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Practice 10.15 Did the type of signal effect response time?
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t crit (15) = +/- 2.447 t obs = -2.24 Fail to reject Ho
The type of signal does not have a significant effect on response time.
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New Step Should add a new page Determine if One-sample t-test
Two-sample t-test If it is a dependent samples design If it is a independent samples with equal N If it is a independent samples with unequal N
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Thus, there are 4 different kinds of designs
Each design uses slightly different formulas You should probably make up ONE cook book page (with all 7 steps) for each type of design Will help keep you from getting confused on a test
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Practice Does drinking milkshakes affect (alpha = .05) your weight?
To see if milkshakes affect a persons weight you collected data from 5 sets of twins. You randomly had one twin drink water and the other twin drank milkshakes. After 3 months you weighed them.
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Results Water Twin A 186 Twin B 200 Twin C 190 Twin D 162 Twin E 175
Milkshakes 195 202 196 165 183
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Hypothesis Two-tailed
Alternative hypothesis H1: water = milkshake Null hypothesis H0: water = milkshake
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Step 2: Calculate the Critical t
N = Number of pairs df = N - 1 5 - 1 = 4 = .05 t critical = 2.776
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Step 3: Draw Critical Region
tcrit = tcrit = 2.776
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Step 4: Calculate t observed
tobs = (X - Y) / SD
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(D) -9 -2 -6 -3 -8 D = -28 D2 =194 N = 6 -28 3.04 = 194 5 5 - 1
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Step 4: Calculate t observed
tobs = (X - Y) / SD 1.36=3.04 / 5 N = number of pairs
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Step 4: Calculate t observed
-4.11 = (182.6 – 188.2) / 1.36 X = 182.6 Y = 188.2 SD = 1.36
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Step 5: See if tobs falls in the critical region
tcrit = tcrit = 2.776 tobs = -4.11
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Step 6: Decision If tobs falls in the critical region:
Reject H0, and accept H1 If tobs does not fall in the critical region: Fail to reject H0
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Step 7: Put answer into words
Reject H0, and accept H1 Milkshakes significantly ( = .05) affect a persons weight.
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What if. . . You were asked to determine if psychology and sociology majors have significantly different class attendance (i.e., the number of days a person misses class) You would simply do a two-sample t-test two-tailed Easy!
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But, what if. . . You were asked to determine if psychology, sociology, and biology majors have significantly different class attendance You can’t do a two-sample t-test You have three samples No such thing as a three sample t-test!
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One-Way ANOVA ANOVA = Analysis of Variance
This is a technique used to analyze the results of an experiment when you have more than two groups
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Example You measure the number of days 7 psychology majors, 7 sociology majors, and 7 biology majors are absent from class You wonder if the average number of days each of these three groups was absent is significantly different from one another
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Hypothesis Alternative hypothesis (H1)
H1: The three population means are not all equal
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Hypothesis Alternative hypothesis (H1) socio = bio
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Hypothesis Alternative hypothesis (H1) socio = psych
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Hypothesis Alternative hypothesis (H1) psych = bio
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Hypothesis Alternative hypothesis (H1) psych = bio = soc
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Hypothesis Alternative hypothesis (H1)
Notice: It does not say where this difference is at!!
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Hypothesis Null hypothesis (H0) psych = socio = bio
In other words, all three means are equal to one another (i.e., no difference between the means)
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Results X = 3.00 X = 2.00 X = 1.00
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Logic Is the same as t-tests
1) calculate a variance ratio (called an F; like t-observed) 2) Find a critical value 3) See if the the F value falls in the critical area
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Between and Within Group Variability
Two types of variability Between the differences between the mean scores of the three groups The more different these means are, the more variability!
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Results X = 3.00 X = 2.00 X = 1.00
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Between Variability S2 = .66 X = 3.00 X = 2.00 X = 1.00
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Between Variability + 5 X = 3.00 X = 2.00 X = 1.00
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Between Variability X = 8.00 X = 2.00 X = 1.00
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Between Variability S2 = 9.55 X = 8.00 X = 2.00 X = 1.00
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Between Group Variability
What causes this variability to increase? 1) Effect of the variable (college major) 2) Sampling error
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Between and Within Group Variability
Two types of variability Within the variability of the scores within each group
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Results X = 3.00 X = 2.00 X = 1.00
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Within Variability S2 =.57 X = 3.00 X = 2.00 X = 1.00
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Within Variability S2 =.57 S2 =1.43 S2 =.57 X = 3.00 X = 2.00 X = 1.00
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Within Group Variability
What causes this variability to increase? 1) Sampling error
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Between and Within Group Variability
Between-group variability Within-group variability
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Between and Within Group Variability
sampling error + effect of variable sampling error
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Between and Within Group Variability
sampling error + effect of variable sampling error Thus, if null hypothesis was true this would result in a value of 1.00
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Between and Within Group Variability
sampling error + effect of variable sampling error Thus, if null hypothesis was not true this value would be greater than 1.00
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