − 0.267 = 0.733
      • P(Type II error) = 0.733

  3. Find the probability of a Type I error in each test. Give answers to 3 significant figures.

    1. (a) \(X \sim N(\mu, 0.9^2)\), sample size 6. H0: μ = 15, H1: μ ≠ 15. Critical region \(\bar X < 14.0\) or \(\bar X > 16.3\).

      • Under H0, \(\bar X \sim N(15, 0.9^2/6)\), standard deviation 0.367.
      • P(Type I error) = P(\(\bar X < 14.0\)) + P(\(\bar X > 16.3\))
      • From the calculator: normal cdf, mean 15, standard deviation 0.367: P(\(\bar X < 14.0\)) = 0.00325; P(\(\bar X > 16.3\)) = 0.000201
      • P(Type I error) = 0.00345

    2. (b) \(X \sim B(25, p)\). H0: p = 0.4, H1: p < 0.4. Critical region \(X \le 5\).

      • Under H0, X ~ B(25, 0.4).
      • P(Type I error) = P(X ≤ 5)
      • From the calculator: binomial cdf, n = 25, p = 0.4, upper bound 5: 0.0294
      • P(Type I error) = 0.0294

    3. (c) The number of events in one day is Poisson with mean 0.8 under H0. The test counts events over 5 days, and the critical region is \(X \ge 7\).

      • Under H0, X ~ Po(4).
      • P(Type I error) = P(X ≥ 7) = 1 − P(X ≤ 6)
      • From the calculator: Poisson cdf, mean 4, upper bound 6: 0.889, so 1 − 0.889 = 0.111
      • P(Type I error) = 0.111

  4. Answer the following.

    1. (a) A school says that 65% of its students walk to school. The head thinks the proportion has fallen since a new bus route opened and will test this with 40 students. State suitable hypotheses, defining p.

      • The head thinks the proportion has fallen, so the test is lower tailed.
      • H0: p = 0.65
      • H1: p < 0.65, where p is the proportion of students who walk to school.

    2. (b) An office receives a mean of 14 emails an hour. A new filter is meant to reduce the hourly mean. State suitable hypotheses, defining λ.

      • A Poisson mean, tested for a decrease.
      • H0: λ = 14
      • H1: λ < 14, where λ is the mean number of emails an hour.

    3. (c) In the walking test, describe a Type I error and a Type II error in context.

      • Type I error: H0 is true but is rejected. The head decides the proportion has fallen when it is still 0.65.
      • Type II error: H0 is false but is not rejected. The head decides there is no evidence of a fall when the proportion has in fact fallen.

In context

  1. A dairy sells yoghurt pots whose mass is normally distributed with standard deviation 12 g. The mean mass is meant to be 250 g. A trading standards officer thinks the mean mass is too low. She takes a sample of 6 pots and tests H0: μ = 250 against H1: μ < 250 at the 5% significance level.

    2202302402502600.020.040.060.08sample mean (g)yH0 true: mean 250True mean 238241.94

    1. (a) Find the critical region. Give the critical value to 3 significant figures.

      • Under H0, \(\bar X \sim N(250, 12.0^2/6)\), standard deviation 4.90.
      • From the calculator: inverse normal, area 0.05, mean 250, standard deviation 4.90: 242
      • Critical region: \(\bar X < 242\)

    2. (b) The true mean is 238.0 g. Find the probability of a Type II error. Use the unrounded critical value.

      • True distribution: \(\bar X \sim N(238.0, 12.0^2/6)\).
      • Type II error: \(\bar X > 241.94\).
      • From the calculator: normal cdf, lower bound 241.94 (the unrounded critical value), upper bound 1E99, mean 238.0, standard deviation 4.90: 0.211
      • P(Type II error) = 0.211

  2. A carpenter cuts planks whose length is normally distributed with standard deviation 4.0 cm. An inspector tests H0: μ = 120.0 against H1: μ ≠ 120.0 using the mean of 9 planks. He rejects H0 if the sample mean is below 117.4 cm or above 122.6 cm.

    1. (a) The true mean length is 123.0 cm. Find the probability of a Type II error.

      • True distribution: \(\bar X \sim N(123.0, 4.0^2/9)\), standard deviation \(4.0/\sqrt{9} = 1.33\).
      • Reject H0 if \(\bar X < 117.4\) or \(\bar X > 122.6\). A Type II error is \(117.4 < \bar X < 122.6\).
      • From the calculator: normal cdf, lower bound 117.4, upper bound 122.6, mean 123.0, standard deviation 1.33: 0.382
      • P(Type II error) = 0.382

    2. (b) The inspector now uses 36 planks and keeps the same significance level. Explain the effect on the probability of a Type I error and on the probability of a Type II error.

      • The significance level is fixed, so P(Type I error) does not change.
      • A larger sample makes the sample mean less spread out, so the two curves overlap less and P(Type II error) decreases.

  3. A bank uses software to decide whether a card payment is fraudulent. For each payment the software chooses between H0: the payment is genuine, and H1: the payment is fraudulent.

