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Chapter 11

# Chapter 11 - 6851F_ch11_173_188 19:38 Page 173 11 11.1(a s>...

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173 11 11.1 (a) (b) Since 0.01, s (0.01) 11.2 (a) 2.015. (b) 2.518. 11.3 (a) 2.145. (b) 0.688. 11.4 (a) df 11, t * 1.796. (b) df 29, t * 2.045. (c) df 17, t * 1.333. 11.5 (c) The t 2 curve is a bit shorter at the peak and slightly higher in the tails (see TI-83 plot below). (d) The t 9 curve has moved toward coincidence with the standard normal curve. (e) The t 30 curve cannot be distinguished from the standard normal curve. As the degrees of freedom increase, the t (df) curve approaches the standard normal density graph. 1 2 3 2 0.0173. s > 2 3 9.3 > 2 27 1.7898. s > 1 n 11.6 (a) 0.228. (b), (c), and (d) Absolute df P ( t 2) difference 2 .0917 .0689 10 .0367 .0139 30 .0273 .0045 50 .0255 .0027 100 .0241 .0013 (e) As the degrees of freedom increases, the area to the right of 2 under the t df distribution gets closer to the area under the standard normal curve to the right of 2. 11.7 (a) 14. (b) 1.82 is between 1.761 ( p 0.05) and 2.145 ( p 0.025). (c) The P -value is between 0.025 and 0.05 (in fact, P 0.0451). (d) t 1.82 is significant at 0.05 but not at 0.01. 6851F_ch11_173_188 17/9/02 19:38 Page 173

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174 Chapter 11 11.8 (a) 24. (b) 1.12 is between 1.059 ( p 0.15) and 1.318 ( p 0.10). (c) The P -value is between 0.30 and 0.20 (in fact, P 0.2738). (d) t 1.12 is not significant at either 0.10 or at 0.05. 11.9 (a) Since the sample size is small ( n 15), the distribution of the CSB vitamin C data should be close to normal. We can check this using a stemplot and a normal probability plot. 1 14 2 2236 3 11 Since there are no outliers and the normal plot is reasonably linear, the assumption of nor- mality seems justified despite the small number of observations. We also must assume that the eight observations represent an SRS from the population of all possible amounts of vitamin C in samples of CSB. Since the eight observations were taken from a production run, this seems like a reasonable assumption provided that the observations were taken at regular intervals. (b) We will use the t -procedure. 22.50, s 7.19, df n 1 7. Using Table C with df 7, we find t * 2.365. The 95% confidence interval is therefore 22.50 (2.365) 22.5 6.0, or (16.5, 28.5). (c) Letting the mean vitamin C content per 100 g, we wish to test H 0 : 40 vs. H a : 40. The t test statistic is and the corresponding P -value (from soft- ware) is 0.0002. Clearly, this result is incompatible with a process mean of 40. We reject H 0 and conclude that the vitamin C content for this run does not conform to specifications (specifically, it is below the specifications). 11.10 (a) The stemplot shown below has stems in 1000s, split 5 ways. The data are right-skewed with a high outlier of 2433 (and possibly 1933). The normal plot shows these two outliers, but otherwise it is not strikingly different from a line. 0 3 0 4444455 0 666667777 0 88899999 1 0011 1 22223 1 44 1 1 9 2 2 2 4 (b) 926, s 427.2, standard error 69.3 (all in mg). (c) Use of the t -procedure is justified here because the sample size is large ( n 38 30) and thus the distribution of will be approximately normal by the central limit theorem. Using Table C x x t 22.5 40 7.2 > 1 8 6.88 1 7.19 > 2 8 2 x 2500 2000 1500 1000 500 0 Calcium intake 3 2 1 0 1 2 3 z score 6851F_ch11_173_188 17/9/02 19:38 Page 174
Inference for Distributions 175 with 30 degrees of freedom, we have t * 2.042. The approximate 95% confidence interval is then 926 (2.042) (69.3), or 784.5 to 1067.5 mg; MINITAB reports 785.6 to 1066.5 mg.

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