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Unformatted text preview: 10.4 30 = 5063460610, = —94l.900226, 32 = —429.060418
,1? = 506.346 — 941.90011 ~ 429.0601:2
SSE 2 15101592375. MSE = 8883271781,: = Root MSE = 94.25114 We expect about 95%. of the y—values to fall within :25 or 1294251141) or i188.50228
units of the fitted regression equation. 0
0 Hil H03 1
Ha 13‘
: 131 B —0 1
.1 = 941.900226 = _3_424 531 27508555925 The test statistic is t = The rejecrion region requires 0:12 = .0512 = .025 in each tail of the t distribution with df =
H — {k + 1) = 20 — (2 + 1) = 17. From Table VI, Appendix B. {035 “I 2.110. The
rejection region is t < 2.110 or t > 2.110. Since the observed value of the test statisric falls in the rejection region (I = —3.424 <
—2.110‘J. H0 is rejected. There is sufﬁcient evidence to indicate 131 i 0 at or = .05. For conﬁdence coefﬁcient .95. tr 2 .05 and 00"2 = .025. From Table VI, Appendix B, with
df = n — (it + l) = 20 — (2 + 1) = 1?, {.025 = 2.110. The 95% conﬁdence interval is: A 133 4; 1.02553? =4 —429.060 : 2.110(379.3257)=~ —429.060 i 801.4322
= {—12304922, 322.3222) We are given 131 3.1, $31 = 2.3, and n = 25. Ha: 13'! > 0
f3 — 0
The test statiStic is I = 1‘ = .221 = 1.35
.53] _.3 The rejection region requires at : .05 in the upper tail of the t distribution with (if = n — (k + 1) = 25 ~ {2 + l) = 22. From Table VI, Appendix B, {05 = 1.717. The
rejection region is r > 1.21?. Since the observed value of the test statistic does not fall in the rejection region it = 1.35 r»
1.212}. H0 is not rejected. There is insufﬁcient evidence to indicate .131 > 0 at o: = .05. We are given 152 = .92, 5&2 = .27, and n = 25. H0: '82 =7 0
H3: ,82 i 0
" — 0
The tesr statistic is t = 62 = 2 = 3.41 $52 .27 10.10 a. The rejection region requires m2 = .05f2 = .025 in each tail of the t distribution with df
= n — (k + l} = 25 — (2 + 1) = 22. From Table VI, Appendix B, {025 = 2.074. The
rejection region is t < —2.074 or t > 2.074. Since the observed value of the test statistic falls in the rejection region (I = 3.41 > 2.074;.
reject HO. There is sufﬁcient evidence to indicate 32 9t 0 at or = .05. For conﬁdence coefficient .90, a = 1 — .90 = .10 and M2 = .1012 = .05. From Table \'1. Appendix B, with df = n  (k + I) = 25 — (2 + l) = 22, :05 = 1.717. The conﬁdentt.
interval is: £31 i r0555] = 3.1 i 19179.3): 3.1 i 3.949=:(~.s49,7.049) We are 90% conﬁdent that ,8, falls between —.849 and 7.049. For conﬁdence coefﬁcient .99, o: = I — .99 = .01 and eti'2 = .OUZ = .005. From Table
VI, Appendix B, with at = n  (k + l) = 25 — (2 + 1) = 22. (005 = 2.319. The
conﬁdence interval is: 332 i rows; 3 .92 i 2.819(.27)= .92 i .T6l=>(.159. 1.681) We are 99% conﬁdent that 52 falls between .159 and 1.68]. To determine if the AD score is positively related to assertiveness level, once age and length
of disability are accounted for, we test: H0: .81 = 0 Ha: :31 > 0 The test Statistic is t = 5.96. The pvalue is p = .0001t2 = .00005. Since the pvalue is iess than or (p = .00005 < .05].
H0 is rejected. There is sufﬁcient evidence to indicate that the AD score is positively related
to assertiveness level, once age and length of disability are accounted for with a = .05. To determine if age is related to assertiveness level, once AD score and length of disability are
accounted for, we test: H0: ,82 z 0 Ha: .82 75 0 The test statistic is I = 0.01. The pvalue is p = .9620. Since the p—value is greater than a (p = .9620 :t .05), H0 is not
rejeCted. There is insufﬁcient evidence to indicate that Age score is related to assertiveness
level, once AD score and length of disability are accounted for with a = .05. To determine if length of disability is positively related to assertiveness level. once AD scon
and age are accounted for. we test: H0: 63 : 0 Ha: .83 > 0 The test statistic is t = 1.91. The pvalue is p = .05T6i’2 = .0288. Since the pvalue is less than 0: (p = .0288 < .053. it
is rejected. There is sufﬁcient evidence to indicate that length of disability is positively rclznm
to assertiveness level, once the AD score and age are accounted for with o: = .05. 10.14 a. 50’) I 130 '1' 51x1 + 132752 + 5313 1“ 134154 + 5&‘5 1' 3676 + 4319‘?
