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Unformatted text preview: Midterm Exam 1
Statistics 252 Version 1, Winter 2007
January 25, 2007 Namuugt it ‘vr Please show all work to receive full credit. There are 4 problems, each with multiple parts. The total
number of possible points is 60. Good luck! 1) For each ofthe following questions, provide choose a method from the following list which might
be used to help answer the question. If none ofthe methods we’ve discussed this term so far will
help answer the question, you should indicate an answer of“not covered.” Note that you might
not use all the choices. (3 points each) Small sample conﬁdence interval for ,u Small sample conﬁdence interval for [u] — #2
'Small sample hypothesis test about ,u Small sample hypothesis test about ,ul — ,le a) +0 b) d) Paired difference ttest for ,ut, Paired difference conﬁdence interval for #1, ANOVA
Not covered Twenty randomly selected students who took the SAT twice are selected. D0 students tend to
improve their SAT score the second time they take the test? ANIN A X A sample of 10 grocery items was randomly selected from Trader Joe’s, and another random
sample of l0 grocery items was selected from Von’s. What is an estimate ofthe difference
between the average prices of all grocery items at Trader Joe’s and Von’s? X \amvm dﬁtcrmwtﬁ, Jrtest for Ma At a selective university, 25 mathematics majors, 25 English majors, and 25 engineering
majors are randomly selected. 15 the average time spent studying per week for all
mathematics majors more than the average time spent studying per week for all engineering
majors at the university? X errd dxttericmt. Conﬁdmcf m‘mrvcil for Md A random sample of 10 small liberal arts colleges is selected, and a male and female
Economics major are randomly selected from each college. Are the average GPA’s ofthe
male and female economics students different? ‘ODT (0%ch
K Twenty households are randomly selected from San Luis Obispo county, and the incomes of
those households are recorded. ls the average income of households in San Luis Obispo
county higher than the average income of all households in California? poude diffﬁrmﬂ t test for»; Ms,
K Twenty—ﬁve Cal Poly students are randomly selected and the amount spent on textbooks by
those students is recorded. How much is the average amount spent by all Cal Poly students
on textbooks? Smﬁm sample conﬁdthm m+crvrh (“or M I . rtmg salary ofeach sibling was recorded. .4
Some descriptive statistics calculated in MlNlTAB are shown in the table below. Variable N Mean StDeV Sibling l 10 43930 11665 Sibling 2 10 43530 11617 Differences 10 400 435 (Sibling lSibling 2) information described in the problem. Should we use an independ t samples t—test or a
paired differences ttest? Briefly, explain why. Haunts) 2 low shame use aimed cu meme t’ttﬁ mama use“
Wow 10 comma: /\ SET of ms To QTWD’W'KT gm of
WVGS “01’ O“? W\ TD emuﬁndii W (M Ho" M (M 1 different. lfyou are going to use an independent samples ttest, then assume equal population . 1% variances. Note: To get full credit for this problem, you must deﬁne the parameters of
“61 M \ “1 interest, state the null and alternative hypotheses, calculate the test statistic, make a decision tﬁgﬁ» g 17% ﬁ on a critical value or p—value), and {gate a conclusion in the co;1::to:he:]uesti0n. (6
t: __ W :2 s b wcwou cu Ton(ci—
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r» tree/W by \O ‘ We COHLMQUL 4mm WK \8
3 O. 012W Afcﬂl rm at mgmﬁcar‘r’r emergence.
o<= 0.05 \Y\ W Stafﬁng annum Sﬂldﬂf’i mfools bf mm swm’xgs 0.012% >0.02.S 2
P 3) In a controlled laboratory environment, random samples of 10 adults and 10 children were tested
by a psychologist to determine the room temperature that each person ﬁnds most comfortable.
For each person, the temperature (in degrees Fahrenheit) at which he or she felt most comfortable
was recorded. The data are summarized below: N Mean Std. Deviation
Adults IO 77.5 4.5
Children 10 74.5 7 5
a) Clearly define in words and symbols the two population means ofinterest for this problem.
1&7 (3 points) a a c
i: ﬁgf were? witjxgmve 4m): mum «em com furs \0\
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\S h : \O b) Construct a 99% conﬁdence interval for the difference between the mean room temperature
1" 5 X :H 5 found most comfortable by adults and the mean room temperature found most comfortable by 7" ‘ children. Assume Wes. Based on your conﬁdence interval in part, what can
0" s Ar .3 62': 2.5 you conclude about the mean room temperature found most comfortable by adults and the
0.7— 3; 6?— mean room temperature found most comfortable by children? (6 points)
i z. .
