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IB Mathematics: Applications and Interpretation HL · 13.4: Hypothesis Testing · 13.4.11

Critical regions for a test of a mean Problem set A Worked solutions

Skills

  1. Find the critical region for each test of a mean. Give critical values to 3 significant figures.

    1. (a) H0: μ = 64; H1: μ < 64; 5% significance level; n = 25; population standard deviation σ = 5.

      • H0: μ = 64, H1: μ < 64
      • Under H0, \(\bar X \sim N(64, 5^2/25)\), so the standard deviation is \(5/\sqrt{25} = 1.00\).
      • From the calculator: inverse normal with area 0.05, mean 64, standard deviation 1.00: c = 62.4.
      • Critical region: \(\bar X < 62.4\)

    2. (b) H0: μ = 2.4; H1: μ < 2.4; 1% significance level; n = 36; population standard deviation σ = 0.3.

      • H0: μ = 2.4, H1: μ < 2.4
      • Under H0, \(\bar X \sim N(2.4, 0.3^2/36)\), so the standard deviation is \(0.3/\sqrt{36} = 0.0500\).
      • From the calculator: inverse normal with area 0.01, mean 2.4, standard deviation 0.0500: c = 2.28.
      • Critical region: \(\bar X < 2.28\)

  2. Find the critical region for each test of a mean. Give critical values to 3 significant figures.

    1. (a) H0: μ = 120; H1: μ > 120; 5% significance level; n = 30; population standard deviation σ = 15.

      • H0: μ = 120, H1: μ > 120
      • Under H0, \(\bar X \sim N(120, 15^2/30)\), so the standard deviation is \(15/\sqrt{30} = 2.74\).
      • From the calculator: inverse normal with area 0.95, mean 120, standard deviation 2.74: c = 125.
      • Critical region: \(\bar X > 125\)

    2. (b) H0: μ = 0.85; H1: μ > 0.85; 2% significance level; n = 20; population standard deviation σ = 0.1.

      • H0: μ = 0.85, H1: μ > 0.85
      • Under H0, \(\bar X \sim N(0.85, 0.1^2/20)\), so the standard deviation is \(0.1/\sqrt{20} = 0.0224\).
      • From the calculator: inverse normal with area 0.98, mean 0.85, standard deviation 0.0224: c = 0.896.
      • Critical region: \(\bar X > 0.896\)

  3. Find the critical region for each test of a mean. Give critical values to 3 significant figures.

    1. (a) H0: μ = 35; H1: μ ≠ 35; 5% significance level; n = 16; population standard deviation σ = 4.

      • H0: μ = 35, H1: μ ≠ 35
      • Under H0, \(\bar X \sim N(35, 4^2/16)\), so the standard deviation is \(4/\sqrt{16} = 1.00\).
      • Each tail has probability 0.025. From the calculator: inverse normal, mean 35, standard deviation 1.00: c1 = 33.0, c2 = 37.0.
      • Critical region: \(\bar X < 33.0\) and \(\bar X > 37.0\)

    2. (b) H0: μ = 210; H1: μ ≠ 210; 10% significance level; n = 40; population standard deviation σ = 12.

      • H0: μ = 210, H1: μ ≠ 210
      • Under H0, \(\bar X \sim N(210, 12^2/40)\), so the standard deviation is \(12/\sqrt{40} = 1.90\).
      • Each tail has probability 0.05. From the calculator: inverse normal, mean 210, standard deviation 1.90: c1 = 207, c2 = 213.
      • Critical region: \(\bar X < 207\) and \(\bar X > 213\)

In context

  1. The volume of coffee in a takeaway cup is normally distributed with mean 340 ml and standard deviation 6 ml. After a new dispenser is fitted, a café owner takes a random sample of 25 cups to test, at the 5% significance level, whether the mean volume has decreased.

    1. (a) State the hypotheses.

      • μ is the population mean volume of a cup of coffee in ml.
      • H0: μ = 340
      • H1: μ < 340 (a decrease, so one tail, the lower)

    2. (b) Find the critical region for the sample mean.

      • Under H0, \(\bar X \sim N(340, 6^2/25)\), so the standard deviation is \(6/\sqrt{25} = 1.20\).
      • From the calculator: inverse normal with area 0.05, mean 340, standard deviation 1.20: c = 338.
      • Critical region: \(\bar X < 338\)

    3. (c) The sample mean volume is 337.6 ml. State the conclusion of the test.

      • 337.6 < 338, so the sample mean is in the critical region: reject H0.

  2. A courier's delivery times, in minutes, are normally distributed with mean 32 and standard deviation 7. After a change of route, the mean time for a random sample of 49 deliveries is 34.6 minutes. The courier tests, at the 1% significance level, whether the mean delivery time has increased.

    1. (a) Find the critical region for the sample mean.

      • H0: μ = 32, H1: μ > 32
      • Under H0, \(\bar X \sim N(32, 7^2/49)\), so the standard deviation is \(7/\sqrt{49} = 1.00\).
      • From the calculator: inverse normal with area 0.99, mean 32, standard deviation 1.00: c = 34.3.
      • Critical region: \(\bar X > 34.3\)

    2. (b) Is the sample mean in the critical region? State the conclusion.

      • 34.6 is in the critical region (the boundary is 34.3): reject H0.

  3. Resistors are labelled 470 Ω. Their resistances are normally distributed with standard deviation 9 Ω. A sample of 30 resistors from a new supplier has mean 473.6 Ω. Test, at the 5% significance level, whether the mean resistance has changed.

    1. (a) Find the critical region for the sample mean.

      • H0: μ = 470, H1: μ ≠ 470
      • Under H0, \(\bar X \sim N(470, 9^2/30)\), so the standard deviation is \(9/\sqrt{30} = 1.64\).
      • Each tail has probability 0.025. From the calculator: inverse normal, mean 470, standard deviation 1.64: c1 = 467, c2 = 473.
      • Critical region: \(\bar X < 467\) and \(\bar X > 473\)

    2. (b) State the conclusion of the test.

      • 473.6 is in the critical region (the boundary is 467 and 473): reject H0.

Stretch

  1. A cable maker claims a mean breaking load of 800 N, with standard deviation 40 N. A buyer tests H0: μ = 800 against H1: μ < 800 at the 5% significance level.

    1. (a) Find the critical value when the sample has 16 cables.

      • Standard deviation \(40/\sqrt{16} = 10.0\)
      • Inverse normal, area 0.05, mean 800, standard deviation 10.0: c = 784

    2. (b) Find the critical value when the sample has 64 cables.

      • Standard deviation \(40/\sqrt{64} = 5.00\)
      • Inverse normal, area 0.05, mean 800, standard deviation 5.00: c = 792

    3. (c) Explain why the critical value in part (b) is closer to 800 N.

      • 784 and 792 are both below 800, and 792 is nearer to it.
      • With more observations the sample mean varies less, so values close to 800 are already unusual under H0.

  2. A test of H0: μ = 100 against H1: μ < 100 uses a sample of 36 observations from a normal population with standard deviation 12. The critical region is \(\bar X < 96\). Find the significance level of the test.

    • Under H0, \(\bar X \sim N(100, 12^2/36)\), standard deviation \(12/\sqrt{36} = 2.00\).
    • Significance level = P(\(\bar X < 96\)) under H0.
    • From the calculator: normal cumulative distribution, lower bound −∞ (a very small number), upper bound 96, mean 100, standard deviation 2.00: 0.0228.

jmaths.xyz · AI HL · 13.4.11 Critical values and critical regions for a test of a mean