Download Hypothesis Testing with Z

Stats 95
Z-Scores & Percentiles
• Each z-score is associated with a percentile.
– Z-scores tell us the percentile of a particular score
– Can tell us % of pop. above or below a score, and
the % of pop. between the score and the mean and
the tail.
Transforming Zscore into
Percentiles
a) DRAW A GRAPH!
b) Calculate z-score
c) Estimate the
percentile of the zscore using
probability
distribution
d) Use z-score chart to
transform z-score
into percentile
e) Use graph to make
sure answer makes
sense
• Draw a Graph!…did I
mention you need to draw
a graph? Yeah, draw a
graph.
Transforming zScores into
Percentiles
• Use a chart like this in
Appendix A of your text (Yes,
you need the textbook) to find
the percentile of you z-score.
• This table gives the distance
between the mean (zero) and the
z-score.
• To calculate cumulative
percentile :
•Of positive z-score 50 + (z)
•Of negative z-score 50+ (-z)
Example: Height
• Jessica has a height of 66.41 inches tall (5’6’’)
• The mean of the population of height for girls is
63.80
• The standard deviation for the population height fir
girls is 2.66
z
( X  )

66.41  63.80

 .98
2.66
• According to z-score table, the percentile associated
with z = .98 is 33.65%
Height Example:
Did I mention? DRAW A GRAPH!!
• Jessica’s z-score for her height is .98,
associated percentage of 33.65%.
• This means
– there is 33.65% of the population is between the
mean and Jessica’s score.
– There is a 33.65% chance of Jess being taller than
the average by this amount BY CHANCE
ALONE
• Mean = 50th percentile, therefore to find the
Jessica’s percentile = 33.65+50 = 83.65%.
• 84% of the population of girls is shorter than
Jessica, and there is a 100-84% = 16%
chance of someone being this tall by
CHANCE ALONE.
Hypothesis Testing
• Identify Population and comparison group, state assumptions
• Define the Null Hypothesis
• Define the Research Hypothesis or Alternative hypothesis
– Define the Research and Control Group
– Define the Dependent and Independent Variables
•
•
•
•
State relevant characteristics of comparison distribution
Determine critical cutoff values
Calculate statistic
Reject or Fail to Reject the NULL Hypothesis
Statistical Significance
• A finding is statistically significant if the data
differ from what we would expect from chance
alone, if there were, in fact, no actual
difference.
• They may not be significant in the sense of
big, important differences, but they occurred
with a probability below the critical cutoff
value, usually a z-score or p < .05
Assumptions for Parametric Tests
• Dependent variable is a scale variable  interval or
ratio
– If the dependent variable is ordinal or nominal, it is a nonparametric test
• Participants are randomly selected
– If there is no randomization, it is a non-parametric test (and
likely a flawed experiment, or one limited in
generalization)
• The shape of the population of interest is
approximately normal
– If the shape is not normal, it is a non-parametric test
Assumptions for Parametric Tests
Hypothesis Testing with z-Scores:
Example
• Ursuline College, 97 students took Major Field
Test in Psychology (MFTP). Is the score of
this group of students statistically different*
from the total population of students who took
the test?
• *is the probability of the difference greater than what we
would expect to happen by chance alone. (i.e., how big is their
horseshoe?)
Hypothesis Testing with z-Scores
Step 1 Populations, Distribution, Assumptions
• Identify Population,
• Assumptions
distributions and assumptions
o Scale is continuous interval
– All students at Ursuline who took
the exam
– All students nationally who took
exam
• Distribution
– Comparison sample is not an
individual but a group mean for
all the students at Ursuline who
took exam therefore comparison
distribution is a sample mean to a
population.
(test scores)
o Random selection
unknown, so we are limited
in generalizing
o Shape should be normal,
sample size of 97 Ursuline
students, substantially
larger than recommended
30
Hypothesis Testing with z-Scores:
H 0 : 1   2
Step 2: State Null and research Hypothesis
H 0 : 1   2
H 1 : 1   2
Population Parameters
Non-directional
H 1 : 1   2
• The Null Hypothesis: Ursuline
• The Null Hypothesis: Ursuline
Students are not better from
rest of the nation.
• Research Hypothesis, U. students
Students are no different
from rest of the nation.
• Research Hypothesis, U. students
are better than population
• One-tailed Test
(Rare!)
are Different (better or worse)
than population
• Two-tailed test
H 0 : 1   2 or M  156.5
H 1 : 1   2 or M  156.5
Hypothesis Testing with z-Scores:
Step 3: Determine Characteristics (Parameters) of the Distributions
• Comparing Sample Mean to
Population
H 0 : 1   2 or M  156.5
H 1 : 1   2 or M  156.5
PoP.  unknown, Huge __ N ( M )  97
  14.6
– Null H0 and Research H1
– Because we have a sample
mean, we must use the
Standard Error instead of the
Standard Deviation of the
Population
– We have Standard Deviation
 M  14.6 / 97  1.482
of Population, from which we
calculate the Standard Error of
– The average of Ursuline class was 156.11.
The average of all other College mean samples Sample Means
from other college classes is 156.5.
– We are saying Null H. assumes the mean of
this class is no different from the combined
mean of all classes.
Hypothesis Testing with z-Scores
Step 4: Determine Critical Values (Set how big a Horseshoe You Will Willingly Accept)
• Convention is to set
cutoffs at probability less
than 5%
– p < .05
– Two-tailed Test, means it
is 2.5% at both Tails
– If it was a One-tailed Test,
5% at one Tail
• Draw a Graph
• We know 50% of the curve falls between the mean and each end. We
know that 2.5% falls between the critical z statistic and the rest of the
tail. Subtract 50%-2.5 = 47.5% . Look in chart, 47.5% is associated with
the z statistics of 1.96
Hypothesis Testing with z-Scores:
Step 5: Calculate Z Statistic
z
(M   M )
M
(156.11  156.5)  .39


