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Unformatted text preview: 147 Midterm Exam 2
Statistics 252 Version 1, Winter 2007
February 15, 2006 Name u”, l“..  . . 7:6"?!
Please show all work to receive full credit. There are 4 problems, each With multiple parts. The total
number of possible points is 60. Read all the given information before attempting a problem. Please write
your answers neatly. You will not receive credit ifyour handwriting is illegible. Good luck! 1) ls income related to happiness? In large city, 85 households were randomly selected, and the
heads ofthe household were classiﬁed by their income level (above or below the US. median
income level), and by their selfreported state of happiness (not happy and happy). Note: Shown
in the table below are the observed counts obtained from the survey, and each cell’s contribution to the chisquare test statistic (in oarentheses): __'.lI!‘.l' (1.506) (.627)
Below Median 15 20
(2.15l) (.896) 5
__—m ’3 a) State the null and alternative hypotheses necessary to determine whether there is an
association between income and happiness. (3 points) “024th wwmess AM mwme \wu WW \MMC 0% «mac _ . Wmswws me \mwcmcwr /
“N w mwmesg and \mwm \WW 0% W \ncms of “Meet: Noam 0mg owe txepender‘w. Income b) Using the information in the table, calculate the appropriate test statistic, make a decision
(justify with a critical value or approximate pvalue), and provide a conclusion in the context ofthe problem. Use a = .05 . (6 points) «it; We. +(40~o.(z<ﬂ)z+ gigglhw
LEW «(OP_ Zti .Cbgia
’ MAW * 2412.491” ”72.75 ’r 407.325 ” 02004 / we reﬁt \l. on “=06;de
7942/ 7C7“ VX1455 commits mar 4cm W mess 7’ 0V3
X .05 $5 '%+\4(0 Wad \HLOYY‘L’. \QW\ of We ad?
c) lfthe number of heads ofhousehold who are happy and have incomes above the median is
Z— changed from 40 to 4, is it still valid to use the chisquare test for independence? Why or why
not? Justify your answer numerically. (5 points) “At woum em we we we dust wouulm NW
C same. $0me Sac. flag; 41.407 ”r 1‘5W9%’r%.’l‘514o , 16 — 650le75
38%;) W0 W0\l\c\ SYN chcr Ho I\.) 2) The following scatterplot, and MlNlTAB output are based on a random sample ofSO homes that
were sold in a midwestern region ofthe United States. Deﬁne x = size ofthe lot on which the home is situated (in 1000’s of square feet), and y = size ofthe home in square feet, Note that lot sizes range from 3,500 to 41,000 square feet, and home sizes range from 370 to 3,] 10 square feet.
Some useful MlNlTAB output is given below: Predictor Coef SE Coef T P
6~cOnstant 855.7 175.0 4.89 0.000
Pu Lot Size 30.119 8.255 3.65 0.001 s = 453.412 R—Sq = 21.7% RSq(adj) = 20.1% a) Calculate the correlation coefﬁcient. Based on its value, brieﬂy describe the relationship between
H size of the lot and size of the home. (4 points) Y‘W‘ 0.465% TVWS Snows A fmvw weak (>08th
mammmwwm x and \/ b) Write out the equation ofthe least squares regression line. (3 points) Q= 93957 +30 Mx / g c) Give precise interpretations ofthe estimated slope and yintercept ofthe regression line in the
I\ context ofthe roblem. (6 oin
p (sf 03mg; ‘0964 is the M010 Sllﬁ of Memes when we \or size
\s M \T‘S SWmlCS’r (erbvf’n‘caliy, o greet) M W my 4 swaéi‘ 160M \8 Thﬁ'ﬁii'xoum of smurf ftﬁﬁ house inwmgm
for man Mdmoml ,0;qu {oat of ma \bT. L000
d) Construct a 95% conﬁdence interval for the true slope ofthe regression line, and give an Wm Based on your answer, would you
conclude there is a signiﬁcant linear relationship between size ofa home and size ofthe lot? Why
or wh not? (8 oints "  y " ’ see  —~§—— = <23. 25 5 0:" “ in 8" ‘ ES“
ﬁt 3?— i tog/feels = 2 ~D7/\
303‘" i Z ilﬂlUVﬁS} 0t= 4% .9 Younclcd ”to 40
57> (i5 435/40. «2303/ 1593. WWW Jmcre \g A eqmﬁcam \mecw mammlmp
WW the size of a mom: and me mi of
JrW’ \ot DCCQUSC Amos \mewm \s so gone forth ‘ ‘ D\N how .
