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**Unformatted text preview: **STAT 35000-Sec 26794: Homework 8
Unit 3: Frequentist Inference Due on December 8 (Thursday) Meghan Tooman 4:30PM—5:45PM KEV ___,_,__._.—.u—=—=——-—-a- August 17, 2016 Instructions:
1) This set of homework covers Chll and Ch13.
2) It is the preparation material for Quiz 8 and Final. 3) Show your work as much as possible. Submit the homework at the beginning of the class on the due
date. 4) The answer key will be posted on Canvas after the lecture class on the due date.
5) Staple your homework! No e—format. Do not slip your homework into the instructor’s ofﬁce. 6) The homework assignments are quite lengthy and I recommend that you get an early start on them
and do a few problems each day. You may also wish to discuss homework with your classmates.
Group discussions and study sessions can be a useful tool for learning. However, outright copying is
unacceptable, as well as pointless, and will be penalized. A good rule of thumb is that it is ﬁne for
you to talk with others about how to do a problem, but then go and write it up yourself, possibly
comparing answers afterwards if you are unsure. Remember that if you copy from a classmate without
understanding it, only your classmate will pass the exam. If blatant copying is detected, all parties
involved (copier and copied) will receive a score of zero for that assignment. This type of behavior will
also be reported to the school. 11.10 Sm = 57, 557 m 7072/9 = 20182222, SW = 51,080 — 6582/9 = 38728880, SW = 53. 258 —
(707)(658)/9 = 1568.4444, 50 = 12.0623 and 51 : 0.7771. (g) 32 _ 3872.8889 — (0.7771)(1568.4444)
_ f (1)) Since 3 = 19.472 and 7.0.025 = 2.365 for 7 degrees of freedom1 then a 95% conﬁdence
interval is = 379.150. (379.150)(57, 557) 12.0623 :1: (2.305) (9}(2012 2922) : 12.0623 :l: 81.975,
which implies M6991 < 60 < 94.04.
(0) 0.7771 a: (2.365)‘/% implies —0.25 < 731 < 1.80.
13.1 The hypotheses are H0 1M1=M2=“'=1I-0.
H1 : At least two of the means are not equal. a = 0.05. Critical region: f > 2.77 with 111 = 5 and v2 = 18 degrees of freedom. Computation: 0 Source of Sum of Degrees of Mean Computed
Variation Squares Freedom Square f
Treatment 534 5 1.07 0.31
Error 62.64 18 3.48
Total | 67.98 23 with P-Value=0.9024.
Decision: The treatment means do not differ signiﬁcantly. 13.2 The hypotheses are f101111=1¢2=---=165, H1 : At least two of the means are not equal. a = 0.05.
Critical region: f > 2.87 with ”01 = 4 and 1:2 : 20 degrees of freedom.
Computation: Source of Sum of Degrees of htlean Computed Variation Squares Freedom Square f
Tablets 78.422 4 19.605 6.59
Error 59.532 20 2.977
Total 137.954 24 with P~value:0.0015.
Decision: Reject H 0. The mean number of hours of relief differ signiﬁcantly. STAT 35000-Sec 26794 (Meghan Tooman 4:30PM-5:45PM): Homework 8 PROBLEM 1 Homework problems from the textbook; 11.16, 13.1, 13.2
Extra problems: Problem 1 Six diﬁerent machines are being conSidered for use in manufacturing rubber seals. The machines are being
compared with respect to tensile strength of the product. A random sample of four seals from each machine
is used to determine whether the mean tensile strength varies from machine to machine. The following are
the tensile-strength measurements in kilograms per square centimeter x10‘1: ‘ Machine
THEM—W
17.5 16.4 20.3 14.6 17.5 18.3
16.9 19.2 15.7 16.7 19.2 16.2
15.8 17.7 17.8 20.8 16.5 17.5
18.6 15.4 18.9 18.9 20.5 20.1 Perform the analysis of variance at the 0.05 level of signiﬁcance and indicate whether or not the mean
tensile strengths differ signiﬁcantly for the six machines. (Note that normality with common variance is
assumed.) 1. Write down the null and alternative hypotheses explicitly with [1.555, where n,- denotes the mean tensile
strength of the product produced from machine 2', z‘ = 1, 2, - -- ,6. 2. Complete the following ANOVA table by ﬁlling out the 9 bold spots (2 already ﬁlled). Source of Sum of Degree of Mean F
Variation Squares Freedom Squares Statistic
Regression SSR=5.34 k—l MSR=SSR/(k-1) F= MSE
Error SSE k(n—1) MSE=SSE/ (km-1)) Total SST=67.98 kn—l 3. Calculate the p value of the F test statistic and make your conclusion in this context. G H“: W“ ’4‘? $143 U“! '74:: auto
H“: ’H ”04+ M Ml Ito moi aqua-Ji- @ SSE 5'- 19?qu "' 63“ 251.10% M5K= 6'5‘4/5' : L01}.
