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HW6-Solutions.pdf

HW6-Solutions.pdf - BIOSTJIOIIAAA Hw 50mm Hw(f M(off(0.70(f...

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Unformatted text preview: BI‘OSTJIOIIAAA Hw 50mm Hwé: (f)? M (off (0.70 (f\? Exercise 9 ' 3. Since the population standard deviation 0 is unknown, we use the t distribution with 13 df rather than the normal distribution. A two-sided 95% confidence interval for p is 3.6 3.6 29.6 — 2.160 —, 29.6 + 2.160 —- ( V14 V14) 01' (27.5, 31.7). b. The length of this intervel is 31.7 - 27.5 = 4.2 Weeks. c. Since the interval is centered around the sample mean :3 = 29.6 weeks, we are interested in the sample size necessary to produce the interval (29.6 — 1.5, 29.6 + 1.5) 01' (28.1, 31.1). We know that the 95% confidence interval is of the form 3.6 3.6 .6 — 1. —— . 1. — . (29 96 VT? 29 6 + 96 fl) To find 11., therefore, we must solve the equation _ 1.96(3.6) 1.5 _ fi 01' 2 n ___ [1.96(3.6) 1.5 = 22.1. A sample of size 23 is required. . (1. Here we are interested in the sample size nece-ary to produce the mterval (29.6 - 1, 29.6 + 1) 01' (28.6, 30.6). The 95% confidence interval takes the form 3.6 3.6) , _ . _., .6 1.96— . (296 196%.: 29 + fl To find n, therefore, we solve the equation _ 1.96(3.6) — T or 49.8. A sample of size 50 is required. Exercise 11 a. Because the population standard deviation is unknown, we use the t distribution with 7 df rather than the normal distribution. The sample mean calcium level is :ic = 3.14 mmol/l and the standard deviation is 3: = 0.51 mmol/l. A one-sided lower 95% confidence bound for the true mean calcium level pc is 3.14 — 1.395(o.51/¢§) = 2.80 mmol/l. b. The sample mean albumin level is in = 40.4 g/l and the standard deviation is 3.. = 3.0 g/l. A one~sided lower 95% confidence bound for the true mean albumin level an is 40.4 — 1.895(3.0/¢§) = 38.4 5/1. c. The lower 95% confidence bound for the mean calcium level does not lie within the normal range of values; this suggests that calcium levels are elevated for this group. There is no evidence that albumin levels difier from the normal range. Exercise 9 a. The null hypothesis of the test is Ho: [1. = 74.4 mm Hg. b. The alternative hypothesis is HA: [17$ 74.4 mm Hg. c. The test statistic is id — on Ud/x/fi , 84 — 74.4 9.1 /\/1_0 = 3.34. II The area to the right of z = 3.34 is less than 0.001, and the area to the left of z = —3.34 is less than 0.001 as well; therefore, 1) < 0.002. d. Since 1) < 0.05, we reject Ho and conclude that the mean diastolic blood pressure for the population of female diabetics between the ages of 30 and 34 is not equal to 74.4 mm Hg. In fact, it is higher. e. Since p < 0.01, the conclusion would have been the same. Exercise 10 a. The null hypothesis is Ho: p 2 7250/mm3, and the alternative hypothesis is HA: p < 7250/mm3. b. Since the population standard deviation is unknown, we use the btest rather than the z-test. The test statistic is - 5 ‘ #0 8/fi 4767 — 7250 3204MB = -3.00. ev- N For a t distribution with 15 — 1 = 14 degrees of freedom, 0.0005 < p < 0.005. Therefore, we reject Ho. c. We conclude that the mean white blood cell count of humans infected with E. cunts is lower than 7250 /m.m3, the mean of the general population. ...
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