    1. (a) Describe a Type I error and a Type II error in this context.

      • Type I: H0 true but rejected: a genuine payment is flagged as fraudulent.
      • Type II: H0 false but not rejected: a fraudulent payment is accepted as genuine.

    2. (b) Which error is more important to avoid? Give a reason.

      • A Type II error means a thief gets money, which is costly and hard to undo.
      • A Type I error blocks a genuine customer, who can usually be put right with a call. A bank can accept more Type I errors to keep Type II small.

  4. A football club says that 60% of its season ticket holders attend each home game. A supporter thinks the figure is too high and counts the attendance of 16 season ticket holders at one game. She rejects the claim if 5 or fewer attend, using H0: p = 0.6 against H1: p < 0.6.

    1. (a) Find the probability that the supporter makes a Type I error.

      • Under H0, X ~ B(16, 0.6).
      • P(Type I error) = P(X ≤ 5)
      • From the calculator: binomial cdf, n = 16, p = 0.6, upper bound 5: 0.0191
      • P(Type I error) = 0.0191

    2. (b) The true probability that a season ticket holder attends is 0.35. Find the probability that the supporter makes a Type II error.

      • True distribution: X ~ B(16, 0.35).
      • Reject H0 if X ≤ 5. A Type II error is X ≥ 6.
      • P(X ≥ 6) = 1 − P(X ≤ 5)
      • From the calculator: binomial cdf, n = 16, p = 0.35, upper bound 5: 0.490, so 1 − 0.490 = 0.510
      • P(Type II error) = 0.510

Stretch

  1. A pizza shop says that 20% of its orders include a dessert. An inspector thinks the proportion is higher and counts the orders with a dessert among 30 orders. The inspector rejects the claim for a critical region \(X \ge c\) with probability of a Type I error less than 0.04.

    1. (a) State H0 and H1 and find the critical region that gives the smallest probability of a Type II error.

      • H0: p = 0.2, H1: p > 0.2
      • Under H0, X ~ B(30, 0.2); critical region X ≥ c, with P(X ≥ c) < 0.04.
      • From the calculator: binomial cdf: P(X ≥ 10) = 0.0611; P(X ≥ 11) = 0.0256
      • The smallest c with probability below 0.04 is 11, which gives the smallest Type II error. Critical region: X ≥ 11

    2. (b) Given that the true probability that an order includes a dessert is 0.4, find the smallest probability of a Type II error.

      • True distribution: X ~ B(30, 0.4).
      • Reject H0 if X ≥ 11. A Type II error is X ≤ 10.
      • From the calculator: binomial cdf, n = 30, p = 0.4, upper bound 10: 0.291
      • P(Type II error) = 0.291

  2. The number of customers entering a kiosk in a ten-minute period is Poisson with mean 11.0. After road works start the owner thinks the mean has decreased. She counts the customers in one ten-minute period and rejects H0 if the count is in a critical region \(X \le c\). The true mean is now 4.0. She wants the probability of a Type II error to be less than 0.2.

    1. (a) State the hypotheses and find the critical region that gives the smallest probability of a Type I error.

      • H0: λ = 11, H1: λ < 11
      • Critical region X ≤ c. Under the true mean, X ~ Po(4): Type II error is X ≥ c + 1, so P(X ≥ c + 1) = 1 − P(X ≤ c) < 0.2, that is P(X ≤ c) > 0.8.
      • From the calculator: Poisson cdf: P(X ≤ 5) = 0.785; P(X ≤ 6) = 0.889
      • The smallest c with P(X ≤ c) > 0.8 is 6, which gives the smallest Type I error. Critical region: X ≤ 6

    2. (b) Find that smallest probability of a Type I error.

      • Under H0, X ~ Po(11).
      • P(Type I error) = P(X ≤ 6)
      • From the calculator: Poisson cdf, mean 11, upper bound 6: 0.0786
      • P(Type I error) = 0.0786

  3. A tray holds 10 cupcakes, r of them chocolate and the rest vanilla. Nadia knows that r is either 3 or 7. She takes 4 cupcakes without looking and tests H0: r = 3 against H1: r = 7. She rejects H0 if at least 3 of the cupcakes are chocolate.

    1. (a) Find the probability that she makes a Type I error.

      • Type I error: reject H0 when r = 3, so X ≥ 3.
      • P(X ≥ 3) = (C(3,3) × C(7,1)) ÷ C(10,4) = 7/210
      • P(Type I error) = 0.0333

    2. (b) Find the probability that she makes a Type II error.

      • Type II error: do not reject H0 when r = 7, so X ≤ 2, which is 1 − P(X ≥ 3).
      • P(X ≥ 3) = (C(7,3) × C(3,1) + C(7,4) × C(3,0)) ÷ C(10,4) = 140/210
      • P(Type II error) = 1 − 140/210 = 0.333

jmaths.xyz · AI HL · 13.4.20 Type II errors and how they link to Type I