Download IB Biology - hhsIBbio09

Recording Data & Statistical Tests
IB Biology 2009-2010
IB Biology
2009-2010
Recording RAW Data
 Be sure all data tables have

DESCRIPTIVE TITLE
 Not “the results” or “data”
 Include some mention of methodology


If handwritten, use ruler to draw gridlines
Units and error recorded in heading of table
 ALL measurements recorded to the same level of
precision ) watch decimal places

IB Biology
Include qualitative data observed during the
investigation
“Complete”
Table 1: Lengths of two plant tissues, potato (Solanum) and apple
(Malus) after soaking in solutions of sucrose of different
concentrations. The initial lengths were 4.0 cm.
It was also noticed before the soaking that the potato tissue floated in the
solution from 0.4 to 1.0 mol. The apple tissue, however, only floated in
the solutions from 0.6 to 1.0 mol. After soaking the tissues became
softer at higher sucrose concentrations but they were quite hard in
the lower concentrations.
IB Biology
“Partial”
IB Biology
“Partial”
IB Biology
“Not at all”
IB Biology
Processing Data
 Processing means mathematical calculation has







occurred
Means are most common method for recording data in
Biology
Other formulas may be used (rate, change, % change,
stoichiometry, etc)
Show sample calculation (except mean)
Don’t forget units and error
 Mean can’t be more precise than raw data
 Use standard deviation as SEM when means used
Make NEW data table for processed data for clarity
You want DATA TABLE AND GRAPH
Graph PROCESSED DATA!!!!!
IB Biology
Standard Deviation
 The standard deviation is a statistic
that tells you how tightly all the various
examples are clustered around the
mean in a set of data.
68% fall +/- 1
 95% fall +/- 2
 99% fall +/- 3

IB Biology
Standard Error
 The standard error of the mean
 This statistic gives us an idea of how “good” our sample
is. The narrower the limits are the closer the mean of the
sample represents the mean of the population
 The standard error of the mean is given by
 Its size indicates how representative your mean is.The
sample mean ±
 gives the 95% confidence limits.
 In other words, we are 95% sure that the mean of the
population lies somewhere between:
 NB We would only use this formula for n>30. If n<30
and we are fairly sure the population we are sampling
from has normal distribution we replace 1.96 by the 95%
point from our t-test tables with n-1 degrees of freedom.
IB Biology
Calculating Standard Deviation
 EXCEL – use built in function
IB Biology
Calculating Standard Deviation
 TI-83/84
Enter data into List in STAT
menu
 Use STAT CALC
 Select “1:1-Var Stats”
 Type L1 (or list used)

IB Biology
The t-test
 Used to determine if two sets of data
are significantly different
Paired t-test (same subjects receive
both treatments)
 Unpaired t-test (compares two groups,
but individuals in groups are different)

IB Biology
Null and Alternate Hypotheses
 Null Hypothesis

There is no significant difference
between the results of the two groups
 The treatment HAD NO effect
 Alternate Hypothesis

There IS a significant difference
between the results of the two groups
 The treatment HAD an effect
 This is usually the hypothesis we stated in
lab
IB Biology
Level of significance (p-value)
 p-value indicates the probability that
the differences between the two groups
is due to chance
 If p<.05 then we reject the null
hypothesis (test supports the alternate
hypothesis)
 If p>0.05, we cannot reject the null
hypothesis (note wording)
IB Biology
t-test in excel
 In Excel the t test is performed using
the formula: =TTEST (range1, range2,
tails, type)

For the examples you'll use in biology,
tails is always 2 (for a "two-tailed" test),
and type can be 1 or 2 depending on the
circumstances.
 Type 1 = paired
 Type 2 = unpaired
IB Biology
Type 2 – unpaired test
 Number of WBC in
infected vs
uninfected patients
IB Biology
Type 1 – paired test
 Pulse rate measured
before and after
eating
IB Biology
Formula – Unpaired t-test
IB Biology
Formula – Paired t-test
IB Biology
Paired T-test on TI-83/84
 [STAT] [1]; cursor to L1 column
heading, press [CLEAR] [ENTER], and
enter the “Before” numbers.
 Cursor to L2 column heading, press
[CLEAR] [ENTER], and enter the
“After” numbers. You want to get the
column of differences, After−Before,
That’s the same as L2−L1, which is
the formula for L3.
 Cursor to L3 column heading. Enter
the formula, [2nd 2 makes L2] [−]
[2nd 1 makes L1] [ENTER].
 NOTE – you need to make print data
table with these differences and
means.
IB Biology
Pulse Rate (+/- 1 BPM)
before and after 100 mg
caffeine consumption
Paired t-test, cont’d
 Use 1 VAR Stats to get mean and std dev


