Transcript Parametric Tests I

January 31 and February 3, 2014
 Some
formulae are presented in this
lecture to provide the general
mathematical background to the topic or
to demonstrate a concept.
 Do
you need to know these formulae?




Hypothesis
•
Statement of belief with respect to population values
•
•
Hypothesis of no relationship
Alternative or research hypothesis
•
Like Z or t, used to determine the position of the mean in the hypothesized distribution of
sample means
Null
Test statistic
Critical region / critical value
The region at the far end of the distribution, also called the tail
One and two tailed tests
When Z or t fall within the critical region, or, are greater than the critical value, we reject
the null hypothesis in favor of the alternative
• The probability that a test statistic falls within the critical region (tail) is alpha.
•
•
•


Significance level
•
Alpha
•
Hypothesis test or the process of figuring out whether or not the test statistic falls within
the critical region (and we reject the null) or below it (and we fail to reject the null)
Test of significance
 One
Sample Mean t test
• Used to compare 1 sample mean with a population
mean.
• Average x from your sample compared to the
population mean – a mean of “zero”
 Confidence
intervals
• 95% / 99%
• Allows us to calculate, with a specified degree of
assurance, that the value of a population parameter
such as mean, was captured
• 1.96 and 2.54 +/- (s/sqrt n)
• Confidence intervals give us a range that is
sometimes very useful
Statistic = difference / measure of variability

Independent
• Subjects in sample 1 have no connection to subjects in sample 2, such
as comparing the BP of men to women. There should be no
connection between the two groups.

Paired
• There is a connection between scores in one group and scores in the
other.
• For example, comparing BP in a group of patients before and after a
specific drug or lifestyle intervention. In this case, we can see that the
changes in BP are connected between the pre and post
measurements.

Calculations for paired and independent samples are different
and yield different results
• The paired t test calculation factors in an expected correlation
between scores – based on this idea of connectedness
• First step is always to determine if the samples are paired or
independent.
 Goal:
• Evaluate the efficacy of a new antihypertensive
medication
 Research question:
 Is the efficacy of the medication the same for males and females?
 Is this a paired or independent sample?
 Why?
 Assumptions
 Efficacy is measured as the mean change from baseline BP three
months after taking the medication.
 Data
 BP is measured in mmHg.
N
Mean
Variance
Standard Standard
Deviation Error
Male
15
120.2
102.3
10.10
2.61
Female
15
108.2
109.89
10.48
2.71
 Step
1
• State your hypothesis:
 Research Hypothesis:
 Mean SBP is higher in males than females
 μ1 > μ2 or this could also be written as μ1 – μ2 > 0
 Null hypothesis:
 Mean SBP is not higher in males than females
 One or two tailed test?
 μ1 ≤ μ2 or μ1 – μ2 ≤ 0
 Why?
 Step
2
• Choose your significance level
 Alpha 0.05 or perhaps 0.01
 Step
3
• Compute t statistic using the following formula
 Step 4
• Given:
 Mean 1 = 120.2
 Mean 2 = 108.2
 n1 and n2 = 15
 From the formula the pooled std deviation = 10.29
 So:
 t = 120.2 – 108.2 - 0 / 10.29 [sqrt (1/15+1/15)]
 t = 12 / 2.757
 t = 3.19
From t statistic table, one tailed t-test,
d.f.=(n1+n2)-2=28, critical value =1.70
t
statistic = 3.19
• This is within the critical region
• This is greater than the critical value
• So, we reject the null that
 μ1 ≤ μ2 or μ1 –μ2 ≤0 at a p-value of <0.05
• How do we know P is < .05?
 We know this because of the calculated and critical
values. Because the calculated value is in the tail of
the distribution, we know that P < .05.

We can also calculate 95% confidence interval for this
independent sample:
• We use the same basic formula except now it reflects:
 two sample means; degrees of freedom of 28 to be used to
determine the critical value for t = .05; uses the pooled standard
deviation
 Mean difference is 12
 Critical value of t at 28 d.f. is 2.0484, alpha .05 (2 tailed)
 Pooled s adjusted for sample size = 3.757
 So:
 12 +/- 2.0484(3.757) = 4.30 < u < 19.70
 This means we are 95% sure (or confident) that the population of males’
blood pressured ranges from a low of 4.3 points higher to as much as 19.7
points higher than females’ blood pressures
 We
now need to draw meaningful
conclusions that are supported by our
statistical analyses
• SBP of males is significantly greater than the SBP
of females
or
• SBP of females is significantly lower than the SBP
of males
• What if I had to do several, say 10 of these t-tests
to get my answer? What kind of error increases?

