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Lecture 9

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Psychology

PSY201H1

Alison Luby

Fall

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PSY201 Lecture 9; Nov. 17, 2011 ○ Rmbr s = SS/(n-1)
○ s = √(SS/df)
Introduction to the t Statistic
The t Statistic: An Alternative to z • sMused more in terms of sample variance, since
• When hypothesis testing: provides accurate & unbiased estimate of the
○ Use a sample mean M to approximate the pop variance σ2
pop mean µ.
○ Standard error, σM, measures how well M
approximates µ ○ Estimated sample error = =
σM= σ/√n ○ Rmbr must know sample mean before
○ Then compare M to µ by computing the computing sample variance, so there is a
relevant z-score test statistic restriction on sample variability
z = (M - µ)/ M ○ Only n - 1 scores are independent & free to
Use to determine whether the obtained vary
dif is greater than expected by chance ○ n - 1 = degrees of freedom (df) for sample
by looking at unit normal table (for a variance
normal distribution)
• Like z-score, t statistic allows researchers to us• t statistic (like the z-score) forms a ratio:
sample data to test hypotheses about an ○ Numerator: Obtained dif btwn M
unknown µ
hypothesized µ
• Particular advantage of t statistic: ○ Denominator: Estimated standard error
○ Doesn’t require any knowledge of the pop (measures how much dif expected by
σwhich we typically don’t know chance)
○ So can be used to test hypotheses about a
completely unknown pop, ie:
Both µ & σ unknown • A large value for t (a large ratio) indicates
Only available info about pop comes obtained difference btwn data & hypothesis >
from sample expected if treatment has no effect (ie. is just
• All required for a hypothesis test with t is a
sample & a reasonable hypothesis about µ error)
○ W large sample, df large, so estimation is
• When we don’t haveσcan estimate it using the very good
sample variability So t statistic will be very similar to z-
○ Much like estimating µ using M score
• Can use t statistic when determining whether
○ W small samples, df small, so t statistic
treatment causes a change in µ provides relatively poor estimate of z
• Sample is obtained from the pop treatment is • Can think of the t statistic as "estimated zscore."
administered to sample
○ As usual, if resulting sample M is significantly
different from original µ, can conclude that
;
treatment has a significant effect
• Like before, hypothesis test attempts to decide • Just like we can make a distribution of z-scores,
btwn: we can make a t distr
○ H o Is it reasonable that the discrepancy ○ Complete set of t values computed for every
possible random sample for a specific
btwn M & µ is simply due to sampling error
and not result of treatment effect? sample size n, or specific degrees of freedom
○ H 1 Is the discrepancy btwn M & µ more than df
expected by sampling error alone? • As df approaches infinity, t distr
ie. Is M significantly different from µ approximates normal distr
st • How well it approximates normal distr
• Critical 1 step for t statistic hypothesis test:
○ Calculate exactly how much dif btwn M & µ is depends on df
reasonable to expect ○ ie. There’s a “family” of t distrs so there’s a
○ But since σ is unknown, impossible to dif sampling distr of t for each possible df
• Larger n larger n – 1 better the t distr
compute standard error of M (σ =Mσ/√n) as approximates the normal distr
done w z-scores
○ t statistic requires to use sample • W small values for df t distr flatter & more
data’s variance, s to compute spread out than a normal distr
estimated standard error, s
M
• s M s/√n ○ ie. obtained dif btwn data & hypothesis (ratio
in numerator) is significantly larger than dif
expected if no treatment effect (standard
error in the denominator)
• When M >> µ (compared to estimated standard
error), obtain large value for t, so can reject H
0
• When mean dif small compared to s , faMl to
reject H0
• Example 1
○ z distr (red) & t-distr (green) (n-1 = 4 = df)
○ t peak lower than z, sides higher than z • Researcher wanted to know whether people who
(more density away from mean) practice yoga have dif (includes greater and
lesser) depression level from general pop; uses
a depression inventory that has mean score of
25 for pop
• 14 indivs who regularly practice yoga complete
the inventory
• Scores: 15, 19, 24, 27, 18, 22, 23, 25, 17, 16, 20,
21, 17, 18
• Do the yoga students show a level of depression
significantly dif from general pop?
• • n = 14; µ = 25; M = 20.14; α = 0.05
○ z distr (red) & t-distr (purple) (n-1= 40 = df) • Hypotheses:
• As you can see, both distrs are unimodal & ○ H o µyoga= 25
symmetrical about the mean (= 0) ○ H 1 µyoga≠ 25
• However, have a dif shaped distrs when • Since we don’t know pop variance, have to use t
small n statistic, so need to find df
• And t has larger and more variable
2 ○ df = n – 1 = 13
variance, s • Now, determine our critical t value from the
○ ie. estimated standard error varies btwn table
samples ○ t = 2.16
s = √(s /n) critical
M • Then, determine the test statistic & reject H of >
As n & df ↑, variability in t distr ↓, tcritical
becomes more normal ○ SS = 5852 – 79524/14 = 171.71
• To evaluate t statistic from hypothesis test:
○ Must select an α level
○ Estimated sample error = s = M =
○ Then find the appropriate value of df
○ Determine the critical t statistic from the t 0.9713
distr table, Appendix B, page 729 SS/13 = 13.21
#s in table are values of t that separate 13.21/14 = 0.94
○ tobtained(20.14-25)/0.9713 = -5.003
the tail from the main body of the distr
Please look at this table to see that as df

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