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Unformatted text preview: Ql. A population has a probability distribution given by the following density function : f($.e):{ ﬂe—“egﬁ if 0<m<6
0 elsewhere where 9 is an unknown parameter. A sample of size n = 1 is taken ; i.e. there is a single sample random variable X1.
a) Show that U = £81 can be a pivotal random variable for the parameter 9. (Hint : First find the cumulative
distribution function FU(u) of U). (10 pts) b) Using the pivotal random variable U in a) find a 90% upper—sided confidence interval for 9 when X1 was measured as :54: 2.1. (10 pts)
. O 2540
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l 74%) ﬁghtbéfcuuz) oéuéi 0‘. _ ( 3 (a.z)2— ’Z./o.2_)3 S— 0. lo) Q2. The following are the burning times (in seconds) of flares of two types : Type I Type II 65 64
81 71
57 83
66 59
82 65
.ﬂ ”"
. ﬁn? 347’ 2
Supposing that the burning times are normally distributed find a 98% two—sided confidence interval for the ratio 312.
(22 pts) 2
n, “L
96 7C. ..
HI 6,13
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e 2 49 6 889
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44 8‘9
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0.7%; 4 93—» 4 W996
W
7’ Q3. Suppose we have a normal population with the variance 02 = 25 and unknown mean y. We want to test the
null hypothesis H0 : p = 20 against the alternative hypothesis H1 : ,u = 28. We decide to adopt the following
rejection rule (RR) : Reject H0 if X > 25. a) We take a sample of size 6 and apply the above rejection rule. Calculate or = P(Type I Error). (10 pts) b) Now suppose we want to repeat the test on another sample not necessarily of the same size. If we adopt the same rejection rule but would like to have ﬂ = P(Type II Error) be at most 0.10 , what should be our
new sample size ? (10 pts) 4) d: PC§>2§lyz¢o)<P(Z> aim )éP(Z>2.A§) M Q
5 wgooé (ablccz. 43‘) $0.5\¢o¢90—(9.492q 20.007] v/in Q4. Suppose that two normal populations have unknown means . The variances are also unknown, but are assumed to be equal, i.e. of = 0%. Consider the test for the following null and alternative hypotheses : H01#1—#2=5 agalT‘St Hlim—Maéts; two independent samples of sizes 711 and 712 are taken from the two populations ; a) Show that apart from the standard test procedure we can also use a confidence interval approach ; i.e. construct
a 100(1—a)% twosided confidence interval for M1 —,u2 and show that the rejection rule (RR) of the standard
test is equivalent to : “‘Accept H0 if 6 is in the confidence interval (10 pts) b) Verify your finding in a) in the following problem 1 A study is conducted to estimate the differences in the mean
exposure to radioactivity in utility workers in the years 2002 and 2007. The following data (in terms of proper
units ) were obtained based on two indpendent samples from the two years : 2002 2007
m = 12 “2 = 10
f1 = 0.94 (32 = 0.62
2121051; — $02 = 060 231:1(1'2J — i2? = 042 Test the hypotheses ; Ho : ,ul — M2 = 0.35 against H1 : M1 — H2 75 0.35
1) by the standard method with a = 0.05; (7 pts) 2) by the confidence interval method using a 95% confidence interval. Have you arrived at
the same decision ? (6 pts) 4) TN 5W‘dcwcl test is ; Rejed H Hf. t7 0‘ J“ ‘—Zm,+v)21 aw alcth Ho RT ~+axq+aﬂ<ﬁi4£¢+¢ 27/ __,_ ~ 6; H’) ,2. Solw 5.
3 2. I 7.
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c0.’1.4.6 1171+“? 0.0968 1
if = 2a 96 . M “940894 «c9\%c799 (N786 Awe/put 14‘)
0075394; 2 H 97 5’51 mid" Pr 7,: MM.” ‘3: 7096 (041%!in £0352 {BO/Rpm l0 Q5. Many consumers believe that automobiles built on Mondays are more likely to have serious defects than those
built on any other day of the week. To support this theory a random sample of 100 cars built on Mondays were
inspected, of them 8 were found to have serious defects. A random sample of 200 cars produced on other week days
revealed 12 with serious defects . a) State proper hypotheses H0 and H1; (4 pts)
b) Test at the a = 0.05 significance level ; (10 pts) C) Repeat the test by the Pvalue method. Do you come to the same conclusion ? (6 pts) zoom 9 ’ )
l r
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 Spring '12
 u
 Statistics

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