Using the same data as Case One, we can now try to determine whether
the experimental conditions led to change in blood pressure.

In a paired sample test, each subject in the treatment group will be used
as its own control.
• This has the benefit of reducing some kinds of experimental error since
variability due to extraneous factors is reduced.
• We will also have fewer degrees of freedom since we will have only 1 sample,
but with two observations
 When calculating a paired t - n always equals the number of pairs
• With fewer d.f. all else equal, the t will be larger, and so the confidence interval
will be larger.

One other issue: do we analyze just males or do we include females in
this analysis too? Pro’s? Con’s?
 Step 1
• Hypothesis (H1): SBP is lower in males after
taking new medication
 μ1 ≠ μ2 or μ1 – μ2 ≠ 0
• Null hypothesis (H0): SBP is not higher in males
than females
 μ1 = μ2 or μ1 – μ2 = 0
 We
will use a two-tailed test since we
don’t know if the SBP will be higher or
lower
 Step
2
• Significant level:
• α=0.05
 Step
3
• Compute t statistic using the following formula:
(next slide)
NOTE:
- d is the mean
difference between
x (before) and y
(after)
-Sd is the estimate
of the standard
deviation of the
differences
- n is always the
number of pairs
From t statistic table, two tailed t-test,
d.f.= n-1 = 14, critical value =2.1448

Calculate t
•

-0.80 / (3.43/sqrt 15) = -0.90
Because t statistic = -0.90 and it falls outside the critical region, which means it is
less than the critical value of +/- 2.1448 we fail to reject the null hypothesis of no
difference

This means there is no difference in SBP between the pre and post measurements
on these paired differences.

What would we report as the P value? Why?

Next, we could calc the 95% confidence interval:



•
•
(120.2 – 121.0 ) +/- t (0.885)
= -0.80 +/- 2.145 (.0885)
= −2.70 < u < 1.10
So, our range is -2.70 < u1 – u2 < 1.10
What does this confidence interval mean? What is the significance of the fact that zero is
contained within the interval? Does this support our conclusion based on the test statistic?
 How
large a sample size do I need to
obtain a statistically meaningful result?
 Factors to be considered:
• How much error can I live with in estimating the
population mean?
• What level of confidence do we need?
• How much variability exists in the data?
 Sample
size can be calculated by
rearranging the formula for Z statistic
• You want to estimate the cholesterol level of a
population within 10mg/dl. You know that σ=20
and you want to stay within 95% confidence
that x is within 10 units of μ.
 Sample size = (1.96)(20)/10=15.36
 Note: 1.96 comes from the Z statistic table corresponding to
95% confidence
you don’t know σ, use s as an estimate and
then use the t distribution for your values
 If
February 3, 2014

True or False:
• Increasing sample size will always improve a study.
• The desired alpha level, the variability in the population and
the size of the difference that is being measured are used to
estimate sample size.
• Very good results can sometimes be obtained with very small
samples.
• Increasing the alpha level from 0.05 to 0.1 will decrease the
estimated size of the sample needed for a study.

True or False
• t-tests are used to compare means or averages in a population.
• Comparing the SBP of men to women is an example of a
dependent sample t-test.

What would be one impact of doing 7 t-tests for
multiple means in a study of SBP? (Multi-choice)
• You would have a better chance of finding significant results.
• You would have a lot less work to do than if you were doing only 1 t-
test.
• You would increase the chances that you found one of the “5” times in
100 that you got your results by chance alone, and not because there is
a real difference in the sample means.
• Your patients arms would hurt from all the blood pressure
measurements.

True or False
• Statistical power is defined as 1-beta error (type II error).
• Statistical power is the probability of getting the right answer (e.g., rejecting
the null hypothesis when its false).
• Statistical power stems from knowing statistics better than others you work with.
• One can think of statistical power as your “confidence” in your results.
• Power is 1 – the chance you got it wrong = the probability you got it right.
• For most studies researchers plan to set at Alpha 0.05, Beta 0.20, and Power at
80%
• While alpha = 0.05 is an absolute according to most statistical experts, power is
not, in other words there is not rock solid cut-off.
• Power analysis is used in sample size planning and can be used for hypothesis
testing.
• To calculate power you need to know: your desired alpha level, an estimate of
how big the effect is in the population, (like the standardized difference
between two means) and an estimate of the variability.
 Develop
an understanding of how we compare
means when we have multiple groups.
 Discuss the concepts of within between groups
differences and between groups differences.
 Learn how to interpret an analysis of variance
model
 Understand the concept of “range tests” or
“multiple comparisons”

