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Unformatted text preview: Stat23 lAssignment 3 Family (last) Name: Section #:
Given (ﬁrst) Name: Grade =
ID: Marks Available: 41 89‘5”.“ Due Date: Thursday Nov 8th 2007, 10 am in the Tutorial Centre MC6094/MC6095. General Rules: 5 . Although it is understood that some measure of collaboration will occur in the assignments, the work you hand in must be your own and diﬁer from those of your compatriots. All help should be cited. Excessive collaboration
(ie. copying) is not acceptable. Please read the university policies on what other things constitute cheating. Late assignments will not be accepted. The tutorial center will be staffed from 8am until 103m on the date the
assignment is handed.in. You must print this assignment and put your solutions ON the assignment pages. The only exception(s) to this
policy will be the course note homework. Course note homework can be attached to the END of the assignment.
Please use the back if you need more space and CLEARLY indicate where the solution has gone. All graphical displays should be properly labelled and titled.
All R graphics output (ie. plots) should be cut and paste into your assignment.
YOU MAY ONLY USE R IN A QUESTION THAT TELLS YOU SO. In all cases show your work. Grades are given for the clarity of your responses. Reading the data table into R (Question 2) . Use the ﬁle on UWACE: Assignment 2 Code, found in the Assignment 2 folder. The Story
The eﬂ‘ects of smoking are usually an interesting problem to investigate statistically. One such study was performed in the mid to late 1970s. At the time a sample of 654 youths (aged 3 to 19) from Boston was taken. The
objective of the study was to measure the eﬂecm of smoking on F EV (Forced Expiratory Volume). FEV is the amount of air you can exhale in one second of a forceful breath. Source: Forced Expiratory Volume, Journal of Statistics Education, Volume 13, Number 2 (July 2005).. NOTE: If a question is Not Marked, I use the notation NM The Story Thus Far....
Last assignment we used MLEs to estimate model parameters. It was assumed that a model to explain the relationship between height (x) and FEV (y) was:
Yx= a+ﬂ(z—E)+R, R~G(0,a’) We found: 3 : 223 any.  6.54in = 0 13 23:: $1 — if
a=y=2m A A 2
A2 2.6:: (3/1 — a + 35 — 3031')
= 654 = 0.19 And, 1 1 2?: (35,. _§)2 2 21242.13 A model used to look at: A random sample of 3 1 8 females, and 336 males were randomly selected and asked
whether or not they smoked. The model used was Y] ~ Bin(nj, 7rj) with j E {0 = females, 1 = males} Wefoundir‘oz—2—and1rlz—=__
n0=318 711 = 336 1. [4] Based on the 2 models.
(a) For the parameter a, use the estimate to suggest an appropriate estimator (use appropriate notation). 54 = Y (b) For the parameter ﬂ, use the estimate to suggest an appropriate estimator ~ 2 i . _ 4 y . ._ y. z' _ a: , _
(use appropriate notation). ﬁ = —————‘=1 a: Y‘ 65 a: — _SW _ _Z (3”’ "7) 1 Z ( z ) (y, y) 2654 (xi — 1")2 Sam Sam Saar i=1 (c) For the parameter no, use the estimate to suggest an appropriate estimator . . ~ Yo
(use appropnate notation). 7r0 = m (d) For the parameter m, use the estimate to suggest an appropriate estimator (use appropriate notation). 7r1 = — (e) [10] Show that the distribution of (3 is a ~ G (a, 0' v654 ) a is a Gaussian random variable since it is a linear combination of Gaussian random variables. E0”) = II Var (7) Therefore the distribution is: a N G (a, II II II 1 654
654 1
QEEO/i) 654 1
— E(a+ﬁ(wi5)+R) 654 agiﬂawxi—mm» 654 1 654
0—35'lﬁizlﬂfci
1 654
a'ﬁf'l‘gaﬂng
a—ﬂf+ﬁf=a 1 654
m 2 Var (Y1)
i=1 654
1 6542 1 654
—6542 Z Var (R)
i=1 1 654
2
6542 Z 0
i=1 1 2
(5—520 gizmwm—EHO) 0' v 654 ZVar(a+ﬂ(xi —E) +R)
i=1 ) (assuming Yi’s are indep) (f) [NM] Show that the distribution of E is approximately given by ,5 N G (,8, ax/I—i) , where 1 " 654 _ 2'
2E1 151' " :13) ,6 is a Gaussian random variable since it is a linear
combination of Gaussian random variables.
