Critical regions for a test of a mean Problem set B One question at a time

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

1. (a) H0: μ = 18.5; H1: μ < 18.5; 1% significance level; n = 64; population standard deviation σ = 2.4.

   • H0: μ = 18.5, H1: μ < 18.5
   • Under H0, \(\bar X \sim N(18.5, 2.4^2/64)\), so the standard deviation is \(2.4/\sqrt{64} = 0.300\).
   • From the calculator: inverse normal with area 0.01, mean 18.5, standard deviation 0.300: c = 17.8.
   • Critical region: \(\bar X < 17.8\)

2. (b) H0: μ = 0.62; H1: μ < 0.62; 10% significance level; n = 25; population standard deviation σ = 0.08.

   • H0: μ = 0.62, H1: μ < 0.62
   • Under H0, \(\bar X \sim N(0.62, 0.08^2/25)\), so the standard deviation is \(0.08/\sqrt{25} = 0.0160\).
   • From the calculator: inverse normal with area 0.1, mean 0.62, standard deviation 0.0160: c = 0.599.
   • Critical region: \(\bar X < 0.599\)

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

1. (a) H0: μ = 72; H1: μ > 72; 1% significance level; n = 36; population standard deviation σ = 9.

   • H0: μ = 72, H1: μ > 72
   • Under H0, \(\bar X \sim N(72, 9^2/36)\), so the standard deviation is \(9/\sqrt{36} = 1.50\).
   • From the calculator: inverse normal with area 0.99, mean 72, standard deviation 1.50: c = 75.5.
   • Critical region: \(\bar X > 75.5\)

2. (b) H0: μ = 4.2; H1: μ > 4.2; 5% significance level; n = 30; population standard deviation σ = 0.5.

   • H0: μ = 4.2, H1: μ > 4.2
   • Under H0, \(\bar X \sim N(4.2, 0.5^2/30)\), so the standard deviation is \(0.5/\sqrt{30} = 0.0913\).
   • From the calculator: inverse normal with area 0.95, mean 4.2, standard deviation 0.0913: c = 4.35.
   • Critical region: \(\bar X > 4.35\)

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

1. (a) H0: μ = 18; H1: μ ≠ 18; 10% significance level; n = 25; population standard deviation σ = 2.5.

   • H0: μ = 18, H1: μ ≠ 18
   • Under H0, \(\bar X \sim N(18, 2.5^2/25)\), so the standard deviation is \(2.5/\sqrt{25} = 0.500\).
   • Each tail has probability 0.05. From the calculator: inverse normal, mean 18, standard deviation 0.500: c1 = 17.2, c2 = 18.8.
   • Critical region: \(\bar X < 17.2\) and \(\bar X > 18.8\)

2. (b) H0: μ = 0.95; H1: μ ≠ 0.95; 2% significance level; n = 45; population standard deviation σ = 0.06.

   • H0: μ = 0.95, H1: μ ≠ 0.95
   • Under H0, \(\bar X \sim N(0.95, 0.06^2/45)\), so the standard deviation is \(0.06/\sqrt{45} = 0.00894\).
   • Each tail has probability 0.01. From the calculator: inverse normal, mean 0.95, standard deviation 0.00894: c1 = 0.929, c2 = 0.971.
   • Critical region: \(\bar X < 0.929\) and \(\bar X > 0.971\)

The journey time of a train between two cities is normally distributed with mean 54 minutes and standard deviation 5 minutes. After a timetable change, a rail company takes a random sample of 36 journeys to test, at the 5% significance level, whether the mean journey time has decreased.

1. (a) State the hypotheses.

   • μ is the population mean journey time in minutes.
   • H0: μ = 54
   • H1: μ < 54 (a decrease, so one tail, the lower)

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

   • Under H0, \(\bar X \sim N(54, 5^2/36)\), so the standard deviation is \(5/\sqrt{36} = 0.833\).
   • From the calculator: inverse normal with area 0.05, mean 54, standard deviation 0.833: c = 52.6.
   • Critical region: \(\bar X < 52.6\)

3. (c) The sample mean journey time is 52.9 minutes. State the conclusion of the test.

   • 52.9 > 52.6, so the sample mean is not in the critical region: do not reject H0.

The mass of a tomato from a grower is normally distributed with mean 82 g and standard deviation 9 g. After a change of feed, the mean mass of a random sample of 40 tomatoes is 84.1 g. The grower tests, at the 2% significance level, whether the mean mass has increased.

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

   • H0: μ = 82, H1: μ > 82
   • Under H0, \(\bar X \sim N(82, 9^2/40)\), so the standard deviation is \(9/\sqrt{40} = 1.42\).
   • From the calculator: inverse normal with area 0.98, mean 82, standard deviation 1.42: c = 84.9.
   • Critical region: \(\bar X > 84.9\)

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

   • 84.1 is not in the critical region (the boundary is 84.9): do not reject H0.

Rolls of fabric are cut to a mean length of 25.0 m. The lengths are normally distributed with standard deviation 1.5 m. A sample of 25 rolls from a new cutting machine has mean length 25.6 m. Test, at the 1% significance level, whether the mean length has changed.

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

   • H0: μ = 25.0, H1: μ ≠ 25.0
   • Under H0, \(\bar X \sim N(25.0, 1.5^2/25)\), so the standard deviation is \(1.5/\sqrt{25} = 0.300\).
   • Each tail has probability 0.005. From the calculator: inverse normal, mean 25.0, standard deviation 0.300: c1 = 24.2, c2 = 25.8.
   • Critical region: \(\bar X < 24.2\) and \(\bar X > 25.8\)

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

   • 25.6 is not in the critical region (the boundary is 24.2 and 25.8): do not reject H0.

A school says that the mean score on a reading test is 55, with standard deviation 10. An inspector tests H0: μ = 55 against H1: μ < 55 at the 1% significance level.

1. (a) Find the critical value when the sample has 25 students.

   • Standard deviation \(10/\sqrt{25} = 2.00\)
   • Inverse normal, area 0.01, mean 55, standard deviation 2.00: c = 50.3

2. (b) Find the critical value when the sample has 100 students.

   • Standard deviation \(10/\sqrt{100} = 1.00\)
   • Inverse normal, area 0.01, mean 55, standard deviation 1.00: c = 52.7

3. (c) Explain why the critical value in part (b) is closer to 55.

   • 50.3 and 52.7 are both below 55, and 52.7 is nearer to it.
   • With more observations the sample mean varies less, so values close to 55 are already unusual under H0.

A test of H0: μ = 50 against H1: μ < 50 uses a sample of 81 observations from a normal population with standard deviation 9. The critical region is \(\bar X < 48.5\). Find the significance level of the test.

• Under H0, \(\bar X \sim N(50, 9^2/81)\), standard deviation \(9/\sqrt{81} = 1.00\).
• Significance level = P(\(\bar X < 48.5\)) under H0.
• From the calculator: normal cumulative distribution, lower bound −∞ (a very small number), upper bound 48.5, mean 50, standard deviation 1.00: 0.0668.