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# Hypothesis testing t-stat

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A. Attitude change is thought to result from the anxiety produced under conditions of cognitive dissonance. The data below represent the anxiety scores of subjects in a dissonance study. Assuming that anxiety is distributed with mean=10 and standard deviation=2 test whether this sample exhibited more anxiety than would be expected by chance alone. Make a clear, concise interpretation.
n=30, x bar=9, SS=96
b. Next, compare (2-tailed) the above (dissonance) to a control group with the following data. Make a clear, concise interpretation. n=30, x bar=11,SS=38

c. To further test these relationships, you cross the dissonance conditions with high and low anxiety conditions. That is, the subjects are assigned to either dissonance or control and either to receive a high or low anxiety stimulus. The DV is attitude change. Interpret these data.
Anxiety- High Dissonance
Anxiety- High Control
Anxiety-Low Dissonance
Anxiety- Low Control
F(3,39)=4.88,p<.05
Anxiety, p<.05
Dissonance, p<.05
Interaction, p<.05

##### Solution Summary

This problem demonstrates the process of testing whether a sample group is significantly different from a population with assumed mean and standard deviation. The problem outlines the process of hypothesis testing and concludes with a discussion of the process.

##### Solution Preview

a)to test:
<br>Ho: M=10
<br>H1: M<10
<br>Standard error Sm=s/SQRT(n)=2/SQRT(30)=0.3651
<br>Then t^=(Xbar-M)/Sm=(9-10)/0.3651= -2.739
<br>Since this is a one-tailed test, we just use 95% confidence level of t-value with df = n-1 = 29. then t*(0.95,29)=1.699
<br>While t-distribution is symmetric to zero, and |t^| > t*, we have to reject Ho.
<br>Then we have to accept H1 that the sample mean is less than 10 with a 95% level of confidence.
<br>
<br>b) To compare two groups of ...

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