E (t) E (fi‘iim  634:?)
Zi=1 :c,’ — z)
1
2?: (x, — if k 25.1 (xiE (m) — 654w m k _ —k654za + ka654a: + We 22‘2"; 90?
[ik (Effix 1? — It 23%)
[3k ((225: x?) I “2?: M)
M ( 634 1:1;( §)— 654332): 13 654
Var (/6 2 mm — 6545?)
i=1 ll Var (B) 654
Var (k2 ((w; — E) Y _
i=1
654
Var (k 2(231 — 5) Yi) i=1 654
szar (2 (xi — E) K) i=1 654
k2 (Z (x" — a)? Var ()9) i=1 654
02192 (2 (xi — E2) 2 £202 11:] Therefore the distribution is: 3 ~ G (is, iii/I?) E (23:: mm — 6545?) k 2'33 xi (a + ﬂ (xi ~ 5))  k654§a
—k654xot + k 26:4 (aria + B22 — ﬁxﬂ)
5,933 2654:“ Var (k:((z1x)((Y—7))>
(nWVO since 2 E Z 654 i=1 (It, ' fi‘i (w? + 52  2M) 654 $2
i=1xi 435452 (g) [4] Fill in the blanks: State the (1 — a) CI you would use fora. a :l: ch5—4 = i] :l: 6%
State the (1 — a) CI you would use for B. B i cx/EE For the next two, use information provided in your textbook. A 1 __ A
State the (1 — a) CI you would use for no. £0 3; c 71.0 ( n "0)
o
A 1 _ A
State the (1 — a) CI you would use form. *1 j; c ”1 ( n 7(1)
1 (h) [6] Find a 95% conﬁdence interval for (1. Be sure to use appropriate notation. You have a value c from the 156542 = 1:65;; 2' C(0, 1) distribution
The 95% value c for this distribution is: I z 1.96 or = 1.963609 The estimate (center of the interval) is: y = 2.64 The variance, 32, used in the interval is: 32 z 0.19 A The interval is therefore (2 decimals please): 3 :l: c k
= 2.64 :l: (1.96) %
= (2.61, 2.67) [NM] Given an appropriate conclusion: We are 95% certain that the true value of a is in the interval, (2.61, 2.67) (i) [NM] Based on the interval you have created in the interval above for a, answer the following True or
False questions. Be sure to properly circle your answer. Any confusion in the ”circles” will result in the
question being marked as incorrect. To get marks for this question, PLEASE COPY YOUR INTERVAL HERE (2.61, 2.67) * With 95% conﬁdence, the value for FEV when we are looking at the average height, is zero. TRUE FALSE * With 95% conﬁdence, that the value for FEV when we are looking at the average height, is 2.6. TRUE lEZIISE * With probability 95%, the value for the FEV when we are looking at the average height, is 2.6. TRUE FALSE * The value for the FEV when we are looking at the average height is, 2.6, TRUE FALSE (j) [NM] Find an 95% conﬁdence interval for ,6. Be sure to use appropriate notation. You have a value c from the t354_2 = 13552 :1 G (0, 1) distribution
The 95% value c for this distribution is: z 1.96 or = 1.963609 A The estimate (center of the interval) is: 0 = 0.13
The variance, 82, used in the interval is: 32 z 0.19 . . n . l
The value of k in this case 1s. —21242.13 The interval is therefore (5 decimals please): 3 :l: cx/EE = 0.13 d: 1.96 (my/0.19 => (0.12414, 0.13586) Given an appropriate conclusion: We are 95% certain that the true value of ﬂ is in the interval, (0.12414, 0.13586) . (k) [4] Based on the interval you have created in the interval above for ﬂ, answer the following True or False
questions. Be sure to properly circle your answer. Any confusion in the ”circles” will result in the question being marked as incorrect. * For marks, PLEASE COPY YOUR INTERVAL HERE (0.12414, 0.13586) * With 95% conﬁdence, the slope of the model is zero. TRUE FALSE * With 95% conﬁdence, the change in FEV for a 1 unit change in the deviation of height is l. x TRUE FALSE >5: With 95% conﬁdence, that the change in FEV for a 1 unit deviation in height is 0. TRUE FALSE (1) [3] Use a test of hypothesis to determine whether the change in FEV for a 1 unit change in the deviation of
