Transcript Chapter 12x

Significance
Tests in
practice
Chapter 12
12.1 Tests about a population
mean


When we don’t know the population standard deviation σ,
we perform a one sample T test instead of a Z test
We get a “T calc” (instead of a Z calc) and the formula is the
same except we put in the standard deviation Sx of our
sample as a replacement for σ
Interpretation and P values
 same
as Z interval, except if you are using
the tables, look up “T star” to find the
critical T value you need to beat. I
recommend the calculator though…
 The P value is from your calculated T to
the tail. If it is a 2 tailed test, remember
this gets doubled!
T test on calc
 Go
to Stat/Tests choose 2: T-test. Enter
either data or Stats (just like Z test)
Example: Diversity (P. 750)

An investor with a stock portfolio sued his
broker b/c lack of diversification in his
portfolio led to poor performance. The table
gives the rates of return for the 39 months that
the account was managed by the broker.
Consider the 39 monthly returns as a random
sample from the monthly returns the broker
would generate if he managed the account
forever. Are these returns compatible with a
population of μ=.95%, The S&P 500 average?
-8.36
1.63
-2.27
-2.93
-2.70
-2.93
-9.14
-2.64
6.82
-2.35
-3.58
6.13
7
15.2
5
-8.66
-1.03
-9.16
-1.25
-1.22
10.2
7
-5.11
-.80
-1.44
1.28
-.65
4.34
12.2
2
-7.21
-.09
7.34
5.04
-7.24
-2.14
-1.01
-1.41
12.0
3
-2.56
4.33
2.35
Step 1: Hypothesis- (Where μ is the mean return for all possible
months that the broker could manage this account)
H0: μ = .95
HA: μ ≠ .95
Step 2: Conditions- Since we do not know σwe must use a onesample t test. Now we check conditions:
SRS- We are told this in problem
Normality: Sample is large enough (39) so CLT applies
Independence: We must treat the 39 monthly returns as
independent observations from the population of months in
which the broker could have managed the account.
 Step


3: Calculations-
Test Statistic: (-1.10 - .95) / (5.99/√39) = -2.14
P value: df are 39-1 = 38. Since we are
performing a 2-tailed test, our P-value is .039
(can be done on calc, or on table- if doing
table, make sure to double the area under the
curve b/c of 2 tailed!)
 Step
4: Interpretation: The mean monthly
return on investment for this client’s account
was -1.1% for this period (this is our x bar). This
is significantly difference at the alpha = .05
level of the S&P 500 for the same period (t = 2.14, P<.05)
Paired T Test – Testing the
differences
Example: Sweet Cola (P. 746)
 Diet
colas use artificial sweeteners which
gradually lose their sweetness overtime.
Trained tasters sip cola and give it a
“sweetness scale rating” of 1 to 10. Then it
is stored for 4 months and tested again.
The differences in sweetness score are
listed (The bigger the differences, the
bigger the loss of sweetness, negative
value means it gained sweetness)
2
.4
.7
2
-.4
2.2
-1.3 1.2
1.1 2.3
 Step
1: Hypothesis- since we care about the
difference, μbefore-μafter = μdiff


H0: μdiff = 0 “The mean sweetness loss for the
population of tasters is zero”
HA: μdiff > 0 “The mean sweetness loss for the
population of tasters is positive. The cola seems
to be losing sweetness in storage
 Step
2: Conditions- Since we do not know the
standard deviation of sweetness loss in the
population of tasters, we must use a onesample t test. Now we check conditions:


SRS- assume the 10 tasters are randomly selected
Normality: 10 is too small a sample so we check
our normal probability plot
 It
is slightly left skewed, but no gaps/outliers so
we proceed with caution
 Independence:
we assume there are more
then 10x10 = 100 tasters in the population
 Step

3: Calculations
Test Statistic- The basic statistics are
X
bardiff = 1.01 and Sdiff= 1.196
 Tcalc= (1.02 – 0)/ (1.196/√10) = 2.70
 P value: degrees of freedom are 10-1 = 9 for
alpha = .05, p = .0123
 Step
4: Interpretation- A P-value this low
gives strong evidence against the null
hypothesis. We reject H0 and conclude
that the cola has lost sweetness during
storage
 *Remember
the 3C’s! (Conclusion,
connection, context)
Robustness and Power
 Recall
that t procedures are robust against
non-normality of the population except when
outliers or strong skewness are present
(skewness is more serious). As the sample size
increases, the CLT ensures that our distribution
of sample means approximates normality.
 Again, calculating power is not in this course,
but remember we can still commit the same
type I / type II errors when rejecting (or failing to
reject) the null.
12.2 Tests about a population
proportion
 Remember,
when the 3 conditions are
met (SRS, Normality, and Indep.), the
sampling distribution of p hat is
approximately Normal with mean equal
to rho and standard deviation = √(pq/n)
Note: In the
standard error
(denominator
of test
statistic) we
use the values
of our
assumed null
hypothesisthis is different
from the
standard error
of the CI
where we
used sample
proportion
values!
One proportion z on calc
 Stat




Test – 1-propZtest put in:
Rho
x = the count (number of successes in your
sample)
n = number of people in your sample
choose one-tailed or 2 tailed hypothesis
calculate
Example (P. 767) work stress

Job stress poses a major threat to the health
of workers. A national survey or restaurant
employees found that 75% said that work
stress had a negative impact on their
personal lives. A random sample of 100
employees from a large restaurant chain finds
that 68 answer “yes” when asked “Does work
stress have a negative impact on your
personal life?”. Is this good reason to think
that the proportion of all employees in this
chain who would say “yes” differs from the
national proportion of ρ= .75?

Hypothesis: we want to test a claim about ρ,
the true proportion of this chain’s employees
who would say that work stress has a negative
impact on their personal lives.


H0: ρ=.75
HA: ρ ≠ .75
Conditions: We should use a one-proportion Z
test if the conditions are met



SRS: we are told in problem
Normality: The expected number of “yes” and
“no” responses are (100)(.75) = 75 and (100)(.25)
= 25 respectively. Both are at least 10 s
Independence: We assume the chain has at
least 10(100) = 1000 employees

Calculations

Test Statistic:
.68 -.75
= -1.62
(.75)(.25)
100
P value = .1052 (if you do normalcdf,
remember to double your value b/c
this is a 2 tailed test! so 2x.0526!)

BUT! MUCH easier to do on Calc though
and not use NormalCDF! Try it

Interpretation There is over a 10% chance of obtaining a
sample result as unusual as or even more unusual than we
did (p hat = .68) when the null is true so we have
insufficient evidence to suggest that the proportion of this
chain restaurant’s employees who suffer from work stress
is different from the national survey result, .75
Confidence Intervals

using the info from that same example, if we were
to calculate a 95% CI (.59, .77)

(.68)(.32)
100
 Note: the CI standard error uses the sample
proportions, but in the test statistic formula we use
the value assumed in the null hypothesis
by hand, using formula:


.68 + / -1.96
Using calc is easier though: choose 1 prop z int
Interpretation: We are 95% confident that
between 59% and 77% of the restaurant chain’s
employees feel that work stress is damaging their
personal lives.