University of Central Florida Psychology Statistics Questions

9/10/22Normal Distribution
Normal Distribution
•
•
•
•
•
•
Standard Deviation
160
Reading ability
Introversion
Job satisfaction
Intelligence
Empathy
Aggression
• C. F. Gauss
• A. Quételet (l’homme moyen)
• US Army Corps of Engineers
Z scores
161
Z-Scores Example 1
Z-Scores Example 2
• Assuming that CAT scores among U.S. college
students are approximately normally
distributed with a mean of 302 (and a made
up standard deviaBon of 10).
• Assuming the same mean (302) and standard
deviation (10), what is the range of scores for
the middle 95% of CAT scores?
• What is the z-score for someone who scored a
312 on the CAT?
162
163
Extra practice
• In the United States, the average IQ is 100,
with a standard deviation of 15. What
percentage of the population would you
expect to have an IQ lower than 85?
164
Extra practice
• z = (x – xbar)/s
• z = (85-100)/15 = -15/15
• z = -1.00
• Which portion of the normal distribution?
• Smaller or Larger?
• p = 0.159
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Extra practice
Extra practice
• z = (x – xbar)/s
• z90 = (90-100)/15 = -10/15
• z90 = -0.66
• p120 = 0.092
• p90 ∞ , sampling distribution -> Normal
186
187
• We use sample mean as an esOmate
of the populaOon mean (µ = 𝑥! ) .
• We use s,n to calculate the SE
• What does LLN predict about SE?
• n -> ∞ , the sampling
distribuOon becomes normal.
• SE = s/√n
• What does CLT predict about SE?
• n -> ∞ , 𝑥 -> μ
• n -> ∞ , 𝑆𝐸 -> 0
188
189
Why should I care about the SE?
• Means of multiple samples taken from the same
population vary (i.e., they differ from each other)
• This variation among means can be considered a
type of error in sampling (i.e., Sampling variability)
• We can use this estimate of error to create
confidence intervals.
• 90% CI = 𝑥 ± 𝑆𝐸 ∗ 𝑍3.45
• 95% CI = 𝑥 ± 𝑆𝐸 ∗ 𝑍3.64
• 99% CI = 𝑥 ± 𝑆𝐸 ∗ 𝑍7.58
190
191
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9/10/22
Example.
Sperm count of
Japanese Quail
• Population mean
of 15 million
• 𝑥 and s vary
among samples
• 95% CI around
each men
• 95% of samples
will contain pop
mean within CIs
• Any sample CI has
a .95 probability
of containing the
true population
value
192
6
9/16/22
Hypothesis Testing
• Karl Popper and Null Hypothesis Testing
Framework
• Psychology was not a science, because the field
made untestable claims à Sigmund Freud
• Falsifiability: Easier to falsify H0 than validate HA
• Confidence Intervals are a way to test H0
• Find p that H0 falls within a Confidence Interval?
• If p -> 0, the it would be absurd to not reject H0 ;
thus validating HA*
• Consider the Claim: Physical activity affects
cognitive performance
197
Claim: CF affected by PA
• Testable?
– Not yet
• DV/Outcome?
– CF
• How operationalize?
– Test scores
– LoM?
• IV/Predictor?
– PA
• How operationalize?
– Cheapest -> one sample compared to known population
• Testable H0
– No difference test scores (𝑥 = μ)
198
CI around μ vs. 𝑥̅
Claim: CF affected by PA
HA is true
∆ = 200; PA affects CF
Population
H0 is true
∆ ≠ 200; Sampling variability
• In one sample of size n, you observe one 𝑥,̅
which leads to one ∆ (𝑥̅ − 𝜇) ≠ 0
• Decision (Null hypothesis significance testing)
– 𝐻!: ∆ ≠ 0; 𝑥 ≠ 𝜇; PA -> CF
– 𝐻”: ∆ = 0; 𝑥 = 𝜇; sampling variability
2 possibilities for the raw
observed effect (∆= 200):
1000
Sampling Distribution
(Theoretical)
Sample
m = 1200
s = 150
n = 25
1200
199
200
Confidence Interval around
population mean
Confidence Interval: As N -> ∞
• Confidence interval
SE = s/√n
• 95% Confidence interval
– UB/LB: 𝜇 ± (1.96 x s/√n)
– Upper Bound: Mean + (z x SE)
– Lower Bound: Mean – (z x SE)
• Our small sample of students
– s = 150 and n=25
– UB/LB: 𝜇 ± (1.96 x 150/√25)
– UB/LB: 𝜇 ± (1.96 x 30)
– UB/LB: 𝜇 ± 58.8
– UB/LB: 1000 ± 58.8
– [941.2 1058.8]
– Does 𝑥 (1200) fall outside 95%CI around population mean?