_, ._._____..1 ‘ 2. . l =
  i S“ 32 {me—wet 3.250 115.13.; “ g: 5.2.402;
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0'47. =0 .005 \NQ can concludll ‘MM 9 C dim" .mﬂ mo Jr mum find
: ' YW "\ ‘3’1 ‘*VW( ‘chzvzx ‘)<j(y ‘(R: ‘QPY\CK ’
df Mintm \; L ) mmﬁaa; \3 loc‘twah 73. \° Mia (berm F tilz’aZEO whuf \WV (“mag WWG € «WRIT ONHGY?“
q, find comforvme \s loci—ween ’l 2.2\°l= ma
(b2 14°F 4) In a study to investigate the impact of smoking on heart rate, 6 randomly selected nonsmokers, 6
light smokers, and 6 heavy smokers undertook sustained physical exercise. Their heart rates, measured in beats per minute (bpm), were recorded after resting for three minutes. The sample
results based on the heart rate measurements are given in the following table: Group Size Mean StDev
Non 6 62.33 7.26
Light 6 63.17 8.16
Heavy 6 81.67 10.76 V L“, a) Define the factor, the factor levels, and the experimental units in this experiment. (4 points) WWO? ' smokma ./
mm?— mms ' mnsmoms, Mntsrmﬂrt , Wm growers
mamWM that; ' SYYlDYXYS 0K n ‘2 (p n 2‘; {p b) Suppose we want to determine ifnonsmokers and light smokers have different average heart
1‘ ' (a2 $3 T _@$ ﬂ rates. Answer this question using a 90% conﬁdence interval. Assume egual variances. Note
1 ‘ 2 I . g‘ 7 1.1L” g1, (b I ‘9 that Si = 59.65. Make sure you give a conclusion in the context ofthe problem.
LI (6 points) h’r‘vtehlgrtawb (0297262241 Ilenm = warn: 77.0761?) (‘734l4‘b17l34‘tﬁ ‘/
we Conchldl am of are (lot. confi (\M‘i’r +m+ 3:198 WACWQH Yam: 0% nOYKMDKMS is wmtcn
dfvnﬁnfl= bﬂo’l=\0 53.4.”me ma 4.51 ‘0er WWW Aw: average 4w. " \flm, WAN Wm 01‘ “girl VS \s between [34.28va
2 Ana 70M Wm pretation of the meaning of the phrase
“90% conﬁdent” (3 points) —a.— There is a 90% chance that the difference between the average heart rates of all non
smokers and the average heart rates ofall light smokers falls in this interval. b: 90% of the differences between the average heart rates of all nonsmokers and all light
smokers will fall in the interval. e.— In repeated sampling, 90% of similarly constructed intervals would contain the true
differences between the heart rates of all nonsmokers and all light smokers.
d.<‘ln repeated sampling, 90% of similarly constructed intervals would contain the differences in the sample mean heart rates of nonsmokers and light smokers.
@ ln repeated sampling, 90% ofsimilarly constructed intervals would contain the true difference between the average heart rates of all nonsmokers and all light smokers. H d) Now we would like to perform oneway ANOVA to investigate the differences in the average heart rates for all three groups. Deﬁne the parameters of interest, and write out the null and
alternative hypotheses (in symbols) for the ANOVA test. (4 points) pLﬁ—HW ONQWAQC heart rates of nonsmoléerg
AM: .mL awemc WWW Yﬂl'lﬁt Of Mm smear: /
#7,», W Mfmflf mart mm 06 mam MMTS “0"MlﬁM15/‘vk’b um: A—l' team mo (rt Wk M‘S
OWN 3 c) Circle the letter that corresponds to the correct inter 6) Carry out the ANOVA test at the .0] level. What is your conclusion? Note: Some useful
3 quantities that you will want to use are SSW“, = 2608.9 and SS,:.,W = 1 175.5. (6 points) _..'
'¢.kp08"l'\rlc5S: ZD/Llj tZXOlK (X: ()"Q\ d‘f’l’l
fill = 0 ‘005 tdligz‘gag ﬂ? (mew: W)
X s (026% (0341 “mm = ZO‘H’l S’le’t‘glb’t \O.’l§a"2i0: it! MW e = (04 .0997
mm \°o HD.M\:M15Ma fsl’lt‘l' : W33: g ‘3‘
pm: at term two Woe me. 1H1 1"""' 1‘.
«Im‘ ' 1 1 54 Appendix B Tabies TABLE VI Critical Values oft \7 ft!)
la
r
0 ta
Degrees of Freedom (100 («so [.025 1.010 (005 [.00] (0005 I 3.078 6.314 63.657 318.31 636.62
2 1.886 2.920 9.925 22.326 31.598
3 1.638 2.353 5.841 10.213 12.924
4 1.533 2.132 4.604 7.173 8.610
5 1 .476 2.015 4.032 5.893 6.869
6 1.440 1.943 3.707 5.208 5.959
7 1.415 1.895 3.499 4.785 5.408
8 1.397 1.860 3.355 4.501 5.041
9 1.383 1.833 3.250 4.297 4.781
10 1.372 1.812 3.169 4.144 4.587
11 1.363 1.796 3.106 4.025 4.437
12 1 .356 1.782 3.055 3.930 4.318
13 1.350 1.771 3.012 3.852 4.221
14 1.345 ._ 1.761 2.977 3.787 4.140
15 1 .341 1.753 2.947 3.733 4.073
16 1.337 1.746 2.921 3.686 4.015
17 1.333 1.740 2.898 3.646 3.965
18 1.330 1.734 2.878 3.610 3.932
19 1.328 1.729 2.861 3.579 3.888
20 1.325 1.725 2.845 3.552 3.850
21 1.323 1.721 2.831 3.527 3.819
22 1.321 1.717 2.819 3.505 3.793
23 1.319 1.714 2.807 3.485 3.76‘
24 1.318 1.711 2.797 3.467 3.745
25 1.316 1.708 2.787 3.450 3.735
26 1.315 1.706 2.779 3.435 3.70"
27 1.314 1.703 2.771 3.421 3.69:!