W08 comma \inewm Wlmdﬂ gin \p Wﬂn‘ﬂowgﬁ UNONQ 1’ OW Y(6\\\Y lS 3W 0 O
AWE , 3) A magazine polled 40 of its readers concerning which of the three vegetables brussel sprouts, La S e 61691993: “l 0%: (term)2 1 (323011 .. \avz. 1 a 075+ 1 +02%
a Mo \‘L En: (QC4) 3 \‘7’ okra, and lima beans  is their least favorite. The results ofthe survey are displayed in the table below. Let pl , p2 , and p3 represent the proportion of all readers who indicated brussel sprouts,
okra, and lima beans, respectively, as their least favorite vegetable. brusselsrouts lima beans 15 12 I3 . . . \ . 2 5 . . a) Suppose it IS believed tha‘ﬁhe proportions of readers hose least favorite vegetable IS brussel
sprouts, okra, and lima beans are .30, .40, and .30, respectively. State the null and alternative
hypotheses to test this claim. (4 points) Ho= \%=o.2>, \1=o‘t , \7>=o.a .
the m \CQS’V two Qvoqomons are not mum to «Mir “\IPDWfSllEel Vanna. b) Now conduct the test for the claim given above. Calculate the appropriate test statistic, make
a decision (justify with a critical value or approximate pvalue), and provide a conclusion ii
the context 9: the problem. You can use the following MlNlTAB output to determine your test statistic. Use a = .01 . (6 points) Contribution
Category Observed Expected to Chi—Sq
1 15 12 0.75000
2 12 16 1.00000
3 13 12 0.08333 EWWX'Q'O'“ 9&0, = 4 .th / \.<b%2$4.2ioa'+ Will ”to woo MW
We 0 swap/i=7, 4) Four brands of batteries are under study: Brand 1, Brand 2, Brand 3, and Brand 4. We are W‘ﬁﬂg—C L3 interested in determining ifthe average lives (in hours) ofthe four brands are different. Five {MVOY \‘Tf Wt“ W
variance was performed. ' .10 batteries of each brand were randomly selected and the lifetimes were recorded, and an analysis 0 a) Deﬁne (in symbols and words) the appropriate population means that are relevant to this experiment, AND then state the appropriate null and alternative hypotheses for performing the
ANOVA test. (3 points) Mr‘ MQYMC Me of wmricu WWW 3&0: Nu: M1: Ma, AM Mr WNW); \itc of brand 2 mum \ ‘
My ”WWW: “t6 0t pram 93 Wm +09 M s out“tea nAxm \(M’r mo of mum \Vtt’ {rt vmm 4' WM /4 =\.aa%// /
th’ﬁ tto m ol=o.0\ and commde Aﬂoat AT WU W090Yﬂ00€ AXE n tam mum to mm in ' simmer
At em 0 W «gamma pouea {w \gtDroxﬁsj 5? \an Wm: NC 000%th mH’n *Wtc provomon btgygpgggts b) Suppose that the evidence was strong enough to reject the null hypothesis at the .0] level.