Z = 1.64 for 90% CI
Z = 1.96 for 95% CI
Z = 2.58 for 99% CI
• 95% Confidence interval
– UB/LB: Mean ± (1.96 x s/√n)
201
202
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9/16/22
CI around μ vs. 𝑥̅
2 possibilities for the raw
observed effect (∆= 200):
2 ways of saying the same thing
HA is true
∆ = 200; PA affects CF
Population
H0 is true
∆ ≠ 200; Sampling variability
If H0 were true: 95% of
sample means from
941-1059 just from
sampling variability.
1000
Sampling Distribution
(Theoretical)
[941.2 — 1000 — 1058.8]
Sample
(Physical Activity
Intervention)
1200
– 95% p a sample mean will fall within this range
just from sampling variability
• CI around sample mean:
– 95% p the population mean falls within this range
just from sampling variability
p of 𝑥 b/t 941-1059 is
0.95
p of 𝑥 of 1200 (i.e.,
𝑥 < 941 or 𝑥 > 1059)
just from sampling
variability is less than
5% (p < 0.05) m = 1200 s = 150 n = 25 • CI around population mean: • Same decision each time with respect to H0 p(∆=200|H0) < 0.05 203 204 Confidence Interval around sample mean • 95% Confidence interval – UB/LB: 𝑥 ± (1.96 x s/√n) CI around 𝑥̅ vs. μ 2 possibilities for the raw effect (∆=200) H A is true (i.e., ∆ ≠ 0; PA affects CF) H 0 is true (i.e., ∆ = 0; Sampling variability) 𝜇 = 1000 • Our small sample of students – UB/LB: 𝑥 ± 58.8 – UB/LB: 1200 ± 58.8 – [1141.2 1258.8] – Does 𝜇 (1000) fall outside 95% CI around sample mean? 95% Confidence Interval In other words, the probability of observing a difference this large (∆ = 200) just from sampling variability is so low (p 𝛼 = 0.32 • Y axis is n (%) of sample • 90% CI (1.64 SE) -> 𝛼 = 0.10
• % -> p
• 𝛼 = (100% – CI%) / 100
• 68% CI (1.00 SE) -> p = 0.68
• 90% CI (1.64 SE) -> p = 0.90
Confidence Interval for any z
• 95% CI?
– 100% – 95% = 5%
– 5% = 𝛼 of 0.05
– But 0.05 in both tails of Normal Distribution!
– 0.05 ÷ 2 = 0.025
– z table: 𝛼 -> z
– z = 1.96
34
DISCOVERING STATISTICS USING SPSS
FIGURE 1.15
0.40
The probability
density function
of a normal
distribution
0.35
Probability = .95
0.30
Density
0.25
0.20
0.15
Probability = .025
Probability = .025
0.10
0.05
0.00
−4.00
212
213
−3.00
−1.96
−1.00
0.00
z
1.00
1.651.96
3.00
4.00
data. That is to go where eagles dare, and no one should fly where eagles dare; but to
become scientists we have to, so the rest of this book attempts to guide you through the
various models that you can fit to the data.
1.7. Reporting data !
1.7.1. Dissemination of research !
Confidence Interval for any z
• 90% CI?
Confidence Interval for any z
• 60% CI?
– 100% – 90% = 10%
– 10% = 𝛼 of 0.10
– But 0.10 in both tails of Normal Distribution!
– 0.10 ÷ 2 = 0.05
– Check table for corresponding z
– z = 1.64
214
– 100% – 60% = 40%
– 40% = 𝛼 of 0.40
– But 0.40 in both tails of Normal Distribution!
