stat252 wint07 midterm1

stat252 wint07 midterm1 - Midterm Exam 1 Statistics 252...

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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 t-test 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, Jr-tes-t 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 l-Sibling 2) information described in the problem. Should we use an independ t samples t—test or a paired differences t-test? 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 emu-findii W (M Ho" M (M 1 different. lfyou are going to use an independent samples t-test, 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— A/WIVT; A ‘ ‘0 ? Lag? ‘ (film) sen, smce ‘Hflf \ovmu @400 €a=43>5 _ 0.017404 WK V 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\ 1 L g fl?” W awva Tet": «Somme “WW? CMHdreo fund commwm \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; Xi—Xz lid}; 41...:— \o \O . / 0(‘0'0‘ of r NH- com 4mm— 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’a-ZEO 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 mam-WM that; '- SYYlDYXYS 0K n ‘2 (p n 2‘; {p b) Suppose we want to determine ifnon-smokers 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 (02972-62241 Ilenm = warn: 77.0761?) (‘73-4l4‘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 non-smokers 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 non-smokers and all light smokers. -d.<‘ln repeated sampling, 90% of similarly constructed intervals would contain the differences in the sample mean heart rates of non-smokers and light smokers. @ ln repeated sampling, 90% ofsimilarly constructed intervals would contain the true difference between the average heart rates of all non-smokers and all light smokers. H d) Now we would like to perform one-way 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'\rlc5-S: 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‘g-lb’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 F-distrib 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 ———l-l- 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.

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stat252 wint07 midterm1 - Midterm Exam 1 Statistics 252...

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