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IB Mathematics: Applications and Interpretation HL · AHL 8: Statistics 2 · AHL 8.12

z-intervals, t-intervals and choosing between them Problem set A Worked solutions

Skills

  1. The lengths of a type of fish are normally distributed with \(\sigma=3.1\) cm. A random sample of 22 fish has mean length 16.8 cm. Find a 95% confidence interval for the mean length.

    • From the calculator: z Interval, Stats, \(\sigma=3.1\), \(\bar x=16.8\), \(n=22\), C Level 0.95
    • \(15.5 < \mu < 18.1\) cm

  2. The times (seconds) of eight runners are 48, 52, 45, 57, 50, 54, 49, 53. The times are normal with known standard deviation 3.5 s. Find a 90% confidence interval for the mean time.

    • From the calculator: z Interval, Data, the list, \(\sigma=3.5\), C Level 0.9
    • \(49.0 < \mu < 53.0\) s

  3. The heights, in metres, of eight trees are 7.3, 8.1, 6.8, 7.9, 8.4, 7.0, 7.6, 8.2. They are normally distributed with unknown variance. Find a 95% confidence interval for the mean height.

    • The variance is unknown: t-interval
    • From the calculator: t Interval, Data, the list, C Level 0.95
    • \(7.17 < \mu < 8.15\) m

  4. Test scores are normally distributed. A random sample of 20 scores has mean 71.4 and unbiased estimate of the population variance 22.3. Find a 90% confidence interval for the mean score.

    • \(s_{n-1}=\sqrt{22.3}=4.722\)
    • From the calculator: t Interval, Stats, \(\bar x=71.4\), \(s_{n-1}=4.722\), \(n=20\), C Level 0.9
    • \(69.6 < \mu < 73.2\)

  5. A random sample of 18 jars has mean mass 212 g and standard deviation 14.8 g, calculated by dividing by n. Assume the masses are normal. Find \(s_{n-1}\), then a 95% confidence interval for the mean mass.

    • \(s_{n-1}=\sqrt{\dfrac{18}{17}}\times14.8=15.23\)
    • From the calculator: t Interval, Stats, \(\bar x=212\), \(s_{n-1}=15.23\), \(n=18\), C Level 0.95
    • \(204 < \mu < 220\) g

  6. A 90% confidence interval for the mean daily number of visitors to a museum is \(23.4<\mu<28.2\), in hundreds. It was found from a random sample of 20 days from a normal population, using \(s_{n-1}\).

    1. (a) Write down the sample mean.

      • \(\bar x=\dfrac{23.4+28.2}{2}=25.8\)

    2. (b) Write down the margin of error.

      • \(\dfrac{28.2-23.4}{2}=2.4\)

    3. (c) Find \(s_{n-1}\).

      • df = 19, 90%: \(t=1.729\)
      • \(2.4=1.729\times\dfrac{s_{n-1}}{\sqrt{20}}\)
      • \(s_{n-1}=6.21\)

In context

  1. A factory makes rods that should be 1.50 m long. A random sample of 16 rods has mean 1.52 m and standard deviation 0.09 m, calculated by dividing by n. Assume the lengths are normal.

    1. (a) Find \(s_{n-1}\).

      • \(s_{n-1}=\sqrt{\dfrac{16}{15}}\times0.09=0.09295\)

    2. (b) Find a 95% confidence interval for the mean length.

      • From the calculator: t Interval, Stats, \(\bar x=1.52\), \(s_{n-1}=0.09295\), \(n=16\), C Level 0.95
      • \(1.47 < \mu < 1.57\) m

    3. (c) The factory will adjust the machine if the mean length is not 1.50 m. Does the interval show that it needs adjusting?

      • 1.50 is inside the interval, so the interval does not show the mean is different from 1.50 m: no adjustment is needed.

    4. (d) An inspector says the rods are longer than 1.45 m on average. Does the interval support this?

      • Every value in the interval is above 1.45 m, so the interval supports the inspector's claim.

  2. State whether to use a z-interval or a t-interval for the mean in each case, and justify your choice.

    1. (a) Heights of players are normal with a known standard deviation of 7 cm. A sample of 15 players is taken.

      • z-interval: the population standard deviation is known and the population is normal.

