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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 ___
t distributions when carrying out
significance tests about .
One Sample t-Test
Draw an SRS of size n from the population.
The one-sample t statistic:
x  0
t
s
n
has the t distribution with n – 1 degrees of
freedom.
There is a slight change in the
procedure for computing the pvalue. The next examples show
this change.
Between .025 and .05
TI 83 / 84 : 2nd VARS 5 : tcdf(1.81,10000,19)
.043
Between .001 and .0025
3.17
P  value  2( Between .001 and .0025)
P  value  Between .002 and .005)
tcdf (3.17,1000,36)  .002
p  value  2(.002)  .004
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
o
98.2 F 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
H 0:  98.6 vs H a :   98.6 where   mean body temperature
Conditions:
SRS : Random sample but if not an SRS, results may not
generalize to the population.
Normality of x dist. : With n  106, CLT gives a dist.
that is approx. Normal
Independence : Reasonable to expect individual temp. to be
independent and since we are sampling w/o replacement N  10n
Calculations:
t
98.2  98.6
 6.611
.6229
106
p  value  2(.0005)  .001
TI83/84 : STAT TESTS
2 : T - Test
Interpretation:
Our very low p - value, less than .0005, indicates a very
small liklihood of getting a sample with with a mean temp.
of 98.2 or smaller when we assume the mean temp. is 98.6.
This suggests the mean body temp. is not 98.6 but actually
less.
If you were to choose a significance level
for this test, what would it be? Why?
.01 We are trying to disprove a commonly held belief, so
we need convincing evidence to disprove the H 0 .
Describe a Type I and Type II error in this
setting.
Type I : Determining that the mean is not 98.6 when in fact it is.
Type II : Failing to reject that the mean is not 98.6 when in fact it
is lower than 98.6.
Construct a 99% confidence interval for the
mean body temperature and interpret it.
x t

s
n
.6229
98.2  2.626
106
(98.04,98.36)
We are 99% confident that the mean body temperature for
the population is between 98.04 and 98.36