t tests Flashcards by D'Mayah Lewis

two ways to generate two gets of scores

repeated measures design
independent groups design

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repeated measures design

every subject is exposed to each treatment condition and scores measured
comparison are between scores for the same individuals under different conditions
analyze using paired t test

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independent groups design

each subject is exposed to a single treatment condition and scores measured
comparisons are between scores from different individuals under different conditions
analyze using the independent t test

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what are some sources of variation between scores?

effects of treatment

individual differences
- differences in baseline score
- differences in responsiveness to treatment

effects associated with uncontrolled variables
measurement error

repeated measures eliminates variations due to individual differences!

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repeated-measures calculated difference (D) scores

Di = yBi — yAi

difference= (condition B) - (condition A)

this converts two sets of scores (conditions A and B) into a single set of scores (D)
- can compare and see if there’s variation between scores or variation between D scores

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is there an effect of treatment?

Is there a difference between scores measured under the two conditions?

no effect of treatment, the sets of scores are identical
Di = yBi - yAi = 0

we wouldn’t expect all D scores to be exactly zero, but the mean of all D scores should approx to 0

null hypothesis: H0: muD=0

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how to test null hypothesis

determine the probability that observed sample mean would have been obtained from a population where muD=0
- p value is the probability of obtaining observed data, if H0 is true

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2 approaches to test H0

use sample data to obtain a sampling distribution (similar to DSM) with a mean of D-bar
- determine location of muD=0 (H0) within this distribution
- calc 95% CIs
position the same sampling distribution with a mean of muD=0 (H0)
- determine location of observed D-bar within this distribution
- hypothesis testing

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GLM for D scores

GLM is:
Di= D-bar + error

based on D scores, not raw scores

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bootstrapping 95% Cl for muD

use observed D scores to generate an infinite hypohtesis population of D scores
randomly sample n=16 D scores (match original sample size) from pop and calculate D-bar
repeat many time, generating new random sample and calculating D-bar
generate a distribution of D-bar
- instead of distribution of sample means (DSM), this is the distribution of mean differences (DMD

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bootstrapping 95% Cl in R

boot.t.test ()
- calculates standard deviation of DMD (standard error)
- defines a precise p value for probability of obtaining observed data, if null hypothesis is true

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how to test H0

DMD estimates the distribution of all sample means that would be obtained from a population that matched our sample

we can located 0 in the distribution and focus on difference between D-bar and 0

subtract value of D-bar from all values in distribution
- calculates difference between D-bar and 0 but now 0 is the mean of the DMD

If D-bar - muD is small == observed data are likely if H0 is true

If D-bar - muD is large == observed data are unlikely if H0 is true

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standardizing D-bar - muD

convert scores to z scores

zyi= (yi-y-bar)/(Sy)

subtract the mean of the DMD (0) then divide by standard deviation of the DMD

standard deviation of DMD represents the average value of the D-bar - muD
- standard deviation seeks to calculate the average deviation score

standardized values tell us the average difference that we would expect between D-bar and mu=0 if H0 is true
- if the observed difference between D-bar - muD is twice as large as the average difference that would be expected due to sampling variation, if H0 is true

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central limit theorem and t statistics

t= D-bar / (Sd-bar) = D-bar / (Sd/ sqrt n)

if t=2, the observed difference between D-bar and H0 is twice as large as the average difference expected due to sampling variation

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t distribution

start with noramly-distributed pop of D scores with muD=0
- pop has normal distribution, assumption of CLT
- pop can have nay standard deviation as the t statistic standardizes the values of D-bar - mud based on the observed value of Sd

define sample size (n)

randomly sample n D scores from poulation and calculate t based on sample (t= D-bar - muD/S d-bar)

repeat one million times and plot distribution

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what does the t distribution represent ?

all values of t that would be expected based on the sample size if H0: muD=0, is true

what does the shape of the t distribution depend on?

sample size!

when calculating t, Sd is being used to estimate the corresponding population parameter
- estimate is less accurate for samples with smaller n
- t statistics will therefore be less precise for smaller n, resulting in some estimates of t that are unusually large or small
- result is t distribution with smaller n have wider tails

applying the CLT t statistic 95% CI

for 95% CI we modify the same t distribution so it has muD=D-bar and s= Sd-bar
- 95% CI defined by the boundaries of the central 95% of this distribution

