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

\(15.5 < \mu < 18.1\) cm

\(49.0 < \mu < 53.0\) s

\(7.17 < \mu < 8.15\) m

\(69.6 < \mu < 73.2\)

\(s_{n-1}=15.23\); \(204 < \mu < 220\) g

1. (a) 25.8
2. (b) 2.4
3. (c) \(s_{n-1}=6.21\)

1. (a) 0.09295
2. (b) \(1.47 < \mu < 1.57\) m
3. (c) 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) Every value in the interval is above 1.45 m, so the interval supports the inspector's claim.

1. (a) z-interval: the population standard deviation is known and the population is normal.
2. (b) t-interval: the population variance is unknown, so it is estimated by \(s_{n-1}\).

1. (a) \(8.10 < \mu < 9.10\) hours
2. (b) Valid: n = 90 is large, so by the central limit theorem the sample mean is approximately normal even though the population is not.

1. (a) \(50.3 < \mu < 54.5\) kg
2. (b) \(48.2 < \mu < 56.6\) kg
3. (c) The 99% interval is wider (8.37 against 4.17): higher confidence needs a wider interval.

1. (a) \(23.4 < \mu < 29.3\) minutes
2. (b) Each time is replaced by the midpoint of its class, so the mean and standard deviation are estimates.

1. (a) Differences (after − before): 3, −2, 4, 4, 4, 6, 0, 5
2. (b) \(1.21 < \mu_d < 4.79\) kg
3. (c) The whole interval is above 0, so the mean lift increases: the programme seems to work.

1. (a) \(97.1<\mu<106.9\)
2. (b) \(99.6<\mu<104.4\)
3. (c) Widths 9.80 and 4.90: the width halves, because the sample is four times as large and the width is proportional to \(1/\sqrt n\).

\(n=78\)