Independent Samples t-Test:
Compare two s__ m__ in a b__-groups design (i.e., each p__ is in only one c__/g__)
Sampling distribution of d__ b__ m__.
What does this mean??
Mean = μ1- μ2
To find the variance, we will create something called p__ variance
sample means, between-groups
participant, condition/group
differences between means
pooled
Hypothesis Tests & Distributions Overview:
z test:
- number of samples: __
- Distribution of __
single-sample t test:
- number of samples: __
- Distribution of __.
paired-samples t test:
- number of samples: __ (__ participants)
- Distribution of m__ d__ s__.
independent-samples t test:
- number of samples: __ (__ participants)
- Distribution of d__ b__ m__.
one, means
one, means
two (same participants), mean difference scores
two (different participants)
differences between means
Hypothesis Test For Independent Samples:
Dr. A. Sleep is interested in sleep’s effect on memory. She gives one group of 12 subjects a set of words to memorize right before sleeping for 10 hours, then has them recall them upon awakening. She gives another group of 10 subjects a list of words to memorize at the beginning of the day & asks for recall 10 hours later.
Step 1: Identify the populations, distribution, and assumptions.
Population 1: People asleep for 10 hours before recall
Population 2: People awake for 10 hours before recall
The distribution: differences between means (rather than a distribution of mean difference scores)
Test: Independent samples - 2 samples with different groups of participants
Assumptions:
DV: scale
Random selection
No evidence of non-normal distribution
Hypothesis Test For Independent Samples:
Dr. A. Sleep is interested in sleep’s effect on memory. She gives one group of 12 subjects a set of words to memorize right before sleeping for 10 hours, then has them recall them upon awakening. She gives another group of 10 subjects a list of words to memorize at the beginning of the day & asks for recall 10 hours later.
Step 2 – Null & research hypotheses
H0: μ1 = μ2
H1: μ1 ≠ μ2
Hypothesis Test For Independent Samples:
Dr. A. Sleep is interested in sleep’s effect on memory. She gives one group of 12 subjects a set of words to memorize right before sleeping for 10 hours, then has them recall them upon awakening. She gives another group of 10 subjects a list of words to memorize at the beginning of the day & asks for recall 10 hours later.
Asleep: 15, 13, 14, 14, 16, 15, 16, 15, 15, 15, 17, 14
Awake: 14, 13, 14, 12, 11, 13, 13, 12, 12, 13
Step 3 – Characteristics of comparison distribution
Need to determine appropriate mean.
Standard Error:
A) calculate corrected variance for each group like normal.
B) pool the variances
C) get variance versions of standard error
D) add the two variances together to get variance of distribution of differences between means
E) take square root to get standard error
Mean: If μ1 = μ2, then μ1 - μ2 = 0, so mean of comparison distribution is zero.
a) asleep group: -average: 14.92 -s2: 1.174 -Nx: 12
awake Group:
- average: 12.7
- s2: 0.9
- Ny: 10
b) Pool the variances:
dfx/df total S2x+dfy/df totalS2y
so (11/20)1.174+(9/20)0.9
–> 0.55(1.174)+.45(0.9)
–> 0.6457+0.405
=1.051
c) Get variance versions of standard error:
- divide pooled variance number by N individually for each.
1. 051/12=0.0876
1. 051/10=0.1051
D) Add the variances together to get variance of distribution of differences between means:
0.0876+0.1051=.1927
E) take square root to get standard error
√ .1927= 0.439
Hypothesis Test For Independent Samples:
Dr. A. Sleep is interested in sleep’s effect on memory. She gives one group of 12 subjects a set of words to memorize right before sleeping for 10 hours, then has them recall them upon awakening. She gives another group of 10 subjects a list of words to memorize at the beginning of the day & asks for recall 10 hours later.
Step 4 - Determine Critical Values:
Two-tailed, alpha=.01
df total=20
Look in t table under .01 column
t critical= -2.846, +2.846
Reject H0 if tobs > 2.846 or < -2.846
Hypothesis Test For Independent Samples:
Dr. A. Sleep is interested in sleep’s effect on memory. She gives one group of 12 subjects a set of words to memorize right before sleeping for 10 hours, then has them recall them upon awakening. She gives another group of 10 subjects a list of words to memorize at the beginning of the day & asks for recall 10 hours later.
Step 5 – Calculate test statistic (tobs)
Subtract the mean of Mx and My and then subtract 0 from those.
Divide by pooled standard error (aka very last number in part 3)
so: (14.92-12.7)-0/0.439
- ->
2. 22/0.439= 5.06
tobs=5.06
Hypothesis Test For Independent Samples:
Dr. A. Sleep is interested in sleep’s effect on memory. She gives one group of 12 subjects a set of words to memorize right before sleeping for 10 hours, then has them recall them upon awakening. She gives another group of 10 subjects a list of words to memorize at the beginning of the day & asks for recall 10 hours later.
t critical= -2.846, +2.846
tobs=5.06
Step 6 – Make Decision
Write in APA
tobs= 5.06 > 2.846 = tcrit
Reject H0
There is a significant difference in word recall between awake & asleep conditions (p < .01)
t(20)=5.06, p<0.01
Interpreting SPSS for independent samples t test:
Look at the s__ (_ tailed) column and specifically the t__ row.
You want a ns result (less than 0.05) to test the a__ of h__ of variance.
If NS (less than 0.05) then r\_\_ the null. If significant (greater than 0.05) f\_\_ to r\_\_.
significance (2), top
assumption, homogeneity
reject
fail to reject
Confidence Intervals for Independent-Sample t-test:
Let’s do a 95% confidence interval for the sleep/wake memory data
CI upper=
- 086(.439) + (14.92-12.7)
- -> .916+2.22= 3.14
CI lower=
- 2.086(.439)+(14.92-12.7)
- -> -.916+2.22 = 1.30
95% CI [1.30, 3.14]
Probability is .95 that interval constructed in this way will contain true difference between means.
Effect Size for the Independent Samples t-test:
Cohen’s d, use s pooled in denominator
BUT…it is variance, we need standard deviation…
Take square root
Then divide this by Mx-My result
size?
interpret in words
pooled variances= 1.051
√1.051 =1.03
next subtract Mx-My and divide this by 1.03
- 92-12.7= 2.22
- 22/1.03= 2.16
large effect
The asleep recall was 2 standard deviations higher than awake recall