– 0.40 ÷ 2 = 0.20
– Check table for corresponding z
– z = 0.84
01-Field 4e-SPSS-Ch-01.indd 34
07/11/2012 6:14:56 PM
215
We could create any size (0-100)* confidence interval
using all the z’s in the z table
Sample
(Physical Activity
Intervention)
m = 1200
s = 150
n = 25
1200
[ 1150.8 —- 1249.2 ]
[ 1141.2 ————- 1258.8 ]
[ 1122.6 ———————- 1277.4 ]
216
Having established a theory and collected and started to summarize data, you might
want to tell other people what you have found. This sharing of information is a fundamental part of being a scientist. As discoverers of knowledge, we have a duty of
care to the world to present what we find in a clear and unambiguous way, and with
enough information that others can challenge our conclusions. Tempting as it may be
to cover up the more unsavoury aspects of our results, science should be about ‘the
truth’. We tell the world about our findings by presenting them at conferences and in
articles published in scientific journals. A scientific journal is a collection of articles
written by scientists on a vaguely similar topic. A bit like a magazine, but more tedious. These articles can describe new research, review existing research, or might put
forward a new theory. Just like you have magazines such as Modern Drummer, which is
about drumming, or Vogue, which is about fashion (or Madonna, I can never remember which), you get journals such as Journal of Anxiety Disorders, which publishes
articles about anxiety disorders, and British Medical Journal, which publishes articles
about medicine (not specifically British medicine, I hasten to add). As a scientist, you
submit your work to one of these journals and they will consider publishing it. Not
everything a scientist writes will be published. Typically, your manuscript will be given
to an ‘editor’ who will be a fairly eminent scientist working in that research area who
has agreed, in return for their soul, to make decisions about whether or not to publish
articles. This editor will send your manuscript out to review, which means they send it
to other experts in your research area and ask those experts to assess the quality of the
work. The reviewers’ role is to provide a constructive and even-handed overview of
the strengths and weaknesses of your article and the research contained within it. Once
these reviews are complete the editor reads them all, and assimilates the comments
z = 1.64 for 90% CI
z = 1.96 for 95% CI
z = 2.58 for 99% CI
LLN
• Sampling distribution
becomes normal when
sample sizes gets larger
• SE = s/√n
• In order for sampling
distribution to be
perfectly normal, n = ∞
217
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9/16/22
Experimental design
z vs. t
• In practice, z is often not used for inferential
statistics
– data are rarely normal enough
– often related to sample size
• t is for degenerate cases of normal distribution
– i.e., when sample is not large enough to be
perfectly normal
218
219
t for CIs
t for CIs
• Confidence Intervals
• (𝑥̅ ± t * SE)
• (𝑥̅ ± t * s/√n)
• t indexed by 3 criteria
• df (i.e., n-1)
• α (total area under curve outside ± t)
• Always 2 sided for CIs
• Confidence Intervals
• (𝑥̅ ± tα, df, 2-sided * SE)
• (𝑥̅ ± tα, df, 2-sided * s/√n)
• t indexed by 3 criteria
• df
•α
• # sides to test
220
34
FIGURE 1.15
221
z for 95% confidence interval
DISCOVERING STATISTICS USING SPSS
How to find the correct t for any CI
0.40
The probability
density function
of a normal
distribution
• All CIs are 2 sided by definition
• Find 𝛂 corresponding with __% CI
0.35
Probability = .95
0.30
Density
0.25
• Calculate df (n-1)
0.20
• Find corresponding t in t table
0.15
Probability = .025
Probability = .025
80%
0.10
90%
95%
98%
99%
0.05
0.00
−4.00
−3.00
−1.96
−1.00
0.00
z
1.00
1.651.96
3.00
4.00
data. That is to go where eagles dare, and no one should fly where eagles dare; but to
become scientists we have to, so the rest of this book attempts to guide you through the
various models that you can fit to the data.
For Z = 1.96 (95% CI)
• 2 side test = 0.050
1.7. Reporting data
• 1!side test = 0.025
1.7.1. Dissemination of research !
222
Having established a theory and collected and started to summarize data, you might
want to tell other people what you have found. This sharing of information is a fundamental part of being a scientist. As discoverers of knowledge, we have a duty of
care to the world to present what we find in a clear and unambiguous way, and with
enough information that others can challenge our conclusions. Tempting as it may be
to cover up the more unsavoury aspects of our results, science should be about ‘the