[STAT] [►] [1] for 1-VarStats, then [2nd 3 makes L3].
Since the sample statistics are automatically pasted
to the t test screen, you don’t need to copy them.
Just enter the μ (hypothesized mean/difference) and
the “>” from your H1.[STAT] [◄] [2] for T-Test
Enter 0 for μo. Skip past x̄, s, and n. Enter >μo
(since you hypothesize caffeine increases HR), then
select CALCULATE or DRAW.
In this case, p< 0.05, so the null hypothesis is rejected… the alt
hyp that caffeine increases HR is supported
IB Biology
Showing work using TI-83/84
 Write what you entered in each list
 Record stats for mean and SD
 State t-test statistic and p-value
 State what this info “tells” you
IB Biology
UNPAIRED t-test on TI-83/84
USE this test when populations samples aren’t
the same individuals
 Select 2-SampTTest
 Enter mean, sample size (n) and






SD for each sample
Select appropriate alt hypothesis
Selected POOLED data
Calculate results
Note, in this
case p>0.05
Null hypothesis
NOT rejected
NO sig difference in
two samples
IB Biology
Problem #1
 As a biochemist working for a pharmaceutical company, your

job is to test new drugs for possible side effects, both
deterimental and beneficial. The chemistry of agent TFK-05W
suggests that it may have the side effect of reducing the
tendency toward obesity. The Zucker rat is an established
genetic model for both obesity and hypertension. As rats of
the obese strain age they gain weight much more rapidly than
do the so-called "lean" Zucker strain. You plan to treat a group
of animals over a period of one year and compare their
average weight with that of a group of untreated animals.
The odds of the drug actually showing this side effect are
small and maintenance of rats for a year is expensive, so you
limited the scope of the study to twelve animals in each group.
The null hypothesis is that treated animals will show an
average weight gain over one year that is no different from the
average weight gain of untreated animals. What statistical test
will you use to compare the two means?
IB Biology
Answer
 You'll run a t test for independent samples. It
doesn't matter that the number of animals in
each data set is the same, nor that they are all
the same type of animal. You sampled 12
treated individuals and 12 different untreated
individuals. There is no special relationship
between a data point from one group and any
particular data point from a second. The
sampling method was independent.
IB Biology
Problem #2
 This study follows from problem #1. Not only did your study suggest

that the agent TFK-05W indeed does affect weight gain, but it also
proved effective and safe (so far) and it is in clinical trials. Because
the drug was designed to treat symptoms that have nothing to do with
obesity, the clinical trials do not focus on that problem and won't
answer the question of whether or not the agent is a potential weight
loss drug. The company, however, has permitted you to test the agent
on a group of 12 people with morbid obesity, who have signed the
appropriate consent forms.
This time the plan is to treat the 12 obese individuals for a year,
having measured their weights on the day treatment was started. The
paid participants will be monitoring their weight regularly, taking the
drug, and are required to keep a daily log of activity and eating habits
so that the experiment can be properly controlled. Nevertheless, the
simplest initial test of the hypothesis that obese individuals treated
with TFK-05W for a year will show an average weight loss is to
compare average weight at the beginning and end of the experiment.
Thus, as with problem #1, you will have two sets of 12 data points
each to compare. What statistical test will you use to compare the two
means?
IB Biology
Answer
 This study calls for running a paired t test. The same
individuals were sampled (weights measured) at the
beginning and at the end of the study. Thus each data
point in the first set can be paired with a data point from
the same individual in the second set.
 Variability among distinct individuals contributes
considerable experimental error to many experiments.
Such error can mask effects, especially small effects,
even if the null hypothesis is indeed false. For example,
if the average individual lost 10 pounds but the standard
deviation at the beginning of the experiment was 55
pounds, the loss might not show up as a significant
difference. By controlling for individual variability the
paired t test can focus on the average change in weight.
IB Biology
Problem #3



