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AP Statistics 13.1.

Apr 12, 2018

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Page 1: AP Statistics 13.1.

7/21/2019 AP Statistics 13.1.

http://slidepdf.com/reader/full/ap-statistics-131 1/22

AP Statistics 13.1

 To Conduct a 2 Sample t Test

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 Two Sample Problems

•  The goal of this inference is tocompare the responses to 2treatments or to compare thecharacteristics of 2 populations.

• We have a separate sample fromeach treatment or each population.

•  The responses of each group areindependent of those in the othergroup.

• P. !" 13.3

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2 sample tests

• 2 Sample tests often are e#perimentswith 2 independent groups $ notmatched pairs.

• Sample si%es do not have to bee&ual'

•  T(pical 2 sample problems P. !1e#ample 13.1

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)otation

Parameter Statistics

Population

*ariable

+ean Standard,eviation

SampleSi%e

+ean Standard,eviation

1 #1 -1 1 n1 s1

2 #2 -2 2 n2 s2

1 x

2 x

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)ull and Alternative/(pothesis• 0n our h(pothesis tests we are

comparing the means of the twogroups

• /o -1  -2 or -1 4 -2  56 there is no

di7erence in their means

• /a -

1 ≠ -

2 or -

1 4 -

2≠ 5

-1 8 -2  or -1 4 -2 8 5

-1 9 -2  or -1 4 -2 9 5 .

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,oes Calcium :educe ;lood

Pressure<

• A stud( on 21 blac= men wasconducted. 15 of these men wererandoml( assigned to get a calcium

pill and 11 were randoml( assignedto get a placebo.

•  The variable of interest was the

decrease in s(stolic blood pressure.• A > value meant a decrease in ;P a

$ value meant an increase in ;P.

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 Test :esults

• ?roup 1 calcium 4@1!1434"1151142

• ?roup 2 placebo 4112414334""24114143

• Summar( Stats

• 0s this outcome good evidence that calcium

decreases ;P in the entire population ofhealth( blac= men more than a placebo does<

?roup Treatment

n s

1 Calcium 15 ".55 !.@3

2 Placebo 11 45.23 ".51

 x

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Bnce Again

• State the null and alternativeh(pothesis

• Chec= the conditions

• S:S

• 0ndependence

)ormalit( of two samples how is thishandled<D +ore about this later.

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/old on a minute

• ;efore we calculate t here are somefacts about the sampling distribution ofthe di7erence

•  The mean of is -1 $ -2  theunbiased estimator.

•  The variance of the di7erence of is

• 0f the 2 populations are normal thenis also normal

21  x x   −

21   x x   −

21  x x   −

2

2

2

1

2

1

nn

σ  σ  +

21  x x   −

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;ut do we ever reall( =nowσ

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B+? how do we calculate ,

of E<

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:eall( how do we calculate, of E<• Fse the smaller of n1 $ 1D or n2 $ 1D or

• Bh Gust use the calculator'

• :un calculator :un' Eind p value t value

d of f ,:AW A P0CTF:H• Conclusion in conte#t of the problem'

• Create a t interval for the data using a 5I

CJ and interpret in conte#t of the problem• As a h(pothesis test are the results the

same as before< Wh( or wh( not<

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:obustness

• Sample si%e strongl( inKuences p valuesof test.

• When the si%es of the 2 samples are

e&ual and the 2 populations beingcompared have distributions with similarshapes then p values are accurate with

n1  n2  ". Small samples• 0n planning a stud( tr( to ma=e n1  n2 .

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+ore on conditions for tprocedure• Fse t procedures if

• n1 > n2 8 35 no problems

•35 L n1 > n2 L 1" loo= foroutliers and severe s=ewness

• 1" L n1 > n2  be ver( war( of

s=ewness and outliers'

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Hver(one out of the pool

• Briginall( when two populations hadthe same variance (ou could MpoolN oraverage the 2 variables. Advantage

was the degrees of freedom n1 > n2 42 and it made calculations easier.

• ,onOt pool unless (ou are as=ed to doso. 0f (ou pool (ou will have to Gustif(

(our pooling this can be done b(stating the variances appear e&ualD.