truth’. We tell the world about our findings by presenting them at conferences and in
articles published in scientific journals. A scientific journal is a collection of articles
written by scientists on a vaguely similar topic. A bit like a magazine, but more tedious. These articles can describe new research, review existing research, or might put
forward a new theory. Just like you have magazines such as Modern Drummer, which is
about drumming, or Vogue, which is about fashion (or Madonna, I can never remember which), you get journals such as Journal of Anxiety Disorders, which publishes
articles about anxiety disorders, and British Medical Journal, which publishes articles
about medicine (not specifically British medicine, I hasten to add). As a scientist, you
submit your work to one of these journals and they will consider publishing it. Not
everything a scientist writes will be published. Typically, your manuscript will be given
223
4
9/16/22
𝜇
224
225
Confidence intervals
to test H0 involving 𝛍
• When 𝛍 falls outside 95% CI around sample mean
• if H0 were true, then the p of observing a ∆ this
large between 𝑥̅ and 𝛍 just from sampling
variability is ≤ 𝛂 (.05)
• When 𝛍 falls inside 95% CI around sample mean
• if H0 were true, then the p of observing a ∆ this
large between 𝑥̅ and 𝛍 just from sampling
variability is > 𝛂 (.05)
226
5
Assignment 2
i
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Class A
grade
xbar
56 67.813
79 67.813
62 67.813
70 67.813
72 67.813
77 67.813
61 67.813
62 67.813
74 67.813
79 67.813
80 67.813
71 67.813
68 67.813
62 67.813
42 67.813
70 67.813
2
xi – xbar (xi – xbar)
-11.813 139.547
11.187 125.149
-5.813
33.791
2.187
4.783
4.187
17.531
9.187
84.401
-6.813
46.417
-5.813
33.791
6.187
38.279
11.187 125.149
12.187 148.523
3.187
10.157
0.187
0.035
-5.813
33.791
-25.813 666.311
2.187
4.783
z p below p above
-1.18
0.12
0.88
1.11
0.87
0.13
-0.58
0.28
0.72
0.22
0.59
0.41
0.42
0.66
0.34
0.91
0.82
0.18
-0.68
0.25
0.75
-0.58
0.28
0.72
0.62
0.73
0.27
1.11
0.87
0.13
1.21
0.89
0.11
0.32
0.63
0.37
0.02
0.51
0.49
-0.58
0.28
0.72
-2.57
0.01
0.99
0.22
0.59
0.41
STUDENT ID: _____________________
Class B
grade
xbar
79 75.938
80 75.938
80 75.938
69 75.938
73 75.938
82 75.938
77 75.938
81 75.938
74 75.938
79 75.938
61 75.938
90 75.938
90 75.938
57 75.938
73 75.938
70 75.938
xi – xbar (xi – xbar)2
z p below p above
3.062
9.376 0.340
0.63
0.37
4.062
16.500 0.450
0.67
0.33
4.062
16.500 0.450
0.67
0.33
-6.938
48.136 -0.780
0.22
0.78
-2.938
8.632 -0.330
0.37
0.63
6.062
36.748 0.680
0.75
0.25
1.062
1.128 0.120
0.55
0.45
5.062
25.624 0.570
0.72
0.28
-1.938
3.756 -0.220
0.41
0.59
3.062
9.376 0.340
0.63
0.37
-14.938 223.144 -1.670
0.05
0.95
14.062 197.740 1.570
0.94
0.06
14.062 197.740 1.570
0.94
0.06
-18.938 358.648 -2.120
0.02
0.98
-2.938
8.632 -0.330
0.37
0.63
-5.938
35.260 -0.660
0.25
0.75
xbar =
67.813
xbar =
75.938
n =
16.000
n =
16.000
SS = 1512.438
SS = 1196.940
df =
15.000
df =
15.000
s2 =
100.829
s2 =
79.796
s =
10.041
10.04
s =
8.933
8.93
Assignment 2
i
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Class C
grade
xbar xi – xbar (xi – xbar)2
71 74.125 -3.125
9.766
70 74.125 -4.125 17.016
78 74.125
3.875 15.016
74 74.125 -0.125
0.016
77 74.125
2.875
8.266
76 74.125
1.875
3.516
62 74.125 -12.125 147.016
78 74.125
3.875 15.016
66 74.125 -8.125 66.016
86 74.125 11.875 141.016
88 74.125 13.875 192.516
77 74.125
2.875
8.266
65 74.125 -9.125 83.266
71 74.125 -3.125
9.766
76 74.125
1.875
3.516
71 74.125 -3.125
9.766
z p below p above
-0.45
0.33
0.67
-0.59
0.28
0.72
0.56
0.71
0.29
-0.02
0.49
0.51
0.41
0.66
0.34
0.27
0.61
0.39
-1.74
0.04
0.96
0.56
0.71
0.29
-1.16
0.12
0.88
1.70
0.96
0.04
1.99
0.98
0.02
0.41
0.66
0.34
-1.31
0.10
0.90
-0.45
0.33
0.67
0.27
0.61
0.39
-0.45
0.33
0.67
STUDENT ID: _____________________
Class D
grade
xbar xi – xbar (xi – xbar)2
68 78.750 -10.750 115.563
76 78.750 -2.750
7.563
86 78.750
7.250 52.563
72 78.750 -6.750 45.563
79 78.750
0.250
0.063
77 78.750 -1.750
3.063
69 78.750 -9.750 95.063
80 78.750
1.250
1.563
91 78.750 12.250 150.063
91 78.750 12.250 150.063
84 78.750
5.250 27.563
70 78.750 -8.750 76.563
82 78.750
3.250 10.563
87 78.750
8.250 68.063
77 78.750 -1.750
3.063
71 78.750 -7.750 60.063
xbar =
74.125
xbar =
78.750
n =
16.000
n =
16.000
SS = 729.756
SS = 867.008
df =
15.000
df =
15.000
s2 =
48.650
s2 =
57.801
s =
6.975
6.97
s =
7.603
7.60
z p below p above
-1.41
0.08
0.92
-0.36
0.36
0.64
0.95
0.83
0.17
-0.89
0.19
0.81
0.03
0.51
0.49
-0.23
0.41
0.59
-1.28
0.10
0.90
0.16
0.56
0.44
1.61
0.95
0.05
1.61
0.95
0.05
0.69
0.75
0.25
-1.15
0.13
0.87
0.43
0.67
0.33
1.09
0.86
0.14
-0.23
0.41
0.59
-1.02
0.15
0.85