AP Statistics Section 12.1 A

Now that we have looked at the principles of testing claims, we proceed to practice.

We begin by dropping the unrealistic assumption that we know the population standard deviation when testing claims

about a population mean. As with confidence intervals, this leads to the use

of ___ distributions when carrying out significance tests about .

t

One Sample t-TestDraw an SRS of size n from the population.

The one-sample t statistic:

has the t distribution with n – 1 degrees of freedom.

nsx

t 0

There is a slight change in the procedure for computing the p-value. The next examples show

this change.

.05 and .025Between

10000,19)tcdf(1.81,:5 VARS 2:84/83 ndTI

.043

17.3

.0025 and .001Between

.0025) and .001 (2 BetweenvalueP

.005) and .002 BetweenvalueP

004.)002(.2

002.)36,1000,17.3(

valuep

tcdf

These P-values are exact if the population distribution is Normal and are approximately correct for

large n in other cases.

Example: Is 98.6oF Wrong? From a random sample of 106 people, the

mean body temperature was 98.2oF with a standard deviation

of .6229. Test the common belief that the mean body temperature is

98.6oF.

Parameter: The population of interest is all people. We wish to test

Conditions:

raturebody tempemean where6.98:H vs6.98 a:0 H

.population the togeneralize

notmay results SRS,an not ifbut sample Random :SRS

Normal approx. isthat

dist. a gives CLT 106,n With :dist. x ofNormality

10nNt replacemen w/osampling are wesince andt independen

be to temp.individualexpect toReasonable :ceIndependen

Calculations:

Interpretation:

less.

actuallybut 98.6not is body temp.mean thesuggests This

98.6. is mean temp. theassume en wesmaller whor 98.2 of

mean temp. a with withsample a getting of liklihood small

verya indicates .0005, than less value,-p lowOur very

611.6

1066229.

6.982.98

t

001.)0005(.2 valuep

Test-T:2 TESTS STAT :TI83/84

If you were to choose a significance level for this test, what would it be? Why?

.H thedisprove toevidence convincing need we

so belief, heldcommonly a disprove to tryingare We01.

0

Describe a Type I and Type II error in this setting.

is.it fact in when 98.6not ismean that thegDeterminin :I Type

98.6. lower than is

it fact in when 98.6not ismean t thereject tha toFailing :II Type

Construct a 99% confidence interval for the mean body temperature and interpret it.

106

.62292.62698.2

n

stx

)36.98,04.98(

98.36 and 98.04between is population the

for raturebody tempemean that theconfident 99% are We