MATH 376 -- Probability and Statistics II

Lab Project 4 Solutions

April 4, 2006

A)

>	read "/home/fac/little/public_html/ProbStat0506/MSP.map";

The data on breaking strength for the two glue application processes:

>	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    = .01,  the rejection region is   z >  = 2.326

>	1 - NormalCDF(0,1,2.326);

>	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    = .01,  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)  The data for the male and female spiders:

>	XListB:=[20.4,21.7,21.9,21.4,21.1,23.6,18.9,22.6,21.3];

>	nops(XListB);

>	XbarB:=Mean(XListB);

>	SB2[X]:=Variance(XListB);

>	YListB:=[20.5,20.4,20.3,21.1,21.2,20.9,21.0,21.3,20.9,20.0,20.4,20.8,20.3];

>	nops(YListB);

>	YbarB:=Mean(YListB);

>	SB2[Y]:=Variance(YListB);

With     being  

>	Fval:=SB2[X]/SB2[Y];

For a level   two-tail F-test (for instance)

>	1-FCDF(8,12,3.5114);

>	1-FCDF(8,12,.2381);

RR:  all F with  F < .2381 or  F > 3.5114 (the union of two intervals).

Here   F = 10.58  is  in RR, so we reject the null hypothesis on the basis of this data.

There is evidence to suggest that the variances are different.

The attained significance level is

>	p:=2*(1 - FCDF(8,12,Fval));

Note:  Must multiply by 2 here since this is a two-tail test.

2)  Since we cannot assume the variances are the same, we use the

t-test for equality of means with the Welch variant for the number of d.f.

>	Welch:=trunc((SB2[X]/9+SB2[Y]/13)^2/((SB2[X]/9)^2/8+(SB2[Y]/13)^2/12));

The test statistic value:

>	T:=evalf((XbarB-YbarB)/(sqrt(SB2[X]/9+SB2[Y]/13)));

The p-value for the test on the means is:

>	2*(1-TCDF(Welch,T));

So we cannot reject null hypothesis ( ) on the basis of this data ( p -value is too large).  The

conclusion is that the mean lengths could be the same based on this data.  

3)  The box-whisker plot confirms our conclusions in parts 1 and 2:

Note that the males (the bottom) have much more variance than the females.

But the medians are not that far apart, so the means are probably not

that different either:

>	BoxWhisker(XListB,YListB);