hw9Spring08Answers

hw9Spring08Answers - STAT 301T HW9 ANSWERS 14.1(a The...

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Unformatted text preview: STAT 301T HW9 ANSWERS 14.1. (a) The standard deviation of f is cr/J 100 i 1.8974. (b) m = 20.8974) i 3.8 (or “i3.8”). (c) The confidence intervals drawn may vary, of course, but they should be 2m = 7.6 units wide. (d) 95%. 274.3 276.2 278.1 280 281.9 283.8 286.7 M <———————-—o——-——->- 14.4. Search Table A for 0.0125 (half of the 2.5% that is ' Standard * not included in a 97.5% confidence interval). This area corresponds to 2* = 2.24. Software gives 2* = 2.2414. Probability = 0.0125 Normal'curve 14.6. (a) The two low scores (72 and 74) are both possible outliers, but 7. 24 there are no other apparent deviations from Normality. (b) State: What is 7 the mean IQ p, of all seventh—grade girls in this school district? 2 69 Formulate: We will estimate 11 by giving a 99% confidence interval. 9 13 Solve: The problem states that these girls are an SRS of the popula— 9‘ 68 tion. In part (a), we saw that the scores appear to come from a Normal i8 £33334 distribution. With J? i 105.84, our 99% confidence interval for ,u. is 11 11222444 105.84 :1: 2.576 (1—5) é 105.84 :1: 6.94 = 98.90 to 112.78 IQ points. L; 39 J37 ' 1.2 8 Conclude: We are 99% confident that the mean IQ of seventh—grade girls 13 02 in this district is between 98.90 and 112.78. 14.7. (a) - (c) The confidence intervals are: n Confidence interval m.e. 1000 22 :1: 1. '960Jfifi6_ — 22 :1: 3.099 = 18.9 to 25.1 points 3.1 250 22 :l: 1. 960— ' :22 :1: 6.198 = 15.8 to 28.2 points 6.2 4000 22 :1: 1. 960 J- 22 :1: 1.550 = 20.45 to 23.55 points 1.55 J25_0 J4 4000 (d) The margin of error (“m.e.” in the table above) decreases with larger samples .(by a factor of 1 A/Z ). 14.8. (a) and (b) The confidence intervals are: Conf. level Confidence interval me. 95% 22 :1: 1. 960713;:- — 22 i 3. 099- ._ 18. 9 to 25.1 points 3.099 90% 22 i 1. .645V,——-'- —— 22 :1: 2. 601: 19. 4 to 24. 6 points 2.601 99% 22 i 2. 576JéT — —'-22 :i: 4.073 = 17.93 to 26.07 points 4.073 (c) The margin of error (“m.e.” in the table above) increases with increasing confidence level. 2 14.9. n = (9.952%) = 2401. (2.576)(1'5) 2 14.10. n = ( 5 ) : 59.72—take n = 60. . . ...
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