28 1.313 1.701 2.763 3.408 3.671
29 1.311 1.699 2.756 3.396 3.659
30 1.310 1.697 2.750 3.385 3.646
40 1.303 1.684 2.704 3.307 3.551
60 1.296 1.671 2.660 3.232 3.460
120 1.289 1.658 2.617 3.160 33‘;
00 1.282 1.645 2.576 3.090 3.39] Source: This table is reproduced with the kind permission of the Trustees of Biometrika froth E. S. Pearson and H. O. Hartley (eds).
The Biometriku Tables for Staristicians, V01. 1, 3d ed., Biomelrika, 1966. .\ Appendix B Tables 1163
TABLE XI Percentage Points of the Fdistrib ution, a = .0] 1
i 1; 99.25 99.30 99.33 99 36 99.37
4 2] . . . 28.71, 824 27.91 27.67 27.49 77.35 .20 18.00 16.69 15.98 15.52 15.21 14.98 14.80 14.66
5 16.26 13.27 12.06 11.39 10.97 10.67 10.46 10.29 10.16
6 13.75 10.92 9.78 9.15 8.75 8.47 8.26 8.10 7.98
7 12.25 9.55 8.45 7.85 7.46 7.19 6.99 6.84 6.77
8 11.26 8.65 7.59 7.01 6.63 6.37 6.18 6.03 5.91
9 10.56 8.02 6.99 6.42 6.06 5.80 5.61 5.47 5.35 10 10.04 7.56 6.55 5.99 5.64 5.39 5.20 5.06 4.94 11 9.65 7.21 6.22 5.67 5.32 5.07 4.89 4.74 4.63 12 9.33 6.93 5.95 5.41 5.06 4.82 4.64 4.50 4.39 13 9.07 6.70 5.74 5.21 4.86 4.62 4.44 4.30 4.19 14 8.86 6.51 5.56 5.04 4.69 4.46 4.28 4.14 4.03 15 8.68 6.36 5.42 4.89 4.56 4.32 4.14 4.00 3.89 16 8.53 6.23 5.29 4.77 4.44 4.20 4.03 3.89 3.78 17 8.40 6.11 5.18 4.67 4.34 4.10 3.93 3.79 3.68 18 8.29 6.01 5.09 4.58 4.25 4.01 3.84 3.71 3.60 19 8.18 5.93 5.01 4.50 4.17 3.94 3.77 3.63 3.52 20 8.10 5.85 4.94 4.43 4.10 3.87 3.70 3.56 3.46 21 8.02 5.78 4.87 4.37 4.04 3.81 3.64 3.51 3.40 22 7.95 5.72 4.82 4.31 3.99 3.76 3.59 3.45 3.35 23 7.88 5.66 4.76 4.26 3.94 3.71 3.54 3.41 3.30 24 7.82 5.61 4.72 4.22 3.90 3.67 3.50 3.36 3.26 25 7.77 5.57 4.68 4.18 3.85 3.63 3.46 3.32 3.22 26 7.72 5.53 4.64 4.14 3.82 3.59 3.42 3.29 3.18 27 7.68 5.49 4.60 4.11 3.78 3.56 3.39 3.26 3.15 28 7.64 5.45 4.57 4.07 3.75 3.53 3.36 3.23 3.12 29 7.60 5.42 4.54 4.04 3.73 3.50 3.33 3.20 3.09 30 7.56 5.39 4.51 4.02 3.70 3.47 3.30 3.17 3.07 40 7.31 5.18 4.31 3.83 3.51 3.29 3.12 2.99 2.89 60 7.08 4.98 4.13 3.65 3.34 3.12 2.95 2.82 2.72 120 6.85 4.79 3.95 3.48 3.17 2.96 2.79 2.66v ‘ 2.56 00 6.63 4.61 3.78 3.32 3.02 2.80 2.64 2.5] Source: From .Vl. Merringlon and C. M.’lhompson."Tablcs 01‘ Percentage Points of the Inverted Bela (F)Dis1rihulion,“ Bionwlrikn. 1943.
33. "/3458. Reproduced by permission of the Biometriku Trustees. DENOMINATOR DEGREES OF FREEDOM (continued) — \
or = .01
l) Fm F
v] NUMERATOR DEGREES 0F FREEDOM v ———ll 9 ...
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This homework help was uploaded on 02/13/2008 for the course STAT 252 taught by Professor Staff during the Winter '05 term at Cal Poly.
 Winter '05
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