CHAPTER 6 Confidence Intervals •6.1 (a) y– = 1269; s = 145; n = 8. The standard error of the mean is SEy- = sn = 1458 = 51.3 ng/gm. (b) y– = 1269; s = 145; n = 30. The standard error of the mean is SEy- = sn = 14530 = 26.5 ng/gm. 6.2 (a) 15/ 25 = 3.0 cm (b) 15/ 100 = 1.5 cm 6.3 y– = 9.520; s = 1.429; SE = 1.429/5 = .6391 ≈ .64 gm/kg. 6.4 (a) 3.06/86 = .33 mm (b) 52 56 60 64 68510152025Tail length (mm)Frequency}sSE} 6.5 (a) We would predict the SD of the new measurements to be about 3 mm because this is our estimate (based on Exercise 6.4) of the population SD. (b) We would expect the SE of the new measurements to be 3/ 500 ≈ .13 mm. 6.6 To convey the homogeneity of the group of rats, the 10 gm should be the SD, since the SD describes variability among the rats. (The SE describes the precision of the sample mean, but this depends on the sample size.)85 6.7 (a) the SE (b) the SD (c) the SE 6.8 and 6.9 See Section III of this Manual. •6.10 (a) y– = 31.720 mg; s = 8.729 mg; n = 5. The standard error of the mean is SEy- = sn = 8.7295 = 3.904 ≈ 3.9 mg. (b) The degrees of freedom are n - 1 = 5 - 1 = 4. The critical value is t.05 = 2.132. The 90% confidence interval for µ is y– ± t.05sn 31.720 ± 2.132(8.7295 ) (23.4,40.0) or 23.4 < µ < 40.0 mg. 6.11 (a) The degrees of freedom are n - 1 = 5 - 1 = 4. The critical value is t.025 = 2.776. The 95% confidence interval for µ is y– ± t.025sn 31.720 ± 2.776(8.7295 ) (20.9,42.5) or 20.9 < µ < 42.5 mg. (b) We are 95% confident that the mean thymus gland weight in the population of chick embryos is between 20.9 and 42.5 mg. 6.12 (a) y– = 28.7; s = 4.5898;SE = 4.5898/6 = 1.87 ≈ 1.9 µg/ml. 28.7 ± (2.571)(1.9) (23.8,33.6) or 23.8 < µ < 33.6 µg/ml. (b) µ = mean blood serum concentration of Gentamicin (1.5 hours after injection of 10 mg/kg body weight) in healthy three-year-old female Suffolk sheep. (c) No. The "95%" refers to the percentage (in a meta-experiment) of confidence intervals that would contain µ. Since the width of a confidence interval depends on n, the percentage of observations contained in the confidence interval also depends on n, and would be very small if n were large. 6.13 (a) This statement is false. The confidence interval allows us to make an inference concerning the mean of the entire population. We know that 59.77 < y– < 61.09. (b) This statement is true. (See part (a).) 6.14 This statement is false. The confidence interval concerns the mean of the population. It does not tell us where individual data points lie. 6.15 The higher the confidence level, the greater the length of the interval. Thus, 90%, 85%, and 80% confidence intervals correspond to (.822,.858), (.824,.856), and (.826,.854), respectively.86 •6.16 (a) y– = 13.0; s = 12.4; n = 10. The degrees of freedom are n - 1 = 10 - 1 = 9. The critical value is t.05 = 2.262. The 95% confidence interval for µ is y– ± t.025sn 13.0 ± 2.262(12.410 ) (4.1,21.9) or 4.1 < µ < 21.9 pg/ml. (b) We are 95% confident that the average drop in HBE levels from January to May in the population of all participants in physical fitness programs like the one in the study is between 4.1 and 21.9 pg/ml. 6.17 A histogram and normal probability plot of the data, shown here, support the use of a normal curve model for these data. -20 0 10 20 30 40-0102030-1 0 1nscores 6.18 (a) 5,111 ± (2.306)(818/ 9 ) (4482,5740) or 4,482 < µ < 5,740 units (b) We are 95% confident that the average invertase activity of all fungal tissue incubated at 95% relative humidity for 24 hours is between 4,482 units and 5,740 units. (c) To check that the data are from a normal population we can make a normal probability plot. 6.19 6.21 ± (2.042)(1.84/36 ) (5.58,6.84) or 5.58 < µ < 6.84 µg/dl. •6.20 y– = 1.20; s = .14; n = 50. The degrees of freedom are 50 - 1 = 49. From Table 4 with df = 50 (the df value closest to 49) we find that t.05 = 1.676. The 90% confidence interval for µ is y– ± t.05sn 1.20 ± 1.676(.1450 ) (1.17,1.23) or 1.17 < µ < 1.23 mm. 6.21 We are 95% confident that the mean Bayley Index of prematurely born infants who receive intravenous-feeding solutions is between 93.8 and 102.1. (Although the center of the interval is 97.95, which is less87 than the general population average of 100, the interval extends above 100, so we cannot be sure that µ is less than 100.) 6.22 y– = 10.3; s = 0.9; n = 101. There are 100 degrees of freedom, so t.025 = 1.984. The 95% confidence interval for µ is 10.3 ± 1.984(.9101) (10.12,10.48) or 10.12 < µ < 10.48 g/dLi. 6.23 1 - .025 = .975. In Table 3, z = 1.96 corresponds to an area of .975. (A t distribution with df = ∞ is a normal distribution.) •6.24 1 - .0025 = .9975. In Table 3, an area of .9975 corresponds to z = 2.81. A t distribution with df = ∞ is a normal distribution; thus, t.0025 = 2.81 when df = ∞. 6.25 The confidence coefficient is approximately 68%, because a t distribution with df = ∞ is a normal distribution, and the area under a normal curve between z = -1.00 and z = 1.00 is approximately 68%. 6.26 (a) Smaller. The area included between t = -1.00 and t = 1.00 is smaller for a Student's t distribution than for a normal distribution. (b) It is not affected, because of the Central Limit Theorem. 6.27 (a) Guessed SD = 20 kg; n must satisfy the inequality 20n ≤ 5 so n = 16. (b) n must satisfy the inequality 40n ≤ 5 so n = 64. The required sample size does not double, but rather is four times as large. •6.28 We use the inequality Guessed SDn ≤ Desired SE. In this case, the desired SE is 3 mg/dl and the guessed SD is 40 mg/dl. Thus, the inequality is 40n ≤ 3 or 403 ≤ n which means that n ≥ 177.8, so a sample of n = 178 men is needed. 6.29 Guessed SD = 80 g (a) The desired SE is 20 g, so n must satisfy 80n ≤ 20 which yields n ≥ 16.88 (b) The desired SE is 15 g, so n must satisfy 80n ≤ 15 which yields n ≥ 28.4, so n = 29. 6.30 Guessed SD = 1.5 in The SE should be no more than .25 in, so n must satisfy 1.5n≤ .25 which yields n ≥ …
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