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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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Equations
First, the sample proportion is defined as….
Equation
$$ \hat{p} = \frac{m}{n} $$
$$ \hat{p} = \text{Sample Proportion} $$
$$ m = \text{Amount of “Wins” or event 1} $$
$$ n = \text{Total Number of Events} $$
$$ \hat{p} = \text{Sample Proportion} $$
$$ m = \text{Amount of “Wins” or event 1} $$
$$ n = \text{Total Number of Events} $$
While the sample proportion is similar to the mean, there is also the standard error for the binomial distribution.
Equation
$$ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} $$
$$ SE = \text{Standard Error} $$
$$ \hat{p} = \text{Sample Proportion} $$
$$ n = \text{Sample Size} $$
$$ SE = \text{Standard Error} $$
$$ \hat{p} = \text{Sample Proportion} $$
$$ n = \text{Sample Size} $$
The z-score used in this case we will be the same, leading to our bounds of….
Equation
$$ \hat{p}-SE*z^* < p < \hat{p}+SE*z^* $$
$$\text{or}$$
$$ \hat{p}-\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}z^* < p < \hat{p}+\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}z^* $$
$$\hat{p} = \text{Sample Proportion}$$
$$n = \text{Sample Size}$$
$$z^* = \text{The Z-Score Related to our Confidence Level}$$
$$SE = \text{Standard Error}$$
$$\text{or}$$
$$ \hat{p}-\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}z^* < p < \hat{p}+\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}z^* $$
$$\hat{p} = \text{Sample Proportion}$$
$$n = \text{Sample Size}$$
$$z^* = \text{The Z-Score Related to our Confidence Level}$$
$$SE = \text{Standard Error}$$
Challenge
A coffee shop is running a free coffee promotion for some lucky winners. Let’s say 25 people out of 135 have won free coffee, what do we think the probability of winning coffee could be within a confidence level of 95%.
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Solution