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Chapter Review Questions 8.48 (A) Construct a 95 pct faith legal separation for the true mean. Sample Mean: 0 + 260 + 356 + 403 + 536 + 0 + 268 + 369 + 428 + 536 + 268 + 396 + 469 + 536 + 162 + 338 + 403 + 536 + 536 + 130 = 6930 / 20 = 346.50 ( try out mean) Standard aberration: 170.378 Sample sizing: N = 20 Degrees of freedom :v = n 1 = 20 1 = 19 (degree of freedom) T critical value for 95%:t0.95 = 2.093 federal agency measure interval: E = 2.093 x 170.378 ÷ ?20 = 356.601154/ ?20 = 356.601154/ 4.472146 = 79.738 95% Confidence Interval= 346.5 79.738 < u< 346.5 + 79.738 = 266.762 < u < 426.238 = 266.8 < u < 426.2 (B) Why power nitrogen be an issue here? This is a down(p) sample and may non include the entire sample. thus we take away to assume equally distributed. (C) What sample size would be requisite to obtain an error of ±10 forthright millimeters with 99 part corporate trust? E = 10 t0.9 9 = 2.576 X = (t critical value * timeworn deviation / E)2 = (2.576* 170.378 /10)2 = (438.893728 / 10)2 = (43.8893728)2 = 1926.2770 = 1927 (D) If this is not a reasonable requirement, suggest wizard that is. You can decrease the confidence interval to 90% or increase the value of E. 8.

64 (A) Construct a 90 percent confidence interval for the proportion of all kernels that would not pop. p = # of kernels un popped / total # of kernels p = 86 / 773 p = 0.111255 = 0.1113 E = 1.96 * ? (0.111255 * 0.8887/773) E = 1.96 * ? (0.111255 * 0.0011497) E = 1.96 * ? (00.000128) E = 1.96 * 0. 0113 E = 0.02217 = 0.0222 Confidence int! erval 90%: CI = (0.1113 0.0222 < p < 0.1113 + 0.022) CI = (0.0891 < p < 0.1335) (B) Check the atomic number 7 assumption. p * n = 0.113 * 773 p * n = 87.349 n (1 p) = 777 * .8887 n (1 p) = 690.52 The normality assumption is good (C) Try the Very ready Rule. Does it cogitation well here? Why, or why not? No, it does not work well here...If you want to get a broad(a) essay, order it on our website: OrderCustomPaper.com

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