height is 1. Your null hypothesis is: H o B = 1 >whereﬂﬂ=Z~t652~G(O,1) Ex/F [NM] The conclusion: We have extremely strong evidence to reject Ho. There is evidence to suggest >73—
3f Thepvalue is: 2Pr<Z>
0—. 13 — l =2Pr<Z>
(2—1242.13)V0 19 _ —2Pr(Z > 290 90)~ the change in FEV for a 1 unit change in the deviation of height is not 1. (m) [6] Find a theoretical (ie. without the nugbers) conﬁdence interval for the.intercept of the model Be
sure to use appropriate notation. Note: E, [3 are independent. Your estimator is: a — [if The theoretical expectation of the estimator will be: E (a — ,3?) = a — Hi The theoretical variance of the estimator will be: Var (r32 — ,3?) = Var (a) + Var (35)
2
_ ‘7— 2 —2
‘ 654 + a I“
The sampling distribution of this estimator will be: G a — x, 25—52 + 02152
Your corresponding conﬁdence interval will be: a  Am 2}: c 603 + 32k?) Where 0 is a value from the trim 2 G(0, 1) table. (11) [NM] Find an 90% conﬁdence interval for the intercept. Your answer to the last problem was: The 90%, c, for this distribution is: The estimate (center of the interval) is: The interval is therefore (2 decimals please): Given an appropriate conclusion: 3 _
 + 02/6222 “—35” 654 a: 1.645 or = 1.647 a — 35 = 2.64 — 0.13(61.14) z —5.31 A_A i 2 —2
a ﬁfic 654 +0 16.7:
0.19
= —5.31 — 1.65 m + (ﬁ) ((1114? => (—5.61,—5.01) We are 95% certain that the intercept is in the interval, (—5.61,—5.01) (0) [NM] Based on the interval you have created in the interval above for the intercept, answer the following
True or False questions. Be sure to properly circle your answer. Any confusion in the ”circles” will result
in the question being marked as incorrect. * * With 90% conﬁdence, the intercept of the model is zero. TRUE We are very conﬁdent, that the intercept is negative. TRUE FALSE 2. Question 2
(a) [4] Find an 95% conﬁdence interval for 1ro. Be sure to use appropriate notation. You have a value c ﬁ'om the G (0, 1) distribution
The 95% value c from this distribution is: = 1.96
The estimate (center of the interval) is: ?o = 531%
ﬁg (1 — ﬁg) The interval is therefore: 11:0 :l: c => (0.09,0.16) Given an appropriate conclusion: We are 95% certain that the true value of 1ro is in the interval, (0.09, 0.16). (b) [NM] State an 95% conﬁdence interval for 7r1. There is no need to show your work. (005,011) (c) [NM] Based on the two conﬁdence intervals, explain whether the proportion of smokers statistically differ
between genders. Explain. Circle one: Yes Explain:
The two intervals overlap, which implies the SP value may lie in this interval. ((1) [NM] Someone suspects that 9% of women smoke. Conduct a~hypothesis test to detexmine if this is
7I’o — 7ro reasonable. To perform this test you will need to use, D = ~ G (0, 1)
7m (1 — 1ro)
no
Your null hypothesis is: H o 7ro = 0.09
Thepvalueis: 2Pr Z) M— whereZ=——E:—7iq—~G(O,l)
7r0(1—7'o «0(1—1r0)
no no
32  0.09 = 2Pr z > —31§———
0.09 (0.91)
V 318
= 2Pr(Z > 2.03)
= 4.2% The conclusion: We have reasonable evidence to reject Ho. ...
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