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Summary Maths Stats 245 Part 2
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These notes give a summary of the theory in Maths Stats 245, with worked examples and detailed explanations.
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ExampleAssuming 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
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