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

\(29.5 < \mu < 33.3\) minutes

\(209 < \mu < 223\) g

\(130 < \mu < 141\) cm

\(40.1 < \mu < 42.3\)

\(s_{n-1}=5.500\); \(83.2 < \mu < 89.6\) ml

1. (a) 189
2. (b) 8.4
3. (c) \(s_{n-1}=13.2\)

1. (a) 0.1567
2. (b) \(1.94 < \mu < 2.14\) kg
3. (c) 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) Every value in the interval is above 1.90 kg, so the interval supports the customer's claim.

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

1. (a) \(13.6 < \mu < 15.0\) km
2. (b) Valid: n = 110 is large, so by the central limit theorem the sample mean is approximately normal even though the population is not.

1. (a) \(1.73 < \mu < 1.91\) m
2. (b) \(1.70 < \mu < 1.94\) m
3. (c) The 98% interval is wider (0.250 against 0.177): higher confidence needs a wider interval.

1. (a) \(82.5 < \mu < 88.7\) g
2. (b) Each mass is replaced by the midpoint of its class, so the mean and standard deviation are estimates.

1. (a) Differences (after − before): 3, 1, 4, 2, 0, 5, 3, 1, 4
2. (b) \(1.27 < \mu_d < 3.84\) points
3. (c) The whole interval is above 0, so the mean score increases: the practice seems to help.

1. (a) \(12.1<\mu<12.7\)
2. (b) \(12.3<\mu<12.5\)
3. (c) Widths 0.526 and 0.263: the width halves, because the sample is four times as large and the width is proportional to \(1/\sqrt n\).

\(n=93\)