# 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

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##### 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.

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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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