MATH 376 -- Probability and Statistics II
Lab Project 3 Solutions
April 14, 2004
A)
> | read "/home/fac/little/public_html/ProbStat/MaplePackage/MSP.map"; |
Warning, the name changecoords has been redefined
> | XList:=evalf([1250,1210,990,1310,1320,1200,1290,1360,1200,1150,1120,1360,1310,1110,1320,980,950,1430,1100,1080,960,1050,1310,1240,1420,1170,1470,1060,1230,1300]); |
> | Xbar:=Mean(XList); |
> | S[X]:=StandardDeviation(XList); |
> | YList:=evalf([1180,1360,1310,1190,920,1060,1440,1010,1000,950,1310,980,1310,1030,960,800,1280,1080,900,1030,930,1050,1010,1310,940,860,1450,1070,840,1100]); |
> | Ybar:=Mean(YList); |
> | S[Y]:=StandardDeviation(YList); |
1)
> | BoxWhisker(XList,YList); |
Here XList (new process) is the bottom, YList (old process) is the top.
The median and 25th %ile are definitely greater for the new process --
guess that
Thus, our null hypothesis is
:
; alternative hypothesis
is
> | nops(XList),nops(YList); |
so a large sample Z-test is appropriate.
Z =
With
, the rejection region is
z >
= 1.645
> | z:=(Xbar-Ybar)/sqrt(S[X]^2/30+S[Y]^2/30); |
This is well into the rejection region, so we reject the null hypothesis
at this level.
c) With
, the data indicate we have good reason to
reject the null hypothesis in favor of the alternative hypothesis
d) The attained significance level is:
> | p:=1-NormalCDF(0,1,z); |
B)
1)
> | XListB:=[21.7,21.0,21.2,20.7,20.4,21.9,20.2,21.6,20.6]; |
> | nops(XListB); |
> | XbarB:=Mean(XListB); |
> | SB2[X]:=Variance(XListB); |
> | YListB:=[21.5,20.5,20.3,21.6,21.7,21.3,23.0,21.3,18.9,20.0,20.4,20.8,20.3]; |
> | nops(YListB); |
> | YbarB:=Mean(YListB); |
> | SB2[Y]:=Variance(YListB); |
With
being
> | F:=8*SB2[X]/(12*SB2[Y]); |
For a level
two-tail F-test (for instance)
> | 1-FCDF(8,12,3.5114); |
> | 1-FCDF(8,12,.2381); |
RR: F > 3.5114 or F < .2381.
Here F = .24329 is not in RR, so we do not reject the null hypothesis on the basis of this data.
2) Since we can work under the assumption that the variances are equal, we use the
t-test for equality of means:
> | S[P]:=sqrt((8*SB2[X]+12*SB2[Y])/20); |
> | T:=evalf((XbarB-YbarB)/(S[P]*sqrt(1/9+1/13))); |
Rejection region, though, is
>
For instance, for
two-tailed t-test
> | 1-TCDF(20,2.0859); |
RR:
> 2.0859
So cannot reject null hypothesis on the basis of this data.
3)
> | BoxWhisker(XListB,YListB); |
> | Welch:=(SB2[X]/9+SB2[Y]/13)^2/((SB2[X]/9)^2/8+(SB2[Y]/13)^2/12); |
> |