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Unformatted text preview: EC 41 UCLA — Sample Problems #9; RE Ch 7 and 10 W I) {an F A pooled two sample t—test assumes the true standard deviations of the two populations are the same. ii) T or®Suppose X is not distributed normally The sampling distribution of X from simple r om samples will approa a normal distribution as the number of samples increases. W09” (M in»? I n when OESFK’V/HTVM ﬁl
iii) T org )lhe Null Hypothesis ofa matched pairs‘Ft‘eEf'isthat the difference (e.g., between a pre and post test) for all obse 'onal units (e.g., teachers) is zero, Ho: xi =0.
ivﬁr F The Null Hypothesis of a two sample ttest assuming unequal variances is usually that the true mean difference between two different populations 1 and 2 is zero, Ho: u}  p2=0; which is equivalent to Ho: pl = u; v r F If I? is further from the value in the null hypothesis, then the pvalue will be smaller, all else constant.
vi r F The pvalue for a two sample ttest with unequal variances given by Excel is typically smaller than that
which we calculate by hand. This may reﬂect the higher degrees of freedom used in Excel’s calculations. 2) A group of people are given tests before and after training. Scores on the pre—test of 4 people randomly selected people
are X. = 30, 30, 40, and 40. Scores on posttests (aﬁer training) for 3 randomly select _ or e to are: X2 = 50, 60, and 70. Assume test scores are normally distributed, but standard deviations may differ. U  2 for the degrees 0 eedom. 33 —[ = 2, a) Form a 95% conﬁdence interval for the difference in population means, it: ~ to ?1_;Ii 751i 5' 5‘ ,> 50—52? I 14.31)") tar. +3” TIT l W; a» 4 m
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$560 s,= m SEW; (.2 1952.79) 64.45249 x) 1)) Perform a Mo sample t—test of the null hypothesis that the scores on averag improve sus an alternative of no
change (equal means). Show your calculated t—test statistic. Are results si ' can at the 5% level? Hi i I“ 1 34/7.
[9 se the smaller of(n.—l) or (mg—l) as the degrees of freedom). 90 5W. 3'1ch #4 by; OH
Do NOT assume the true stande deviation of gorges before is the same as that of the scores after. 2 2 59:23 13.12: i 6:ch p _: EYES 5‘6”!th
FtMlL‘l/lﬂlﬂl ‘Y—Lgl'ab‘? = i 3873 S49; 1420 .DS— ng.01§ @811!)th “5F ”'50? from {niacin/ﬁr ; 0 $— .
3) Recall the data from #9, Sample problems 8: \____—/3’__l_2_32__
Store 1 2 3 4 5 6 7 8 9 10 f“ r 550
Before .10 11 19 12 7 8 5 20 20 14 'ZE‘X’B $3 ‘
After 15 17 13 11 9 15 4 25 26 1919.1)‘24 $4 ., 5.7a a) Find the 95% Conﬁdencc Interval for the mean difference (After  Before) in sales, Assume the 10 stores “after” need
not be the same as “before “ so you do NOT use matched pairs procedure. Deﬁne difference as after minus before, so the mean difference is positive. interval? i £1 b) The connection between sales and individual store was lost/So now we have l0 before observations and 10 aﬂer
observations from two distributions that are approximately normal. Use a two sample Host to ﬁnd a tcmugmd and range for
the pvalue when Ho: sales remained the s ' es increased. As in part “a”, domulation
Stan ar evr ations are t 6 same before and after (so you will not compute a “pooled" estimate of standard deviation). bum: {91191,  Q : 0H1: one Jailed
E32 tMﬁ‘ W #3404)? ll) )0 ANOVA d)_‘ SS MS F Four Observations Regression 1 1122.002 1122.002 10.01707 U.S. com Ln'eld Residual 2 224.018 112.009 Year (X) Yield (Y) Total 3 1346.02 1965 74.1 _ l 975 86.4 Standard 1985 118 Coeﬁicients Error {Star Pvalue 1995 1 13.5 Intercept 2368.04 937.1596 —3.06035 0.092237
Year 1,498 0.473305 3.164976 0.086999 Regression Statistics
R Square 0.83357r Standard Error 10.53343 4) Consider the results above for the regression cj‘k Yield of com (Y) on Year (X).
a) Write the estimated regression equation.
x; — 2 5 6 a +0. 5) )6 GP \/ = ~22691091x 7‘6")
b) Interpret the slope coefﬁcient. What is the relationship between year and yield? date; if) ; (1C: “4.235 ; 3/;le mam/2g, #1 /. 5” each/1411‘. A Ymr
21.; 0) Calculate the sample standard deviations fOr x and y: 5x = i 2 0i 5y: d) What Is the sample correlation coefﬁcient r? (square root R2 is it ositive? can verify with equation on pg.1 14) i 9356‘”? 2,913 2p I'll/“6‘5 i“?— lerg'zcﬂizgj / e) Use the technique on page 591 to calculated a t test statistic 19m for the sample correlation 75. l 6
How does this compare to the ttest statistic on the slope coefﬁcient on the X variable, Year? 0 EH3 We ;
LLCHh W 3/5 t) Interpret this ~ do you reject or accept the null hypothesis that the true correlatiore p= 02]: 0010030 OZ: Ai'v?‘ BEIEH‘ H3" P20 1210190,; ”025 LOW—112:; ”Lewis aliﬂ‘tiaD/My 110 1145/3743 'F 2_q; [1.303 4023 < 9.193%; < .03
‘0’; (P1511611 (10
3,) What 15 the PTGIgCtidm) cro 'Id 111 005" (’a‘l’i l'f’li pm
A = 1295 2.909 141 019m magma ML / :2
0M 0 AL “Hist
45 _ 5315620119) P ~ —;~,09’M44 .097 Mill/Lelia: 6/0/69 hag. harm/MM! om \/ ...
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 Spring '07
 Guggenberger

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