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Psychology 9 Quantitative Methods in Psychology Jack Wright Brown University Section 20

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Psychology 9 l.jpg

Psychology 9

Quantitative Methods in Psychology

Jack Wright

Brown University

Section 20

Note. These lecture materials are intended solely for the private use of Brown University students enrolled in Psychology 9, Spring Semester, 2002-03. All other uses, including duplication and redistribution, are unauthorized.


Agenda l.jpg
Agenda

  • The t distribution

    • One sample inference

    • Two sample inference

  • Announcements

    • Bring standard normal and t tables to class

    • Assignment: Chapter 10

    • Exam: next Tuesday

    • Review: part of Thursday


Inferences about means when n small l.jpg

X - m

sx

Inferences about means WHEN N SMALL…

Normally

distributed

tdf =

A variable that

Is no longer

Normal.

Distributed

As t, with N - 1

“degrees of freedom”

Or df.


Using the t distribution l.jpg
Using the t distribution

  • When m is not known, logic of confidence interval using t is the same as before, but we use t instead of z

  • For confidence intervals using normal:

    • CIa = X +/- Za* sx

    • CIa = X +/- Za* sx (with large n)

  • For confidence intervals using t:

    • CIa = X +/- tdf,a* sx (with small n)

^

^


Using the t distribution5 l.jpg
Using the t distribution

  • When m is known, or when some hypothesis about an assumed m is being test, logic of evaluating t is also the same

  • 1. obtain empirical t:

    • tobserved = (X - m)/sx

  • 2. Proceed to evaluate as with z, except that use t distribution with N - 1 df

    • as before, for “exact” approach: report probability of obtaining observed t:

      • t(df) = observed t, p < inferred probability.

      • eg: “t(22) = 3.43, p < .01”

    • as before, for NHST, “accept” or “reject” null according to pre-established a


Illustrative problem using t l.jpg
Illustrative problem using t

  • A firm claims their prep course increases GRE scores by 8 points, noting that 10 people who took the course had mean = 608, with s=10. Assume all you know is that m for the population = 600. Using a two-sided approach, evaluate the probability of obtaining a result this extreme or more extreme, in either direction.


Our t problem l.jpg
Our t problem

  • Given: mean = 608; m = 600; s = 10; n = 10.

  • SEM = s/sqrt(N) = 10/sqrt(10) = 3.16

  • Observed t = (608 - 600)/3.16 = 2.53


Slide8 l.jpg
T(9)

So, report t(9) = 2.53,

p < .04.

tobs = 2.53

total area =

.016 + .016 = .032.

p = ~.016

p = ~.016

-2.53

-2.53


Our problem l.jpg
Our problem…

  • As we know, we could also evaluate our result using NHST

    • set a in advance (e.g., .05, two-tailed)

    • Locate result in region of acceptance or rejection

    • “accept” or “reject” null

    • See next…


Slide10 l.jpg
T(9)

tobs = 2.53

acceptance region

-2.26

tcrit = 2.26


Note on using t tables l.jpg
Note on using t tables

  • See overhead

  • Walk through


Other notes on t l.jpg
Other notes on t

  • As we know, as n increases, t approaches standard normal

  • Most t tables stop at df=100

  • Switch to normal at that point

  • However, computers keep computing

  • So, you will see “t(500) = 2.02, p < .05”, etc.

  • Key here is to know what you are doing and what this really means


Close to actual example of t l.jpg
Close to actual example of t

  • “Narragansett--High school principal David G. Andrews, while cautioning that SAT scores are not designed to evaluate teachers, schools, or school systems, has released the latest test results that show Narragansett students’ scores are higher overall than in the previous year…Andrews said the average score climbed 10 points…”


Close to actual example of t14 l.jpg
Close to actual example of t

  • Relevant facts:

    • Mean = 542

    • N = 79 students

    • Assume s = 100

    • Infer true mean for last year was 532

    • Evaluate this report


Close to actual example of t15 l.jpg
Close to actual example of t

  • SEM = s/sqrt(n) = 100/sqrt(79) = 11.25

  • t(78) = (mean - m)/SEM

  • = (542 - 532)/11.25

  • = .89

  • Df of 60 is closest conservative df to desired df=78 (see table)

  • So, t critical for 95% CI with df=60 is 2.00

  • CI(.95) = 542 +/- 2.00*11.25

  • CI(.95) = 542 +/- 22.4

  • Note what this means


Close to actual example of t16 l.jpg
Close to actual example of t

  • Alternatively, evalute probability of obtaining this result

  • From table for df=60, probability of |t| = .89 or greater is at least .20, or p > .20

  • More exactly, from (much) better table or, more often, from computer, we get:

  • See next…


Slide17 l.jpg
t(78)

So, report:

“t(78) = .89, p > .3.”

tobs = .89

total area =

.188 + .188 = .38

p = ~.188

p = ~.188

-.89

+.89


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