# Component process model of memory - Download as PowerPoint

Document Sample

```					    Introduction to the t statistic

• Overview
– The previous chapter used a z statistic to
determine whether an observed mean differed
from the population mean
– Z = (M - )/M
– Limitation: necessary to know the population SD
– a t-statistic can be used in situations in which the
population SD is not known
Introduction to the t statistic

• The t statistic
– The t statistic is similar to the z statistic; however,
unlike the z statistic, it estimates from the data
the population variability
– Recall SS = ∑ (X – sample mean)2
– Sample variance = s2 = SS/ (n-1)
– Sample standard deviation = s = √(SS/n-1)
Or s = √SS/df
Introduction to the t statistic

• The t statistic
–   When the population variance is known.
–   Standard error = σM = σ/n.5 = √ σ2/n
–   Estimated standard error sM = s/n.5 = √ s2/n
–   Note: sM is an estimate of the standard error
–   Used when σ is unknown
–   Computed from the sample variance
–   Estimates distance between M (sample mean) and 
Introduction to the t statistic

• The t statistic
– The t statistic = t = (M-)/ sM
– where sM = √ s2/n
• Summary
– the t statistic used to test hypotheses about an
unknown  when  is unknown. Structure of t
similar to z
Introduction to the t statistic

• Similarity of z and t statistic
– Z = (M - )/M = (M - )/√ 2 /n

– t statistic = t = (M-)/ sM = (M-)/ √ s2/n

– Note: 2 statistics are similar except that s2 is used with t
and 2 is used with z; thus, the t statistic approximates
the z statistic to the extent that the sample variance
approximates the population variance
Introduction to the t statistic

• Degrees of freedom
– Degrees of freedom, df = n-1
– The larger the df the better the sample variance
approximates the population variance
• t distributions
– By taking all possible samples of a given size (n)
from a population it is possible to construct a t
distribution
Introduction to the t statistic

• t distributions
– t distributions are symmetrical just like normal
distributions
– As the sample size increases the shape of the t
distribution becomes more similar to normal
distribution (see Figure 9.1)
– As this figure shows, the t distribution is flatter
and more spread out
– Mean of t distribution is 0
Introduction to the t statistic

• t distributions
– In order to test hypotheses it is necessary to
determine proportions for different t values, just
as one does with z distributions
– Appendix B.2 has t distribution table
– Note: proportion (1 or 2 tailed) is at top; left-most
row has df; the body of the table has the t-
distribution value
Introduction to the t statistic

• t distributions
– E.g., study involving 20 people evaluates hypothesis with
population mean memory performance is affected by a
memory training program
– Results find a t-value of 2.86. What proportion of scores
in t-distribution have scores greater than 2.86
– Sol’n: df = 20-1 =19. look up value in table. Answer .005
Introduction to the t statistic

• Hypothesis testing with t distributions
– Logic is the same as with a z score. There is a known
population mean. Purpose of study is to determine
whether a treatment changes population mean
significantly.
– Step 1. state null and alternative hypothesis
 Population mean is unchanged as a result of treatment H0
 set alpha level
– Step 2. Locate critical region in t-distribution
– Step 3. Collect data and compute test statistic
– Step 4. Evaluate null hypothesis
Introduction to the t statistic

• Example
– Purpose of study is to determine whether birds avoid a
chamber in which eye-spot patterns have been painted.
Bird placed in a chamber for 60 minutes. 1 compartment
has eye-spot pattern and other (plain) compartment does
not.
– DV = time spent in plain compartment
 H0  = 30 minutes
 H1  ≠ 30 minutes
– Alpha = .05 two tailed. why two tailed?