 .26
1.482
1.482
• Draw a Graph!
Hypothesis Testing with z-Scores:
Step 6: DECIDE Already!
• Decision is either to Reject
the Null Hypothesis or to
Fail to Reject the Null
Hypothesis
• With a Z statistic of -.26,
there was a approx. 40% of
getting this score by chance
alone, p = .3976.
• Compare to the cutoffs
– z statistics: -.26 is not more
extreme than -1.96
– p: .3976 >.05
• We fail to reject the Null H.
• The probability of Ursuline
producing the score of
156.11 is greater than .05
– p >.05
– The Ursaline mean was not
smaller, than we would expect
from chance alone, than the
population mean.
H 0 : 1   2
H 1 : 1   2
H 0 : 1   2
H 1 : 1   2
• What problems did Fisher have to tackle in the
designing the experiment that could determine if
the Lady could distinguish between milk-then-tea
and tea-then-milk cups of tea? She could guess
randomly with a 50% probability of being correct.
She could also make an error is spite of being able to
distinguish, getting 9 our of 10. How many cups to
present, in what order, how much to instruct the lady,
and determine beforehand the likelihood of possible
outcomes.
What was the nature of research like before
experimental methods became established? They
were idiosyncratic to each scientist, lesser scientists
would produce vast amounts of data would be
accumulated but would not advance knowledge, such
as in the inconclusive attempts to measure the speed
of light. They described their conclusion and
selectively presented data that demonstrated or was
indicative of the point they wanted to make, without
clear procedures to replicate and not with full
transparency.
•
• In examining the fertility index and rainfall
records at Rothamsted, what two main errors did
he conclude about the indexes and how did he
express results about the rainfall? He examined the
competing indexes, and showed when reduced to their
elemental algebra, they were all versions of the same
formula. And he found that the amount of rainfall was
greater than the type of fertilizer used, that the effects
for fertilizer and rainfall had been confounded.
•
• What did Fisher conclude scientists needed to
start with to do an experiment, and what would a
useful experiment do? Scientists need to start with a
mathematical model of the outcome of the potential
experiment, which is a set of equations in which some
of the symbols stand for numbers that will be
collected as data, and other symbols stand for the
overall outcomes of the experiment. Then a useful
experiment allows for estimation of those outcomes.