Using MINITAB, the output is: The regression equation is v = 0.998  0.0221. x1 + 0.156x2 — 0.0172x3  0.009533% + 0.421x5
+ 0.417x6 — 0.155s? Predictor Coef StDev T P
Constant 0.9981 0.2475 4.03 0.002
x1 4.022429 0.005039 4.45 0.001
x2 0.15571 0.07429 2.10 0.060
x3 —0.01719 0.01136 —1.45 0.175
x4 0.009527 0.009619 0.99 0.343
)6 0.4214 0.1003 4.18 0.002
x6 0.4171 0.5.377 0.95 0.361
x? ~0.1552 0.1436 1.04 0.319
S = 0.4365 Rqu = 77.111; RSqtadi) = 62.5% Analysis of Variance Source DF 55 HS F P
Regression ? 7.9578 1.1363 5.29 0.00?
Residual Error 11 2.3632 0.2148 Total 18 10.3210 Source 0F Seq 55 x1 1 1.4016 x2 1 1.9263 x3 1 0.1171 '. at. 1 0.0446 w x5 1 4.0771 :6 1 0.1565 x? 1 0.2345 Unusual Observations Obs x1 y Fit StDev Fit Residual St Resid
14 80.0 0.120 ~0.628 0.328 0.748 2.23R R denotes an observation with a large standardized residual. The least squares model is 5; = .9981 — .0224x1 + 155302 — .0172x3 «.009er4
+ .4214x5 + .417116 #1552», 30 .9981 = the estimate of the y—intercept.
[3’1 = — .0224. We estimate that the mean voltage will decrease by .0224 kwfcm, for each additional increase of 1% of II. the disperse phase volume (with all other variables held
constant). .03 = .1557. We estimate that the mean voltage will increase by .1557 kwlcm for each
additional increase of 1% of x2. the salinity (with all other variables held constant). 03 =  .0172. We eStimate the the mean voltage will decrease by .0172 kwtcm for each additional increase of 1 degree of x3, the temperature in Celsius (with all Other variables held
constant). 64 = * .0095. We estimate that the mean voltage will decrease by .0095 kwfcm for each
additional increase of 1 hour of at... the time delay (with all other variables held constant). 105 = .4214. We estimate that the mean voltage will increase by .4214 kwl’cm for each
additional increase of 1% of 1:5. surﬁcant concentration (with all other variables held constanll. 36 = .4171. We estimate that the mean voltage will increase by .4171 kwl’cm for each
additional increase of 1 unit of 3:6, span: Triton (with all Other variables held constant). 37 = —.1552. We estimate that the mean voltage will decrease by .1552 kwfcm for each
additional increase of 1% of x7, the solid particles (with all other variables held constant). 10.22 a. To determine if crime prevalence is positively related to density. we test: Ha:161> 0 The test statistic is r = 3.88. The pvalue is p '< .019 = .005. Since the pvalue is so small, there is strong evidence to
indicate that the crime prevalence is positively related to density for a > .005. No. The tests are not independent of each other. If we conduct a series of ttests to determine
whether the independent variables are contributing to the predictive relationship, we would very likely make one or more errors in deciding which terms to retain in the model and which
to exclude. To test the utility of the model, we test: H0351=132=53="'=313=0
Ha: At least one 61 ¢ 0,1' = i. 2, 3, , 18 The test statistic is: F = R2”: = .411x13 = 1139.,
(1 _ R2)/[n _ (k + 1)] (1 — .411)f[3l3 — (18 + 1)] The rejection region requires a = .05 in the upper tail of the F distribution with V} = k = 18 and r2 = n e (k + 1) = 313 — (18 + 1) = 294. From Table IX. Appendix B, F05 = 1.57. The rejection region is F > 1.57. Since the observed value of the test statistic falls in the rejection region (F = 11.39? > 1.57}, H0 is rejected. There is sufﬁcient evidence that the model is useful in predicting crime
prevalence at a = .05. ...
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 Spring '07
 Safarzadeh

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