ANOVA
• Allows comparison of data from three or more independent groups

Suppose we have 3 groups of patients
• Children < 18
• Adults 19-64
• Seniors > 65
 And, we want to know if the BP of these three groups are significantly different
 We could do several t tests and use logic to conclude what we want to know
 Increases experiment-wise error by repeated t-tests

Null hypothesis:
 μ1 = μ2 = μ3 or μ1 – μ2 – μ3 = 0
 K is the number of groups which in this case is 3
 (t-tests are just to compare one mean against another)

Alternative hypothesis:
 At least one of the means μ is not equal to the others
 This
procedure offers us a way to do multiple
tests between groups while controlling for the
error introduced by multiple tests.
 The
more times you perform a test, the more
likely you are to find one of those pesky 5 times
in 100 that you got your answer by chance
alone.

Observations are independent
 as in independent t-test

Observations in each group are normally distributed
 In other words, they would have a bell shaped curve

Variances of each of the groups is homogeneous
 Each group has about the same variance

Note:
 ANOVA is rather robust
 This means, in statistical terms, that ANOVA is relatively insensitive to
violations of normality and homogeneity assumptions as long as the
sample size is large and nearly equal for each group
 ANOVA is perfect for mean comparisons with N> 25 per group;
 Have done it with as few as 6 for one pharmacologist!

Goal
• To find out if there is a difference between our three group means:
children, adults and seniors.

How?
• Use a test statistic that will somehow compare the means of these three
groups
 F Statistic = between groups variance / within groups variance
 There are F tables just like t and Z
 Computationally, F = mean square between / mean square within

If the between-group variance is enough bigger than the withingroup variance there will be significant differences
Just in case you were curious….
 Variance
has two components:
• Variance within groups (d.f.= N- k)
• Variance between groups (d.f. = k -1)
• These two variance estimates are used to
calculate the F statistic
3
main steps
• State your hypotheses
• Calculate F test statistic
• Determine critical region based on α and reject
the null hypothesis if the F statistic is greater
than critical value
 Question: Is
there a significant difference in
weight gain among the children fed four
different brands of cereal?
 Step
1
• Alternative H1: One or more of the means are
different from the others
• Null: μ1 = μ2 = μ3 = μ4 (no differences in means)
 Weight
gain of
children fed on four
different brands of
cereal (N=20, 5
children per group)
 Does this data look
like there will be a
difference between
the group means?
A
B
C
D
1
7
9
8
1
7
6
6
1
7
5
4
1
7
3
1
1
7
2
1
Have your computer calculate the F
statistic, which is the ratio of the between
to within groups variance – you will get a
table that looks like this:
Source of Sum of d.f.
Variance Squares
Mean
F
Critical F P
Squares Ratio (from
Value
table at
alpha
0.05)
Between
Groups
93.75
3
31.25
Within
Groups
68.00
16
4.25
Total
161.75
19
7.35
3.24
.0026
 From
F statistic distribution, critical value
of α=0.05 for F3,16=3.24 (our F was over 7)
 Because calculated F statistic is >3.24
and falls within the critical region, we
reject H0
 Conclusion:
• There is a significant difference in weight gain
among the children that were fed the four
different brands of cereal
 ANOVA
only tells us that there is a
difference between all the means.
 Multiple t tests between the various pairs
of means are not appropriate because
the probability of incorrectly rejecting
the hypothesis increases with the number
of t-tests performed
 Must
use a post-hoc test to find out which
of the means is (are) different
 This is called a multiple comparisons
test.
 Some examples are:





Tukey
Tukey-Kramer
Scheffe
Bonferroni
Dunnett’s
 Uses
a formula to determine mathematically if
each mean difference is greater than an
anticipated critical value calculated like a test
statistic.
 This procedure identifies which means are
actually different from each other.
 In our example, this is 3.73, so we compare
each pair of mean differences to this number, if
the difference is greater than 3.73 we know that
those two means are different
Pair
Critical Value
A–B
Mean
Difference
6
A–C
4
3.73
A–D
3
3.73
B–C
2
3.73
B–D
3
3.73
C–D
1
3.73
3.73
 ANOVA
is used for mean comparisons
when more than two groups are
compared
 ANOVA only tells you whether there is a
significant difference between two
groups. It doesn’t tell you which groups
are different
 Tukey or other multiple comparison test
allows you to determine which means are
different.