Embryonic cells (stem cells) from a single human blastula are genetically equivalent. Any of
them has the potential to form any kind of tissue that is normally found in an adult human
body. Exploitation of stem cells for therapeutic purposes has the potential to revolutionize
medicine and expand the average human lifespan considerably. Stem cells have very
complex cultural requirements. So far your stem cell lines require the addition of fetal
bovine serum to the culture medium, and the exact composition (and efficacy) of animal
sera varies from lot to lot, To exercise the greatest control over your experiments, it would
be valuable to be able to culture your cells in a synthetic medium that includes only those
components that are essential to support survival and growth of your cultures.
You have developed a synthetic medium that keeps your cells going for several days, but
not indefinitely. You think that you can extend the life of your cultures by adding an
expensive hormone to the medium. To test the hypothesis that stem cells cultured with
medium 2 will survive longer than stem cells cultured with medium 1, you will set up twenty
cultures from ten original embryos, growing them in complex medium to the point at which
each culture contains about 100 cells. You will then remove the original medium and feed
the cultures from now on with synthetic medium. One culture from each original embryo
will receive medium 1 and the other medium 2. For your data you will record the time at
which each culture declines to the point of having only 50% of its original viable cells
remaining. The null hypothesis is that this average "survival time" will be the same for
cultures treated with either synthetic medium. Alternative hypotheses, of course, are that
feeding with one or the other medium will enhance survival time by comparison.
What statistical test will you run on the two sets of 10 data points each?
IB Biology
Answer
 This case can be thought of as a set of replicate experiments.


In each experiment one culture from a single source was fed
one medium and a second culture from the same source was
fed the other medium. The experiment was replicated 10 times,
using 10 different sources. Since each replicate experiment
consists of a pair of data points linked by the common origin
of the respective cultures, you have a set of 10 pairs of data
(two sets of paired data).
A paired t test is appropriate for the same reasons it was
appropriate for problem #2. The paired method controls for
experimental error that might be contributed by the 10
different sources.
Why not conduct all of the replicate experiments on cultures
from a single source, eliminating all experimental error that is
contributed by individual variability? Then we run the risk that
the result won't hold for cultures from other embryos. We want
to know if the medium we are testing will work for most or all
cultures, not only for cultures from one particular embryo.
IB Biology
Problem #4



You suspect that a cause of decline of your stem cell cultures is a failure to
produce sufficient superoxide dismutase to rid the cells of oxygen free radicals.
You have an assay for the enzyme, but to conduct the assay you must destroy
the culture. From a single source of stem cells you can prepare about thirty
cultures that remain healthy about 10 days in your synthetic medium after
growing to a sufficient number of cells to permit you to run your assay. From
then on, they decline rapidly.
You prepared thirty cultures from the same source and sampled half of them
when alll thirty cultures had reached the point at which the assay was feasible.
You then sampled the other half 10 days later. Your null hypothesis is that
superoxide dismutase activity will not be significantly different between the two
sets of cultures. What statistical test will you run to determine whether or not
the difference is significant?
Normally, this would be a rather poor experimental design, because all of the
cultures are identical. Why conduct replicate sampling on the exact same
culture? The issue is that enzyme assays are notoriously inaccurate. The
chances of mulitple comparisons yielding dubious results are much smaller
than for a single comparison. The p value that we obtain will give us a fairly
accurate estimate of the level of confidence with which we can interpret the
result.
IB Biology
Answer
 This time your samples are all coming from the same
population of cultures, presumably all identical except
that half of them were sampled at one time and half at
the other time. All of the data points are linked by the
fact that they were obtained from cultures from a
common source. However, there is no special one to one
correspondence between any one data point in one set
and a unique data point in the other. There is no basis
for a paired t test, so we must run a test for independent
samples.
 The assay itself is the variable in this example. If the
assay was 100% accurate and reliable, we would only
have needed one sample at each time. On the other
hand, any significant difference should be considered
preliminary until the experiment can be repeated on at
least one or two more sets of cultures.
IB Biology
Problem #5
 Referring to problem #1, one of your
rats died of natural causes during the
study, leaving 11 animals in one group
while you still had 12 animals in the
other. Does this turn of events ruin the
experiment? If you do conduct the
analysis, how will you modify it, if at
all?
IB Biology
 You lost one animal, but because each
data set represents an independent
sample it is not necessary that the
numbers of data points be equal. You
conduct the t test for independent
samples, comparing a set of 11 data
points with a set of 12.
IB Biology
Problem #6
 Referring to problem #2, one of your
human subjects died of a massive heart
attack halfway through the study. The
death was clearly not related to the
drug treatments. Please answer the
same questions as for problem #5.
IB Biology
 This time you lost both data points that
were to be contributed by the deceased
individual. You now have 11 data points
in each set. It shouldn't be a problem
unless the others start dropping off as
well.
IB Biology
Problem #7
 In problem #2 it was stated that Zucker
rats treated with the agent TFK-05W
were significantly less obese than
untreated rats. Does this mean that
when the t test was run it returned a
probability (p) value of > 0.05 or a p
value of < 0.05?
IB Biology
P would be <.05
 The p value is the probability that the null hypothesis is

