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Graphing Data 4
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Lecture1.1
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Lecture1.2
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Lecture1.3
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Lecture1.4
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Mean and Standard Deviation 5
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Lecture2.1
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Lecture2.2
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Lecture2.3
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Lecture2.4
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Lecture2.5
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Distributions 6
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Lecture3.1
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Lecture3.2
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Lecture3.3
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Lecture3.4
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Lecture3.5
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Lecture3.6
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Correlation and Linear Regression 7
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Lecture4.1
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Lecture4.2
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Lecture4.3
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Lecture4.4
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Lecture4.5
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Lecture4.6
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Lecture4.7
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Probability 3
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Lecture5.1
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Lecture5.2
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Lecture5.3
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Counting Principles 3
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Lecture6.1
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Lecture6.2
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Lecture6.3
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Binomial Distribution 3
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Lecture7.1
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Lecture7.2
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Lecture7.3
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Confidence Interval 7
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Lecture8.1
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Lecture8.2
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Lecture8.3
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Lecture8.4
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Lecture8.5
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Lecture8.6
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Lecture8.7
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Proportion Confidence Interval 3
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Lecture9.1
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Lecture9.2
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Lecture9.3
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Hypothesis Testing 5
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Lecture10.1
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Lecture10.2
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Lecture10.3
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Lecture10.4
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Lecture10.5
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Comparing Two Means 5
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Lecture11.1
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Lecture11.2
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Lecture11.3
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Lecture11.4
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Lecture11.5
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Chi-squared Test 3
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Lecture12.1
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Lecture12.2
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Lecture12.3
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Same Unknown Standard Deviations
If we assume the standard deviations of these two samples are the same, we can use the pooled sample variance. The equation for the pooled sample variance is….
Equation
$$s_{p}^2 = \frac{(n_{1}-1)s_{1}^2+(n_{2}-1)s_{2}^2}{n_{1}+n_{2}-2} $$
Let’s plug this into the equation.
Equation
$$t = \frac{(\bar{x_{1}}-\bar{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{s_{p}^2}{n_{1}}+\frac{s_{p}^2}{n_{2}}}} $$
This reduces to….
Equation
$$t = \frac{(\bar{x_{1}}-\bar{x_{2}})-(\mu_{1}-\mu_{2})}{s_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}} $$
The degrees of freedom now is N
1
+N
2
-2.
Challenge
Do the problem from before again with this new method.
Next
Solution