# University of Central Florida Test Performance of Students Questions

9/21/22Confidence 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)
227
228
Testing H0
95% CIs do
NOT overlap
• 1st test of H0: ____ % Confidence Intervals
• 𝑥 vs 𝛍
• ∆=0
• 95% CI -> 𝛂 = 0.05 -> p 𝛂 = 0.10 -> p 𝛂 = 0.01 -> p CF
Increasing Complexity/Confidence
3 types of t tests
• IV/Predictor
• Physical Activity
DV/Outcome
Cognitive Function
à
à
• Time points (dv)?
Test scores
LOM = Continuous
Normally Distributed
𝑥 𝑠
Group Assignment
LOM = Binary
233
# samples?
– Cross-sectional,
– Cross-sectional,
– Cross-sectional,
– Pre-post,
– Pre-post,
– Pre-post,
One sample t (𝑥 𝑣 𝜇)
1 group/sample
2 groups (bt subjects) Independent samples t
>2 groups (bt subjects) One way ANOVA
1 group (wn subjects) Dependent samples t
2 groups (bt/wn subjects)RM Factorial ANOVA
>2 groups (bt/wn subjects)RM Factorial ANOVA
– Longitudinal,
– Longitudinal,
– Longitudinal,
RM ANOVA
1 group (wn subjects)
2 groups (bt/wn subjects) RM Factorial ANOVA
>2 groups (bt/wn subjects)RM Factorial ANOVA
234
3 types of t tests
• One sample t-test (vs. 𝜇)
• Comparing one group 𝑥̅ to fixed value
• Must have 𝜇 to carry out this test
• Independent samples t-test
• Comparing two group 𝑥’s
̅ to each other
• Dependent samples t-test
• Also known as paired samples t-test
• Comparing 𝑥’s
̅ of one group before/after Tx
235
• Used 𝑡! to find Confidence Intervals (CI)
• 𝑥̅ ± (𝑡! * SE)
• If 𝜇 fell outside CI around 𝑥,̅ then
* 𝐻” < 0.05 •𝑝 ∆ • Reject 𝐻" • If 𝜇 fell inside CI around 𝑥,̅ then * 𝐻" > 0.05
•𝑝 ∆
• Fail to Reject 𝐻”
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• Instead of calculating CI’s and looking for
overlap to test H0
• Could Calculate 𝑡#\$% =

‘( ∆
• Then directly compare 𝑡#\$% to 𝑡!
t=0
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9/21/22
POPULATION (𝛍,𝛔)
Sampling
Variability
Sampling Distribution
Theoretical Distribution of 𝑥
𝑥! (= 𝜇); SE
texp vs t𝛼
t𝛼=1.96
(p=0.05)
Samples
𝑡012 < 𝑡+ Sample (𝑥,s,n) t=0 Sample (𝑥,s,n) Sample (𝑥,s,n) Sample (𝑥,s,n) Sample (𝑥,s,n) Sample (𝑥,s,n) Sample (𝑥,s,n) Sample (𝑥,s,n) Sample (𝑥,s,n) 239 𝑡012 > 𝑡+,-.-/
If 𝑡012 ≥ 𝑡+,-.-/, then 𝑝 ∆ 𝐻- ≤ 0.05
Reject H0
If 𝑡012 < 𝑡+,-.-/, then 𝑝 ∆ 𝐻- > 0.05
Fail to Reject H0
𝑡+,-.-/ ≥ 1.96
240
𝑡/01 23 𝑡4
Formula for one sample t
• one sample t-test:
𝑡\$%& 𝑑𝑓 =
Difference between
2 means
𝑡=

𝑆𝐸∆
𝐻” : 𝑡 = 0, 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 ∆ = 0
(Average difference
expected from sampling
variability if H0 were
true)
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−𝑆𝐸∆ 𝑆𝐸∆
3 t-tests all have same conceptual form:
&

𝑥̅ − 𝜇
=𝑠
𝑆𝐸∆
. 𝑛
Standard Error
𝑥 = sample mean
μ = population mean
s = sample standard deviation
n = sample size
𝑜𝑛𝑒 𝑠𝑎𝑚𝑝𝑙𝑒 𝑡 (𝑣𝑠 𝜇) =

‘(∆

\$*+

, #
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Formula for (two) independent
samples t-test
Experimental design
𝑥̅ A = 1000 (sd=250, n=50)
𝑥̅ B = 1300 (sd=250, n=50)
It’s computationally
burdensome to derive a
sample of differences.
Easier to use the difference
between the means (𝑥\$ − 𝑥% )
to estimate the
mean difference (∆)
246

𝑆𝐸∆
𝑡012 =
1000
• Independent samples t-test:
1300
𝑥. − 𝑥/ → ∆
Need 𝑡&!’ to calculate 𝑝 ∆ 𝐻(
𝑡&!’ =
mean difference of 300
(sd=290)

𝑆𝐸∆
xbar1: mean of sample 1
xbar2: mean of sample 2
Find way to estimate 𝑆𝐸∆
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9/21/22
Formula for (two) independent
samples t-test
• Independent samples t-test:
Formula for two (independent)
samples t-test
• when samples sizes are unequal or equal:
Differences
between groups
Pooled Variance
• ONLY WORKS WHEN SAMPLE SIZES FROM
BOTH GROUPS ARE EQUAL (i.e., when n1 = n2)
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249
Experimental design (PA-> CF)
Experimental design
Compare two groups and observe Δ in SAT scores
2 reasons observe Δ
• H 0:
Mean from control group is 1000 (sd=100)
• ∆=0
• ∆ 91 :; cognitive function
2 reasons for observed difference in sample mean
SAT scores (Δ=300, nTX = 50, nCX = 50)
• H 0:
Experimental design
Mean from school1 is 1000 (sd=250, n=50)
Mean from school2 is 1300 (sd=250, n=50)
300
• ∆=0
• ∆ 91 :;

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