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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 B Worked solutions

Skills

  1. The journey times on a route are normally distributed with \(\sigma=5.2\) minutes. A random sample of 28 journeys has mean 31.4 minutes. Find a 95% confidence interval for the mean journey time.

    • From the calculator: z Interval, Stats, \(\sigma=5.2\), \(\bar x=31.4\), \(n=28\), C Level 0.95
    • \(29.5 < \mu < 33.3\) minutes

  2. The masses (g) of seven loaves are 210, 224, 198, 231, 217, 205, 226. The masses are normal with known standard deviation 11 g. Find a 90% confidence interval for the mean mass.

    • From the calculator: z Interval, Data, the list, \(\sigma=11\), C Level 0.9
    • \(209 < \mu < 223\) g

  3. The heights, in centimetres, of seven young plants are 132, 141, 127, 138, 145, 130, 136. 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
    • \(130 < \mu < 141\) cm

  4. Reaction times (in hundredths of a second) are normal. A random sample of 25 has mean 41.2 and unbiased estimate of the population variance 9.7. Find a 90% confidence interval for the mean.

    • \(s_{n-1}=\sqrt{9.7}=3.114\)
    • From the calculator: t Interval, Stats, \(\bar x=41.2\), \(s_{n-1}=3.114\), \(n=25\), C Level 0.9
    • \(40.1 < \mu < 42.3\)

  5. A random sample of 14 cartons has mean volume 86.4 ml and standard deviation 5.3 ml, calculated by dividing by n. Assume the volumes are normal. Find \(s_{n-1}\), then a 95% confidence interval for the mean volume.

    • \(s_{n-1}=\sqrt{\dfrac{14}{13}}\times5.3=5.500\)
    • From the calculator: t Interval, Stats, \(\bar x=86.4\), \(s_{n-1}=5.500\), \(n=14\), C Level 0.95
    • \(83.2 < \mu < 89.6\) ml

  6. A 95% confidence interval for the mean time, in minutes, that a customer waits at a bank is \(180.6<\mu<197.4\). It was found from a random sample of 12 customers from a normal population, using \(s_{n-1}\).

    1. (a) Write down the sample mean.

      • \(\bar x=\dfrac{180.6+197.4}{2}=189\)

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

      • \(\dfrac{197.4-180.6}{2}=8.4\)

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

      • df = 11, 95%: \(t=2.201\)
      • \(8.4=2.201\times\dfrac{s_{n-1}}{\sqrt{12}}\)
      • \(s_{n-1}=13.2\)

In context

  1. A bakery's loaves should have mass 2.00 kg. A random sample of 12 loaves has mean 2.04 kg and standard deviation 0.15 kg, calculated by dividing by n. Assume the masses are normal.

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

      • \(s_{n-1}=\sqrt{\dfrac{12}{11}}\times0.15=0.1567\)

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

      • From the calculator: t Interval, Stats, \(\bar x=2.04\), \(s_{n-1}=0.1567\), \(n=12\), C Level 0.95
      • \(1.94 < \mu < 2.14\) kg

    3. (c) The baker will change the recipe if the mean mass is not 2.00 kg. Does the interval show that it needs changing?

      • 2.00 is inside the interval, so the interval does not show the mean is different from 2.00 kg: no change is needed.

    4. (d) A customer says the loaves are heavier than 1.90 kg on average. Does the interval support this?

      • Every value in the interval is above 1.90 kg, so the interval supports the customer'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) Reaction times are normal and the standard deviation is not known. A sample of 12 times has \(s_{n-1}=0.06\) s.

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

    2. (b) Masses of bags are normal with a known standard deviation of 4 g. A sample of 12 bags is taken.

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

  3. The distances driven by delivery drivers in a day have standard deviation 3.9 km, but the distribution is skewed. A random sample of 110 drivers has mean 14.3 km.

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

      • From the calculator: z Interval, Stats, \(\sigma=3.9\), \(\bar x=14.3\), \(n=110\), C Level 0.95
      • \(13.6 < \mu < 15.0\) km

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

      • Valid: n = 110 is large, so by the central limit theorem the sample mean is approximately normal even though the population is not.

  4. The heights of seedlings are normal with \(\sigma=0.24\) m. A random sample of 20 seedlings has mean height 1.82 m.

    1. (a) Find a 90% confidence interval.

      • From the calculator: z Interval, Stats, \(\sigma=0.24\), \(\bar x=1.82\), \(n=20\), C Level 0.9
      • \(1.73 < \mu < 1.91\) m

    2. (b) Find a 98% confidence interval.

      • From the calculator: z Interval, Stats, \(\sigma=0.24\), \(\bar x=1.82\), \(n=20\), C Level 0.98
      • \(1.70 < \mu < 1.94\) m

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

      • Widths: 0.177 and 0.250. More confidence means a wider interval.

  5. The masses, in grams, of 50 eggs are in the table. Assume the masses are normal.

    Mass x (g)60 ≤ x < 7070 ≤ x < 8080 ≤ x < 9090 ≤ x < 100100 ≤ x < 110Frequency41118125

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

      • Midpoints 65, 75, 85, 95, 105: from the calculator, 1-Variable Stats with the frequencies: \(\bar x=85.60\), \(s_{n-1}=10.96\)
      • t Interval, Stats, C Level 0.95: \(82.5 < \mu < 88.7\) g

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

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

  6. Nine players' scores in a quiz were recorded before and after a practice session. Before: 14, 18, 11, 20, 16, 13, 17, 12, 15. After: 17, 19, 15, 22, 16, 18, 20, 13, 19. Assume the differences are normally distributed.

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

      • after − before: 3, 1, 4, 2, 0, 5, 3, 1, 4

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

      • From the calculator: t Interval, Data, the list of differences, C Level 0.95
      • \(1.27 < \mu_d < 3.84\) points

    3. (c) Does the interval suggest that practice improves the score? Explain.

      • The whole interval is above 0, so the mean score increases: the practice seems to help.

  7. The lengths of bolts are normally distributed with \(\sigma=0.6\) cm. A random sample of 20 bolts has mean length 12.4 cm.

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

      • From the calculator: z Interval, Stats, \(\sigma=0.6\), \(\bar x=12.4\), \(n=20\), C Level 0.95
      • \(12.1<\mu<12.7\)

    2. (b) A second random sample of 80 bolts also has mean length 12.4 cm. Find the 95% confidence interval from this sample.

      • From the calculator: z Interval, Stats, \(\sigma=0.6\), \(\bar x=12.4\), \(n=80\), C Level 0.95
      • \(12.3<\mu<12.5\)

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

      • Widths: 0.526 and 0.263, a ratio of 2.
      • \(\sqrt{80/20}=2\)

Stretch

  1. A population is normal with \(\sigma=2.8\). Find the smallest sample size for which the 99% z-interval for the mean has width at most 1.5.

    • Width \(=2\times z\times\dfrac{\sigma}{\sqrt n}\), with \(z=2.576\)
    • \(\sqrt n\ge\dfrac{2\times2.576\times2.8}{1.5}=9.616\)
    • \(n\ge92.48\), so \(n=93\)

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