95%CI = D-bar +/- (tcrit x Sd-bar)

using R to find 95% CI

find tcrit
qt( p=0.025, df=15) — lower
qt(p=0.975, df=15) — upper
insert values into equation
95% CI= D-bar +/- (tcrit x Sd-bar)

paired test in R

t.test (study 1$b, study1$a, paired= TRUE)
t.test (study 2$b, study2$a, paired= TRUE)
- running a paired t test for differences between a and b

results:
t statistic, df, p=value, 95%CI interval, mean difference (D-bar)

bootstrapping t test in R

boot.t.test( study1$b, study1$a, paired=TRUE, R=100000)
boot.t.test( study2$b, study2$a, paired=TRUE, R=100000)
- from MKinfer package

results:
p value, standard deviation, 95%CI interval

independent groups design

compare two sets of scores generated through independent groups design
ex. x= grouping variable (CTRL, DRUG), y= all measured scores

fitting GLM with independent groups designs

recode xi with 0 or 1
- CTRL= 0
- DRUG= 1

use the model that predicts the value of y:
y-hat= b0 + b1x1

GLM is a linear model
- b0 is intercept, b1 is slope

hypothesis testing

CTRL referred to as 0 group with DRUG as 1 group
H0: no difference between groups
- no difference between the means of the populations
H0: u1-u0= 0
H1: u1-u0 =/ 0

bootstrapping with independent t test

group 0 is expanded to infinite hypothetical pop group 1 is expanded to infinite hypothetical pop n0 scores are sampled at random from hyp pop 0 n1 scores are sampled at random from hyp pop 1 calc means of each sample, then calc difference between sample means repeat many times then plot distribution of the difference between sample means (DDSM)

CLT and independent t tests

NHST: mean of t distribution is 0 standard deviation of t distribution is the standard error - have to standardize the observed value of (y-bar1 - y-bar0) by diving by standard error

paired vs independent t statistic equation

paired: t= (observed difference between Dbar and H0)/ (average difference between D-bar and H0 expected due to sampling variation) = D-bar -muD/Sd-bar independent t= (observed difference between (y-bar1-ybar0) and H0)/ (average difference between (y-bar1-ybar0 )and H0 expected due to sampling variation) = (y-bar1-ybar0)- (mu1-mu0)/ (S(y-bar1-ybar0))

s(ybar1-ybar0)

with paired t statistic, we had one set of scores (D) with independent there's variation in (y-bar1-ybar0) due to variation in both y1 and y0 variance sum law: if we sum multiple values of y, the variance of the sum will be the sum of the variances of each value of y - variance of (y-bar1-ybar0) will be the sum of the variance of y1 and y0

independent t statistic

t= (ybar1-ybar0) - (mu1-mu0)/ sqrt( (Sy1^2/n1) +(Sy0^2/n0))

Welch's t statistic

t=(y-bar1-ybar0) / sqrt( (Sy1^2/n1) +(Sy0^2/n0))

pooled variance Sp^2

according to H0, there's no difference between our two samples - samples came from the population if two samples came from same population, Sy1^2 and Sy0^2 will both estimate the same parameter of the population variance generate a single sample static to estimate population variance

pooled variance equation

s2p = Sum(y1- y-bar1)^2 + Sum(y- y-bar0)^2/(n1-1) + (n0-1)

two sample vs welch's t-test

two-sample t test is appropriate if samples have similar variance welch's t test is appropriate if sample have very difference variances - if using Welch's t test, the estimate of (sigma^2) will be inaccurate, so t statistic will be more prone to error - to compensate the Welch's t test uses t distributions with fewer df than two-sample t test to convert test statistic to a p value - this makes it more difficult to achieve statistical significance, counteracting the increased type I error rate that would result from the increase potential error

independent test in R

t.test( y~x, data= independent) - (outcome ~ predictor) this is the formula - default, R runs the Welch's t test (doesn't assume equal variances) to run regular two group t, modify var.equal argument - t.test( y~x, data= independent, var.equal=TRUE) bootstrap - boot.t.test (y~x, data= independent, R=100000)