1. (a) The population consists of all the times the process could be run. It is conceptual.
(b) The population consist of all the registered voters in the state. It is tangible.
(c) The population consist of all people with high cholesterol levels. It is tangible.
(d) The population consist of all concrete specimens that could be made from the new formulation. It is conceptual.
(e) The population consist of all bolts manufactured that day. It is tangible.
3. (a) False
(b) True
5. (a) No. What is important is the population proportion of defectives; the sample proportion is only an approx-
imation. The population proportion for the new process may in fact be greater or less than that of the old
process.
(b) No. The population proportion for the new process may be 0.12 or more, even though the sample proportion
was only 0.11.
(c) Finding 2 defective circuits in the sample.
7. A good knowledge of the process that generated the data.
9. (a) A controlled experiment
(b) It is well-justified, because it is based on a controlled experiment rather than an observational study.
, 2 CHAPTER 1
Section 1.2
1. False
3. No. In the sample 1, 2, 4 the mean is 7/3, which does not appear at all.
5. The sample size can be any odd number.
7. Yes. If all the numbers in the list are the same, the standard deviation will equal 0.
9. The mean and standard deviation both increase by 5%.
11. The total height of the 20 men is 20 × 178 = 3560. The total height of the 30 women is 30 × 164 = 4920.
The total height of all 50 people is 3560 + 4920 = 8480. There are 20 + 30 = 50 people in total. Therefore the
mean height for both groups put together is 8480/50 = 169.6 cm.
13. (a) All would be divided by 2.54.
(b) Not exactly the same, because the measurements would be a little different the second time.
15. (a) The sample size is n = 16. The tertiles have cutpoints (1/3)(17) = 5.67 and (2/3)(17) = 11.33. The first tertile
is therefore the average of the sample values in positions 5 and 6, which is (44 + 46)/2 = 45. The second tertile
is the average of the sample values in positions 11 and 12, which is (76 + 79)/2 = 77.5.
(b) The sample size is n = 16. The quintiles have cutpoints (i/5)(17) for i = 1, 2, 3, 4. The quintiles are therefore
the averages of the sample values in positions 3 and 4, in positions 6 and 7, in positions 10 and 11, and in
positions 13 and 14. The quintiles are therefore (23 + 41)/2 = 32, (46 + 49)/2 = 47.5, (74 + 76)/2 = 75, and
(82 + 89)/2 = 85.5.
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