    2. (b) Times are normal, but the standard deviation is not known. A sample of 15 times has \(s_{n-1}=4.2\) s.

      • t-interval: the population variance is unknown, so it is estimated by \(s_{n-1}\).

  3. The times spent on homework in a school have standard deviation 2.4 hours, but the distribution is skewed. A random sample of 90 students has mean 8.6 hours.

    1. (a) Find a 95% confidence interval for the mean time.

      • From the calculator: z Interval, Stats, \(\sigma=2.4\), \(\bar x=8.6\), \(n=90\), C Level 0.95
      • \(8.10 < \mu < 9.10\) hours

    2. (b) State, with a reason, whether the interval is valid when the population is not known to be normal.

      • Valid: n = 90 is large, so by the central limit theorem the sample mean is approximately normal even though the population is not.

  4. The masses of sacks are normal with \(\sigma=6.5\) kg. A random sample of 16 sacks has mean mass 52.4 kg.

    1. (a) Find a 80% confidence interval.

      • From the calculator: z Interval, Stats, \(\sigma=6.5\), \(\bar x=52.4\), \(n=16\), C Level 0.8
      • \(50.3 < \mu < 54.5\) kg

    2. (b) Find a 99% confidence interval.

      • From the calculator: z Interval, Stats, \(\sigma=6.5\), \(\bar x=52.4\), \(n=16\), C Level 0.99
      • \(48.2 < \mu < 56.6\) kg

    3. (c) Which is wider? Explain why.

      • Widths: 4.17 and 8.37. More confidence means a wider interval.

  5. The times, in minutes, that 60 commuters spend travelling are in the table. Assume the times are normal.

    Time x (minutes)0 ≤ x < 1010 ≤ x < 2020 ≤ x < 3030 ≤ x < 4040 ≤ x < 50Frequency51221148

    1. (a) Find an approximate 95% confidence interval for the mean.

      • Midpoints 5, 15, 25, 35, 45: from the calculator, 1-Variable Stats with the frequencies: \(\bar x=26.33\), \(s_{n-1}=11.42\)
      • t Interval, Stats, C Level 0.95: \(23.4 < \mu < 29.3\) minutes

    2. (b) Explain why the interval is approximate.

      • Each time is replaced by the midpoint of its class, so the mean and standard deviation are estimates.

  6. Eight athletes' maximum lift (kg) was measured before and after a training programme. Before: 68, 72, 65, 80, 75, 70, 66, 78. After: 71, 70, 69, 84, 79, 76, 66, 83. Assume the differences are normally distributed.

    1. (a) Find the difference for each athlete (after − before).

      • after − before: 3, −2, 4, 4, 4, 6, 0, 5

    2. (b) Find a 90% confidence interval for the mean difference.

      • From the calculator: t Interval, Data, the list of differences, C Level 0.9
      • \(1.21 < \mu_d < 4.79\) kg

    3. (c) Does the interval suggest that the programme increases the lift? Explain.

      • The whole interval is above 0, so the mean lift increases: the programme seems to work.

  7. The scores on a reaction test are normally distributed with \(\sigma=15\). A random sample of 36 people has mean score 102.

    1. (a) Find a 95% confidence interval for the mean score.

      • From the calculator: z Interval, Stats, \(\sigma=15\), \(\bar x=102\), \(n=36\), C Level 0.95
      • \(97.1<\mu<106.9\)

    2. (b) A second random sample of 144 people also has mean score 102. Find the 95% confidence interval from this sample.

      • From the calculator: z Interval, Stats, \(\sigma=15\), \(\bar x=102\), \(n=144\), C Level 0.95
      • \(99.6<\mu<104.4\)

    3. (c) Compare the two widths and explain the change.

      • Widths: 9.80 and 4.90, a ratio of 2.
      • \(\sqrt{144/36}=2\)

Stretch

  1. A population is normal with \(\sigma=4.5\). Find the smallest sample size for which the 95% z-interval for the mean has width at most 2.

    • Width \(=2\times z\times\dfrac{\sigma}{\sqrt n}\), with \(z=1.960\)
    • \(\sqrt n\ge\dfrac{2\times1.960\times4.5}{2}=8.820\)
    • \(n\ge77.79\), so \(n=78\)

jmaths.xyz · AI HL · AHL 8.12 z-intervals, t-intervals and choosing between them