**Hypothesis Testing: One Proportion**

**Introduction**

As the name implies, one proportion tests deal with situations in which you are analyzing a single proportion. A few symbols need to be defined before we dive in:

- \(p_o\) refers to the null proportion, or the given proportion that is assumed to be true.
- \(q_o\) refers to 1 minus the null proportion, i.e. \(1 - p_o\)
- p-hat ( \(\widehat {p}\)) refers to the sample proportion that you will use to disprove the null.
- \(n\) refers to the sample size

**Example**

*With a fair coin, the proportion of heads is 0.5. However, when you flip your coin 100 times, you only get 20 heads (making a proportion of 0.2). Is this data sufficient to prove that your coin is not fair?*

__Step 1__: Name Test: 1-Proportion Hypothesis Test

__Step 2__: Define Test:

H_{0}: p = 0.5

H_{A}: p < 0.5

__Step 3__: Assume \(H_0\) is true and define its normal distribution. Then check the conditions.

1. The data is from a simple random sample.

2. \(N > 10n\)

3. \(np_o > 10 \) and \(nq_o > 10 \)

__Step 4__: Using the normal distribution, calculate the test statistics and p-value.

**Test Statistic (**

**1 Proportion)**

**: \(z = {\widehat{p} - p_o \over \sqrt{p_o q_o \over n}}\)**

Test Statistic:

\(z = {0.2 -0.5 \over \sqrt{(0.5)(0.5) \over 100}}\) → \(z = -6\)

P-Value:

The p-value will be found by using the **normal cdf **function on your calculator:

- lower limit: -999
- upper limit: \(z\)
- distribution center: 0
- standard deviation: 1
- All together, it looks like this: normalcdf (-999, \(z\), 0, 1)

*Note: If it was a right-sided test and the test statistic was positive (z > 0), then your lower limit would be the test statistic (z) and your upper limit would be 999.

In this case, we do normalcdf (-999, -6, 0, 1) to get a p-value of 9.9 \(\times\) 10^{-10} (approximately zero).

__Step 5__: Analyze your results and determine if they are statistically significant.

We calculated a p-value of approximately zero. This p-value is less than the assumed significance level of 0.05. Therefore, we reject the null hypothesis. The data supports the claim that the proportion of heads is less than 0.5.