State a conclusion in the context of the problem. (3 points) we COHCNQQ. at we ot=o.o\ mm 4mm at WOW
mo 06 wig Worms. or Vaﬁtnes have chttfrm W, Me gyms. H c) The sample means for each brand and the Tukey—Kramer output from MINlTAB are \J presented below. Construct a simpliﬁed graph that quickly explains these results AND
provide a brief interpretation of the results. (6 points) Wart“ ”Han e\ 2 , @000 a gamma 4 = 6'70 £74.40 Ag=l00~4ro M5 wisao
\NQ COnCM‘t WYﬁgg \Uig Significantly d’til‘crm’r (from b‘mﬁa 2 pm
not Wane: 3. or +5 lovaml 2. \s r mgmfmnﬂy dﬁtererrt l>rom Wands 3mm 45 WW“ 3 \3 “OT evtmﬁcamw ditterent from brand 4 3 d) True or’ircle the appropriate answer): If we ,had {101 rejected the null hypothesis, then
we shou 0 ill perform multiple comparison procedures. (3 points) can omv perform WY Kramer
Level Mean StDeV WYWH W9 \((5€C\— JerL YlUll \(WVWSK v N
Brand 1 5 95.20 3.35
Brand 2 5 79.40 3.85
Brand 3 5 100.40 4.56
Brand 4 5 95.20 4.09 Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons Individual confidence level = 98.87% Brand 1 subtracted from: Lower Center Upper ————————— + ————————— + ————————— + _________ +
Brand 2 —23.017 —15.800 —8.583 (———*————)
Brand 3 2.017 §I2 I 2,417, (__*____)
Brand 4 —7.217 0.009 7.217) (———*———)
————————— +————————+—————————+—————————+
—15 0 15 30
Brand 2 subtracted from:
Lower Center Upper ————————— + ————————— + ———————— + ————————— +
Brand 3 13.783 21.000 28.217 (—_——*____)
Brand 4 8.583 15.800 23.017 (————*———)
————————— +—————————+—————————+—————————+
—15 O 15 30 Brand 3 subtracted from: Lower Center Upper ————————— + ————————— + ————————— + ————————— + Brand 4 12.417 —5.ZQO 2.017, (*) 1156 Appendix B Tables TABLE VII Continued Degrees of
Freedom X .2050 X1125 X1010 X .2005
1— —4
1 2.70554 3.84146 . 5.02389 6.63490 ' 7.87944
2 4.60517 5.99147 7.37776 9.21034‘ 10.5966
3 6.25139 7.81473 9.34840 11.3449 12.8381
4 7.77944 9.48773 11.1433 13.2767 14.8602
5 9.23635 1 1.0705 12.8325 15.0863 16.7496
6 10.6446 12.5916 14.4494 16.81 19 18.5476
7 12.0170 14.0671 16.0128 18.4753 20.2777
8 13.3616 15.5073 17.5346 20.0902 21.9550
9 14.6837 16.9190 19.0228 21.6660 23.5893
10 15.9871 18.3070 20.4831 23.2093 25.1882
11 17.2750 19.6751 21.9200 24.7250 26.7569
12 18.5494 21.0261 23.3367 26.2170 28.2995
13 19.8119 22.3621 24.7356 27.6883 29.8194
14 21.0642 23.6848 26.1190 29.1413 31.3193
15 22.3072 24.9958 27.4884 30.5779 32.8013 1
16 23.5418 26.2962 28.8454 31.9999 34.2672
17 24.7690 27.5871 30.1910 33.4087 35.7185
18 25.9894 28.8693 31.5264 34.8053 37.1564
19 27.2036 30.1435 32.8523 36.1908 38.5822
20 28.4120 31.4104 34.1696 37.5662 39.9968
21 29.6151 32.6705 35.4789 38.9321 41.4010
22 30.8133 33.9244 36.7807 40.2894 42.7956
23 32.0069 35.1725 38.0757 41.6384 44.1813
24 33.1963 36.4151 39.3641 42.9798 45.5585
25 34.3816 37.6525 40.6465 44.3141 46.9278
26 35.5631 38.8852 41.9232 45.6417 48.2899
27 36.7412 40.1133 43.1944 46.9630 49.6449
28 37.9159 41.3372 44.4607 48.2782 50. 33
29 39.0875 42.5569 45.7222 49.5879 52.3356
30 40.2560 43.7729 46.9792 50.8922 53.6720
40 51.8050 55.7585 59.3417 63.6907 66.7659
50 63.1671 67.5048 71.4202 76.1539 79.4911!
