Summary Maths Stats 245 Part 2

Example




Assuming muzzle velocities are approximately normally distributed




Ho 1 3000 2 0,025

Ha m Rejectionregion te too2s 2.365 idf

30g
T Y Mo observed Tis 2959 3000 2.966
391
Jg


falls inside RR M 3000 Me3000



Whatisthe
p valueforthisexample


PValue P 21205s Ho
P throbs Ho
P TC 2.966 Ho Idf
need to interpolate

PCTC2.3651 0,025

P TC 2.998
interpolating gives
p to 2.9661 0,1046
go


i PValuecommutyfor
2 0,1046 250,1046 Homustbe rejected
any

,Small sampletest to comparetwo means u Ma



Suppose 41,112,42 Yn and XiXaXa Xn are two random samples selected from twonormal

populations Where the mean
andvariance are Mi Mi and 0 03
Assume the sample means and variances are Ya Yaand 53,53


If we use sp ni 1 si na is for thepooled estimator of02then
p n 2



we use thetest statistic T Y Ya Mi ud n
tn tn
sp nt t ha

small sample test comparingtwo means




Anitha 2

,Example




Ho Mi Mz o 905 t 92 to025 2.363

Ha hi Mz yo I RR It 12.365


Test statistic T Y 42 Ca Ma rt Ibdf nitric
sp fi tha


sp ni 1 sit A2 1522 22.24 Sp 4.716
hitN2 2
under

Observed
t Yi Ya Cen nd 35.22 y g
Sp hit the 4.716 It
1.65

Valuedoes not fall in RR donot reject Ho


Findthe Pvaluefortheabove PValue 2P 25 Zobs Ho

LP T 1.65 Ibdf
Linear interpretation
P Ts1.3371 0,1
PCT 1.7461 0,05 gives PCT 1.65 0,0592


26,0592 0,1184 2 0,05 donotrejectto

, Testing hypotheses concerning variances
Assume random sample of Yi Ynfrom a normal distribution withunknownmean u and
unknown variance o

We now lookat testing Ho 0 0 for some fixed value of versus Ha




n Idf




Rejection regions




Notsymmetrical mustuse

of act ora'ssitar




P Values RejectHo if Pvalueis La