Introduction to the t statistic

• Example
– Results: n=16 birds; M = 39 minutes on plain
side; SS = 540
– Locate critical region
Df = 15; alpha = .05, two tailed; t (15) ± 2.131
– Calculate t
– t = (M-)/ sM = (M-)/ √ s2/n
S2 = SS/(n-1) = 540/15 = 36
√ s2/n = √36/16 =√2.25 = 1.50
Introduction to the t statistic

• Example
– t = (M-)/ sM = (M-)/ √ s2/n
= (39-30)/1.50 = 6.0
– Make decision
Reject null hypothesis. Birds spent significantly
more time in the plain compartment than in the
eye-spotted compartment
Introduction to the t statistic

• APA format
– Report descriptive statistics used M = 39
minutes, SD = 6
– Report inferential statistics
– t(15) = 6.0, p < .05
– t (df) = t score, significance level
Introduction to the t statistic

• Importance of looking at your data
– Always look at your data descriptively before
looking at it inferentially
– If you graph your data and the raw scores are
bunched far away from the population mean you
can be confident that the sample mean differs
significantly from the population mean
– See Figure 9.6
Introduction to the t statistic

• Assumptions of the t test
– 1. sample must consist of independent
observations
– How to ensure: use random sampling
– 2. population sampled must be normal
– This assumption is not critical. That is, data show
that the t statistic is robust when this assumption
is violated
Hypothesis test with 2 independent
samples
• Overview
– Previously we have shown that a z statistic could
be used to determine whether an observed mean
differed from the population mean
– Z = (M - )/M
– Limitation: necessary to know the population SD
Hypothesis test with 2 independent
samples
• Overview
– then we showed that a t-statistic can be used in
situations in which the population SD is not
known
– Limitation: the particular type of t statistic can
only be used when there is a single sample
– In psychology and other disciplines one often
wants to determine whether performance
between two samples (or groups) differs
Hypothesis test with 2 independent
samples
• Overview
– This chapter will present a test that can be used
in this situation
– e.g., compare performance of two different
training procedures administered to two different
groups
– e.g., compare performance of two groups trained
in the same way but tested in two different ways
Hypothesis test with 2 independent
samples
• Note:
– Procedure in this chapter applies to a situation in
which the data come from two separate samples
– Hence this procedure is called an independent-
measures research design or a between-subjects
design
– Subsequently we will discuss a procedure in
which the comparison is made between two sets
of data collected from the same sample
Hypothesis test with 2 independent
samples
• Independent measures or between-subjects
design
– Experimental design that uses a different sample
for each condition or treatment
Hypothesis test with 2 independent
samples
• Goal
– To determine whether there is a significant
difference between the means of two different
populations (or two different treatment
conditions)
– Notation: a subscript 1 will designate all
parameters of population/treatment 1, while a
subscript 2 will designate parameters of
population/treatment 2
Hypothesis test with 2 independent
samples
• Hypothesis
– H0 : 1 - 2 = 0 in other words the population means are
equal
– H1 : 1 - 2 ≠ 0
• Further notation
– t test used to test the mean of a single sample will be
referred to as the single sample t statistic
– t test used to test the mean difference between two
independent samples will be called the independent
measures t statistic
Hypothesis test with 2 independent
samples
• General formula for t statistic
– t = (sample statistic –population parameter)/
standard error
– Recall APA uses M to designate sample mean
• Single sample t
– t = (M - )/standard error
• Independent means t test
– t = ((M1 – M2) – (1 - 2))/standard error
Hypothesis test with 2 independent
samples
• Standard error in t statistic
– Measures how accurately the sample statistic measures
the population parameter
– In the case of a single sample t statistic, the standard
error sM represents the error between a sample mean M,
and the population mean 
– In the two sample t statistic, standard error represents the
error between the sample mean difference (M1 – M2) and
the population mean difference (1 - 2)
Hypothesis test with 2 independent
samples
• Standard error in t statistic
– Recall each mean estimate, M1 and M2, has associated
with it an error, which can be estimated by the standard
error of the mean,
– sM = √s2/n (formula for single sample standard error)
– Note: errors add even though you are subtracting the two
sample means. This makes sense because there are now
two sources of error
– This means that sM1-M2 = √(s12/n1+ s22/n2)
– Caveat: this formula is appropriate only when n1 = n2
Hypothesis test with 2 independent
samples
• Standard error in t statistic
– When n1 ≠ n2 one must used the pooled estimate
of the variance
– Recall s2 = SS/ df
– Pooled estimate of variance = Sp2 = (SS1 +
SS2)/(df1 + df2)
– Note: Sp2 is the average of the 2 variances.