60 74.3970 79.0819 83.2976 88.3794 91.951“
70 85.5271 90.5312 95.0231 100.425 104.215
80 96.5782 101.879 106.629 112.329 116.321
90 107.565 113.145 118.136 124.116 128.299
100 118.498 124.342 129.561 135.807 140.169 1154 Appendix 8 Tables
TABLE VI Critical Values of t “M f0)
0:
J
I
1) 1“
Degrees of
“9990'“ {100 1.050 1.025 1.010 (.005 1.001 1.0005 1 3.078 6.314 12.706 31.821 63.657 318.31 636.62 2 1.886 . 2.920 4.303 6.965 9.925 22.326 31.598 3 1.638 2.353 3.182 4.541 5.841 10.213 12.924 4 1.533 2.132 2.776 3.747 4.604 7.173 8.610 5 1.476 2.015 2.571 3.365 4.032 5.893 6.869 6 1.440 1.943 2.447 3.143 3.707 5.208 5.959 7 1.415 1.895 2.365 2.998 3.499 4.785 5.408 8 1.397 1.860 2.306 2.896 3.355 4.501 5.041 9 1.383 1.833 2.262 2.821 3.250 4.297 4.781
10 1.372 1.812 2.228 2.764 3.169 4.144 4.587
11 1.363 1.796 2.201 2.718 3.106 4.025 4.437
12 1.356 1.782 2.179 2.681 3.055 3.930 4.318
13 1.350 1.771 2.160 2.650 3.012 3.852 4.221
14 1.345 1.761 2.145 2.624 2.977 3.787 4.140 J
15 1.341 1.753 2.131 2.602 2.947 3.733 4.073
16 1.337 1.746 2.120 2.583 2.921 3.686 4.015
17 1.333 1.740 2.1 10 2.567 2.898 3.646 396‘
18 1.330 1.734 2.101 2.552 2.878 3.610 3.92:
19 1.328 1.729 2.093 2.539 , 2.861 3.579 3.883
20 1.325 1.725 2.086 2.528 2.845 3.552 3.850
21 1.323 1.721 2.080 2.518 2.831 3.527 3.819 v
22 1.321 1.717 2.074 2.508 2.819 3.505 3.793
23 1.319 1.714 2.069 2.500 2.807 3.485 3.767
24 1.318 1.711 2.064 2.492 2.797 3.467 3.745
25 1.316 1.708 2.060 2.485 2.787 3.450 3.725
26 1.315 1.706 2.056 2.479 2.779 3.435 3.70.
27 1.314 1.703 2.052 2.473 2.771 3.421 3.6% 1
28 1.313 1.701 2.048 2.467 2.763 3.408 3.671
29 1.311 1.699 2.045 2.462 2.756 3.396 3.650
30 1.310 1.697 2.042 2.457 2.750 3.385 3.636
40 1.303 1.684 2.021 2.423 2.704 3.307 3.551
60 1.296 1.671 2.000 2.390 2.660 3.232 . 3.4“. 120 1.289 1.658 1.980 2.358 2.617 3.160 3.373 00 1.282 1.645 1.960 2.326 2.576 3.090 L 3.39] Snun’c; This table is reproduced with the kind permission of the'I'ruslccs of Bionrclrika from [i. 8. Pearson and 11.0. Hartley (eds).
The Biomrm‘ka ’I'ahlcs for Srarisricians, V01. 1, 3d 011., Biumclrika, 1966. ...
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