2005-Fall-Final_s

2005-Fall-Final_s - More PastPaper:...

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Unformatted text preview: More PastPaper: http://ihome.ust.hk/~cs_gxx ISMT 111 Fall 05 Final Exam Suggested solution Question 1: (a) The Hypothesis statement is H0: μ=23 vs H1 μ>23. Naturally we need to assume that that either the data are normally distributed in which case we assume that the CLT applies. The sample size n=56 is large enough and we can use the Z-test and the test statistic is Z= 56 (23.5 – 23)/10.2=.367. We can use the normal table to find the critical value of 1.645 at alpha =.05. In this case since .367<1.645 we cannot reject the null hypothesis and must conclude that the evidence indicates that their claim is false. (b) The 95% confidence interval is approximately 23.5 ± 2.67 (c) To find A, solve the equation 56 (23.5 – A)/10.2 ≥ 1.645, the solution is A ≤ 21.258. Hence set A=21. Note: Question (c) can be adjusted for different values of the standard deviation. Question 2: (a) Using the formula for the differences where the population variances are assumed equal but unknown and the populations are normally distributed. 2K=16.14+6.14 or K=11.14, but K=T.025 SE with 10 degrees of freedom T.025=2.228. Hence the Standard Error is 5. (b) T.05 is 1.812, x1 − x 2 = (16.14 − 6.14) / 2 = 5 . Thus the 90% CI is given by 5 5 ± 1.812 =[-4.06, 14.06] (c) The observed test statistic is -1, which is larger than -1.812 hence we cannot reject the null hypothesis. (d) Since SE is 5 and the hypothesized difference is 10 it follows that the difference of the sample means must be 5 (based on the result of T = - 1). But this can also be calculated as in (b) Question 3: (a) E (Y ) = 1 + 0.6 × 5 + 0.04 × 20 = 4.8 ; s.d . = 1 (b) P (Y < 4) = P ( Z < −0.8) = 0.2119 . gx x/ (c) E (Y ) = 4.8 s.e. = 1 / 24 =0.204 om e. us t.h k/ ~c s_ (d) P (4.5 < Y < 5) = P (−1.47 < Z < 0.98) = 0.8365 − 0.0708 = 0.7657 5 :// ht tp ˆ (f) P (0.05 < p < 0.1) = P(−3.35 < Z < −2.84) = 0.4996 − 0.4977 = 0.0019 ih (e) P(Y < 4.5) = P( Z < −0.3) = 0.3821 s.d . = 0.3821(1 − 0.3821) / 24 = 0.09918 More PastPaper: http://ihome.ust.hk/~cs_gxx Question 4: H 0 : μ = 6 vs H 1 : μ < 6 (a) Let μ be the time reduced by the new system Type I error: The actual time reduced is not less than 6 minutes, but we regard the time reduced is less than 6. Type II error: We think that the time reduced is not less than 6, but in fact it is actually less than 6. (b) ⎛ α= α = P(reject H 0 | H 0 is true) = P( x < 5.7) = P⎜ ⎜ x −6 ⎝ 1.5 100 < 5.7 − 6 ⎞ ⎟ 1.5 100 ⎟ ⎠ = P( z < −2) = 0.0228 (c) p-value = P ( x < 5.9) = P( x −6 1.5 / 100 < 5.9 − 6 1.5 / 100 ) = P( z < −0.667) = 0.2514 (d) Let n be the sample size required 2 × n = ⎛ 2 × 2.576 × 1.5 ⎞ = 1493.0496 ≈ 1494 ⎟ ⎜ 0.2 ⎠ ⎝ Question 5: . Question 6: a) b1 = [8340 − 8(134 / 8)(410 / 8)] /[3020 − 8(134 / 8) 2 ] = 1.8988 , b0 = (410 / 8) − 1.8988 × (134 / 8) = 19.4451 , cost = 19.4451+1.8988 year b) 1 − (24000 − 19.4451 × 410 − 1.8988 × 8340) /[24000 − 8(410 / 8) 2 ] = 0.94 c) observed t: 9.3583,for 6 degrees of freedom t 0.025 = 2.447 gx x/ d) the interval 38.43 ± 22.789 6 ht tp :// ih om e. us t.h k/ ~c s_ e) Cost =19.13+1.89 year , ...
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This note was uploaded on 02/27/2010 for the course ISOM ISOM111 taught by Professor Anthonychan during the Spring '09 term at HKUST.

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