Independent Samples T-Test |
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95% CI for Mean Difference |
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t |
df |
p |
Mean Difference |
SE Difference |
Lower |
Upper |
Cohen's d |
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Reading Level |
1.995 |
69 |
0.003 |
0.878 |
0.289 |
0.293 |
1.493 |
0.721 |
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The test is: Conduct a two-independent samples t test to determine if people with high IQs differ reading level compared to people without a high IQ.
Using this table, first state if this test is directional or non directional. write the test results for a two-independent samples t-test. State which value you focused on to decide whether the test result is significant or not. Then, evaluate the test results and decide whether you should reject the null, then write a hypothesis testing conclusion.
Lastly, show how the Mean Difference can be calculated.
Calculated test statistics is 1.995
And p-value for the test is 0.003
We can decide the significance of the test from p value
alone.
p-value=0.003 implies there is .3% probability that null
hypothesis is correct. Hence we reject null
hypothesis.
General rule is that if p-value is less than given significane
level we reject null hypothesis. Since here p-value is very small
we can trivially reject null hypothesis.
Hence we conclude that people with high IQs differ
reading level compared to people without a high
IQ.
Mean Difference:
Suppose sample mean for
reading level for high IQ people is
and for without high IQ people is
Then mean difference is :
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