Hence it must be between the two variance
estimates
Hypothesis test with 2 independent
samples
• Standard error in t statistic
– Two sample standard error = sM1-M2
– sM1-M2 = √(sp2/n1+ sp2/n2)

• t statistic for independent measures
– t = ((M1 – M2) – (1 - 2))/ sM1-M2
• df for the t statistic = df1 + df2 = n1- 1 + n2 -1
Hypothesis test with 2 independent
samples
Sample Pop’n     Std error
data   parameter
t single    M               √(s2/n
sample
t indepen.   M1- M2   1 - 2    √(sp2/n1+ sp2/n2)
measures
Hypothesis test with 2 independent
samples
• Single sample variance
– S2 = SS/df

• Two independence sample variance
– Sp2 = (SS1 + SS2)/(df1 + df2)
Hypothesis test with 2 independent
samples
• Hypothesis test with independent measures t
statistic
– See related example 10.1
• Purpose: to determine whether method of loci
instructions improve memory for concrete nouns.
Design: Between subjects design: group 1 (no
specific instructions), group 2 (method of loci); n=
10 participants in each group
Hypothesis test with 2 independent
samples
• Hypothesis
– H0 : 1 - 2 = 0 memory performance does not
differ between the two treatment conditions
– H1 : 1 - 2 ≠ 0 imagery makes a difference
• Locate critical region
– Set alpha = .05
– Df = df1 + df2 = 10 – 1 + 10 – 1 = 18
– Look up t (18) value ± 2.101
Hypothesis test with 2 independent
samples
• Calculate test statistic
– Results
– Group 1 (no imagery): n = 10, M = 19, SS = 40
– Group 2 (imagery): n = 10, M = 26, SS = 50
– Sp2 = (SS1 + SS2)/(df1 + df2) = (40 + 50)/ (9+9)
–      = 90/18 = 5.0
– sM1-M2 = √(sp2/n1+ sp2/n2) = √(5/10 + 5/10 = 1
Hypothesis test with 2 independent
samples
• Calculate test statistic
– t = ((M1 – M2) – (1 - 2))/ sM1-M2
= (19 – 26) / 1
= -7

• Make decision
– M1 = 19; SD = √SS/df = √40/9 = √4.44 = 2.11
– M2 = 26; SD = √SS/df = √50/9 = √5.55 = 2. 36
– t (18) = -7 , p < .05, two tailed. Data suggest that memory
performance under imagery conditions is higher than under no
imagery instructions
Hypothesis test with 2 independent
samples
• Using descriptive statistics to check plausibility of
inferential conclusion
– M1 = 19; SD = √SS/df = √40/9 = √4.44 = 2.11
– M2 = 26; SD = √SS/df = √50/9 = √5.55 = 2. 36
What these data are saying is that most of the scores
associated with Group 1 fall between 17 and 21, whereas
most of the scores associated with Group 2 fall between
23-24 and 28-29. Since these distributions barely overlap,
it is plausible to conclude that the two means differ
significantly from each other
Recall in z distribution ± 1 SD contains about two thirds of
the data
Hypothesis test with 2 independent
samples
• Using descriptive statistics to check plausibility of
inferential conclusion
– M1 = 19; SD = √SS/df = √40/9 = √4.44 = 2.11
– M2 = 26; SD = √SS/df = √50/9 = √5.55 = 2. 36
• Note: in order to check plausibility of conclusions it
is important to take into consideration
– Magnitude of difference between the means
– The size of the SD
– The sample sizes of the two distributions
Hypothesis test with 2 independent
samples
• Assumptions
– Observations are independent
– Two populations must be normally distributed
– Two populations must have equal variances
Hypothesis test with 2 independent
samples
• Assumptions
– The first two assumptions were already
discussed in the one-sample t test
– The third assumption is referred to as
homogeneity of variance
– Although modest violations of homogeneity do
not affect interpretation of data, more severe
violations make interpretation difficult
Hypothesis test with 2 independent
samples
• Testing for homogeneity of variance
(Hartley’s F-max statistic)
– Rationale: if population variances are
homogeneous, then sample variances should
also be similar
– Thus, ratio of the two sample variances should
be close to 1
Hypothesis test with 2 independent
samples
• Testing for homogeneity of variance
(Hartley’s F-max statistic)
– Procedure: compute sample variance for each
sample: recall s2 = SS/ df
– Select largest and smallest s2 and compute ratio
F-max = s2 (largest)/ s2 (smallest)
Note: f-max should be close to 1 if there is
homogeneity; compare value of F-max to Table B.

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 9 posted: 7/4/2012 language: English pages: 40