K’l -'-' Lo'l 3' 5 H W (am 0' 12> ““55 3» £02.qu \‘B r: 3345
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® ?—\10\.\U\£ 7-"- QCAQ (D.3\\qq) 51%): one. >_05‘. Page20f5 There “is not Unoudsh evidence W 0dr \easi OWL ‘PopuJLa-Hm WM
-\—mei\c shrew-Hm \s d'\$¥evmt bewaem JML Lo magi/Limp. STAT 35OUO-Sec 26794 (Meghan Tooman 4:30PM—5145PM): Homework 8 PROBLEM 2 Problem 2 It is difﬁcult to determine a person‘s body fat percentage accurately without immersing him or her in water.
Researchers hoping to ﬁnd ways to make a good estimate immersed 4 male subjects, then measured their
waists. X: Waist (in.) 175 181 200 159
Y: Body Fat (95) 6 21 15 6 Him: a = 178375.33 =16.94,y = 12, 5., z 7.35, 33y : 225/3.
a) Compute the correlation coeﬂicient 7‘ between the body fat and waist.
b) Based on the calculated 7" in part 1), comment the association between the body fat and waist.
c) Calculate the least square regression line equation for the prediction.
) cl Conduct a hypothesis test to see whether the linear association is signiﬁcantly positive with signiﬁcance
level 0.05. Please state the hypotheses, compute the p value and make the conclusion in the context. 5x 5:) 10-44(q-gﬁ) senoa‘l lo.) Thu-L ~10 '3— Mod-Modtebé Dir-rave; quasi—Five .lewo modest-law
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1L3: “3%.15 “l“ 0-2-qu 5‘ be 3 \Q. - Olen-{0129.15} : ~3Ll-"9r'5 d” Ho'- ﬁO 0.2L9H-l
Ho" F“ >0 i: 4.35 W?" Love 2. Pavaluu. 5 'Ltolc (Laug‘qq‘z) : o.\q‘3‘+ 70.0%“ Tel-Q in; ﬁxed Ho-
:M Mr ﬂawam\ .m'xdmcb 09‘ 0.. spacings: nuance. Mmﬁmm MW waist “ALL low £33 Mag.
Page3of5 “WM STAT 35DOU—Sec 26794 (Meghan Tooman 4:30PM—5:45PM): Homework 8 PROBLEM 3 Problem 3 The article “Characterization of Highway Runoff in Austin, Texas, Area” (J. of Envir. Engr., 1998: 131-137)
gave data on rainfall volume (m3) and runoff volume (m3) measured on 30 different days for a particular location. We would like to determine how the runoff volume is affected by rainfall volume. Summary
statistics of data are: E = 53.2, SE = 38.3,? : 42.9, sy : 32.1, n : 30. 1. In view of the question we want to answer, what is the predictor variable, and what is the response
variable? wdkickvr =~ Wﬂﬂgm New
magnum» 7 AWE}? 00W 2. Suppose that the correlation coefﬁcientbetween the rainfall volume and the runoff volume is r : 0.89.
Find the least squares regression line equation. z.\
‘0‘: 0.2501 (Egg-7Q a. oﬂq‘b‘i loo 1 Lam — ones‘ﬂhem .: Smurf (37- 3.2m? -‘r omen b0 3. What is proportion of variation in the runoff volume explained by the rainfall volume? ‘23- 0.9991 : 07mm ’ swam: «1 m qum lw M mm 3.0
_ ' WWW trim
MW \OVE ﬁlm W
A ' oW‘ 4. For a day with the rainfall volume 72 m3 and the runoﬁ" volume 53 m3, what is the residual of this
day for the prediction using the above least Square regression line? ' ‘55: 6.1m? at onus‘ii’W—W ‘— 5L0-‘l7-‘5 ?\£$\o\wc«l —. “ﬂow seams : ~ some” Page 4 of 5 STAT 35000—Sec 26794 (Meghan Tooman 4:30PM—5145PM): “Homework 8 PROBLEM 3 5. Answer the following questions by ﬁlling in the blanks based on the interpretations of a and {3. a When the rainfall volume increases by 1 m3, on average the runoff volume will
MW lo? 0- '71LL\5°1 m5 0 When the rainfall volume is 0 m3, on average the runoff volume is 9) .9..\k0ﬂ€ m5. 6. Find the 95% conﬁdence interval for the slope ,6 of the regression line. Does the 95% 0.1. support the
claim that ,8 > 0‘? ,. :L' . 01:7.
‘0‘ : sch/2- SEA , onweﬂ Z.o‘-\E> LO 3 : onqs‘l ‘5- 0H1“ : (0.5% , (3-8an) W SEW: ﬁx ,Lo wait/k W . J10} O W de‘ﬂﬁ
) Page 5 of 5 ...

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