accurate. The higher the p value, the greater is the probability
that there is no significant difference between means. A
probability of 0.05 (1 in 20 chance) or less is considered
sufficient evidence on which to reject the null hypothesis.
Rejecting a null hypothesis means we accept an alternative.
The result with the Zucker rats was that the treated group
weighed less, so we accept the alternative hypothesis that the
drug reduced weight gain. The other alternative, which was not
supported by the data, would be that the drug caused
additional weight gain.
Of course, working with probabilities there is always a chance
that the results of an experiment are simply wrong.
Realistically, though, experimental results are seldom wrong
due to an improbable distribution of samples. They are usually
wrong because of a bad experiment, especially when an
experiment is not well controlled.
IB Biology
Problem #8
 For the third problem the difference
between means was 12 hours, that is,
the average half life of one culture was
12 hours longer than the average half
life of the other one. The t test returned
a p value of 0.33. What is your
conclusion regarding the original
hypothesis?
 Would you be correct in stating that the
result is significant or insignificant?
IB Biology
p > 0.05 so don’t reject H0
 The difference in sample means may have been 12
hours, but apparently there was enough variability
among cultures that the difference was not significant.
With p > 0.3 it is unlikely that you will get a significant
difference even by testing more cultures.
 Unless a difference is supported by probability it is not
considered significant at all. Think about it. Maybe the
range over which cultures lasted was quite wide.
Perhaps the difference in sample means would be
reduced to zero or even reduced if you just switched two
data points.
 You would not be correct to say that the result is
insignificant. The result is that medium 2 has no
apparent effect on longevity of a culture, and that finding
is indeed significant. The difference in means was
insignificant, not the result itself.
IB Biology
Problem #9
 In problem #4, average superoxide dismutase
activity was 30% lower after 10 days than it
was in the beginning. The t test gave a p
value of 0.07. What result do you report? It
turns out that this study is very important,
and if you indeed find that a decline in
superoxide dismutase activity is a primary
cause of cell death then good things will
happen to your career. How will you proceed
from this point?
IB Biology
Repeat, but don’t falsify results
 With a p value of 0.07 you are so close to finding a significant

difference that it is sorely tempting to drop a data point in
favor of your hypothesis, or maybe "round off" to 0.05.
Scientific integrity requires that you treat the data as they
stand, however. You do not have sufficient evidence with
which to reject the null hypothesis.
How to proceed? Because it is so important to you that you
come to a conclusion, an appropriate course of action is to
repeat the experiment. Analysis will be more complicated
because the second experiment will be conducted on a
different set of cultures. To keep it simple you could repeat the
experiment twice and if the results are consistent, pool all of
the data points. By the way, with 45 data points in each set you
probably would no longer need a t test. You could base your
analysis on the normal distribution and simply look at overlap
between standard deviations in order to determine a p value.
IB Biology