## One Proportion Z Test

**Overview:**

Hypothesis tests allow you to test a statistical hypothesis about a proportion in a population.

**When to Use a One Prop Z-Test**

You use a one prop z-test when you are analyzing one sample, and you are making a hypothesis based on proportions

**Steps to Conduct a One Prop Z-Test**

1. Test the conditions

2. State the hypotheses

3. Perform the mechanics

4. State the conclusion

**Conditions**

1. Random: The sample must be randomly selected to reduce bias.

2. Success/Failure: We must expect to have at least 10 success and 10 failures.

3. 10% Condition: If we want to make a conclusion about our population, we can’t sample more than 10% of the population.

**Hypothesis**

Each hypothesis test should have two hypotheses. The first is the null hypothesis (H0), which is assumed value of the parameter. The other hypothesis is the alternative hypothesis (HA), which is the values that are reasonable for p considering H0 is false. The alternative hypothesis will be one of the following:

**Mechanics**

To perform a one prop z-test, use your calculator and follow these steps:

1. Push the STAT button, tab over to TESTS and select 5: 1-PropZTest

2. Enter your hypothesized parameter value (H0) for p0, the observed value for x, the sample size for n, and the alternative hypothesis (HA) for prop.

3. Tab down to calculate and hit enter. The p value will be displayed next to p=

**Conclusion**

Your conclusion should follow this basic format:

With a p-value of (insert p value here) I (reject/fail to reject) the null hypothesis. There (is/is not enough) evidence to conclude that (Insert HA here).

Whether the null hypothesis is accepted or rejected is based on the p-value. Remember this phrase:

**If p is high, let it fly… if p is low, reject that null.**Generally a low p-value is one that is below .05.

**Example:**

The quality control manager of a large ice-cream manufacturer claims that 80 percent of his 1,000,000 customers are very satisfied with the distribution of chunks in the ice-cream. To test this claim, the local newspaper surveyed 100 customers, using simple random sampling. Among the sampled customers, 73 percent say they are very satisfied. Based on these findings, can we reject the quality control manager’s hypothesis that 80% of the customers are very satisfied? Use a 0.05 level of significance.

*Solution:*

*The quality control manager’s hypothesis is the null hypothesis. In other words, Ho:p=.8*

We can set our alternative hypothesis to HA:p<.8

Push the STAT button, tab over to TESTS and select 5: 1-PropZTest, and enter the data for this problem:

Po:.8

X:73

n:100

prop<Po

We can set our alternative hypothesis to HA:p<.8

Push the STAT button, tab over to TESTS and select 5: 1-PropZTest, and enter the data for this problem:

Po:.8

X:73

n:100

prop<Po

You should, among other results, get a p-value of .0401. This is below the .05 level. Therefore our conclusion would look something like this:

With a p-value of

**.0401**I

**reject**the null hypothesis. There

**is**evidence to conclude that

**the percent of customers who are very satisfied with the distribution of chunks in the ice-cream is less than 80%.**