Download Quiz#8 Key

Biometry (BIOL4090) Quiz #8.
Student name: __________KEY____________
This 30-minute quiz is worth 5 points. Show all your work to get partial (full) credit. You may
use a calculator, but not a smart phone. You may also leave calculations as ratios if necessary. I
have extra paper, if you need some. Write your name on every page and staple them together
with this cover page.
1a) Write down the equation of the dependent (or paired) t –test and explicitly show how
numerator and the denominator if your null hypothesis was that mean1 = mean2 (+0.25).
NOTE:
if the null hypothesis is true, µD = 0
1b) Write down the equation of the independent (or regular) t –test and explicitly show the
numerator and the denominator if your null hypothesis was that mean1 = mean2 (+0.25).
NOTE:
if the null hypothesis is true,
Mean1 = Mean2, and t = 0
1c) Show the formula of the pooled sample variance, and explain what the terms mean (+0.25):
n1 = sample size sample 1
n2 = sample size sample 2
s21 = variance sample 1
s22 = variance sample 2
2a) Using the data in the matrix below (6 paired samples), calculate the paired difference and
then, use the formula from 1a to calculate the paired t statistic. Hint: Calculate the mean
difference and the S.D. of the difference, and t statistic. Show your calculations in the space
provided below to get full (partial) credit and enter your answers below:
real
40
35
50
55
65
55
photo
30
35
45
40
50
35
difference
deviations
from mean
deviations
squared
10
0
5
15
15
20
Sum = 65
-0.8
-10.8
-5.8
4.2
4.2
9.2
Sum = 0
0.7
117.3
34.0
17.4
17.4
84.1
Sum = 270.8
Mean difference (+0.25):__65 / 6 = 10.8____
S.D. of the difference (+0.25):_ sqrt (270.8 / 5) = sqrt (54.2) = 7.4_
If your null hypothesis states there is no difference between the paired “real” and “photo” values,
what is the expected population level mean difference
(+0.25):__0______
Paired t statistic (Show your work for full credit, +0.25):__(10.8 – 0 )_/ (SD / sqrt(6))___
T = 10.8 / (7.4 / 2.4) = 10.8 / 3.1 = 3.5
For reference, this is the SPSS output for this test:
2b) Using the data in the matrix below (12 independent samples), use the formula from 1b to
calculate the independent t statistic for this test. Hint: Group 1 (real) and group 2 (photo).
Calculate the mean and variance of both groups (+0.125 for each pair). Show your calculations
in the space provided below to get full (partial) credit and enter your answers below:
real
40
35
50
55
65
55
Mean:
photo
30
35
45
40
50
35
Mean:
50
39.17
SD:
SD:
10.95
7.36
Calculate the pooled variance (+0.25):________
n1 = 6
n2 = 6
s21 = (10.95) 2 = 120
s22 = (7.36) 2 = 54.17
Sp 2 = [ (5 * 120) + (5 * 54.17) ] / (6 + 6 - 2) = [ 600 + 270.83 ] / 10 = 870.83 / 10 = 87.08
Calculate independent t statistic (+0.25): _2.01__
(mean 1 – mean 2) = 10.83 – 0 = 10.83
(Sp 2 / n1) = 87.08 / 6 = 14.51 and (Sp 2 / n2) = 87.08 / 6 = 14.51
t = 10.83 / [ sqrt (14.51 + 14.51) = 10.83 / sqrt (29.02) = 10.83 / 5.38 = 2.01
Calculate the effect size (+0.25): _0.54_
t 2 = 2.01 * 2.01 = 4.04
df = 12 – 2 = 10
t 2 + df = 14.04
sqrt (4.04 / 14.04) = sqrt (0.29) = 0.54
For reference, this is the SPSS output for this test:
3) Interpret these analyses:
Was the difference between the paired data significant (Y/N)? Explain Why / Why Not? (+0.25):
Yes, because the p = 0.031 value is significant (p < 0.05)
If you look at the 95% Confidence Interval, how can you determine whether the mean paired
difference is significant? Explain (+0.25): The result is significant, because the 95% C.I. does not
overlap “0”. It ranges from 0.7 to 13.
Are the group variances equal or
different? Explain how you can tell?
(+0.25)
The group variances are not different.
No, because the p = 0.386 value
is not significant (p > 0.05)
Was the difference between the independent data significant (Y/N)? Explain Why / Why Not?
(+0.25): No, because the p = 0.107 value is not significant (p > 0.05)
If you look at the 95% Confidence Interval, how can you deter mine whether the mean difference
is significant ? Explain (+0.25): The result is not significant, because the 95% C.I. does overlap
“0”. It ranges from -1.6 to 15.
4a) Explain why the result of the paired and the independent t-test differed if these were the same
data? Hint: explicitly explain what is achieved by pairing the data for the analysis (+0.25)
The paired results are able to address unsystematic variability across individuals, by comparing
individuals against themselves (comparing their response to a “real” and a “photo” spider). On
the other hand, the independent t-test cannot account for that variability, which becomes noise.
With added noise, the signal (difference in the group means) is no longer significant.
4b) What is the problem with identifying people who are afraid / who are not afraid of spiders
and placing them in two separate treatments (spider photo vs real spider)? Explain what would
be a better approach for assigning people to the two treatments - to avoid this problem (+0.50)
Assigning people into groups based on how scared they are would introduce a bias, since the two
treatment groups would not be the same. This bias (or systematic error), would lead to a wrong
answer: we would either over-estimate or under-estimate the difference between the photo and
the real spider, depending on what treatment the “afraid” people were assigned to. To avoid this
problem, we can randomly assign the people into the two groups (which creates noise rather than
a bias) or we can test everybody twice and compare their results using a paired test (which
accounts for the individual level variability).