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%.