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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 Ztest 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) pvalue = 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.
 Spring '09
 AnthonyChan

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