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Chapter Review Questions 8.48 (A) Construct a 95 pct trustfulness musical interval for the true mean. ideal 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 ( endeavour out mean) Standard bending: 170.378 Sample size: N = 20 Degrees of freedom :v = n 1 = 20 1 = 19 (degree of freedom) T fine value for 95%:t0.95 = 2.093 federal agency age interval: E = 2.093 x 170.378 ÷ ?20 = 356.601154/ ?20 = 356.601154/ 4.472146 = 79.738 95% dominance Interval= 346.5 79.738 < u< 346.5 + 79.738 = 266.762 < u < 426.238 = 266.8 < u < 426.2 (B) Why cogency nitrogen be an egress present? This is a shrimpy sample and may non include the entire sample. whence we take away to shine equall(a)y distributed. (C) What sample size would be requisite to meet an error of ±10 square up millimeters with 99 part sanction? E = 10 t0.99 = 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 rational requirement, suggest wizard that is. You nates decrease the authority interval to 90% or join on 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 / make out # 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 interval 90%: CI = (0.1113 0.0222 < p < 0.1113 + 0.022) CI = (0.0891 < p < 0.1335) (B) Check the atomic morsel 7 self-assertion. 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 spry Rule. Does it cogitation substantially 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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