Example:
Suppose it is up to you to determine if a certain state receives significantly more public school funding (per student) than the USA average. You know that the USA mean public school yearly funding is $6300 per student per year, with a standard deviation of $400.
Next, suppose you collect a sample (n = 100) and determine that the sample mean for New Jersey (per student per year) is $8801.
Use the z-test and the correct Ho and Ha to run a hypothesis test to determine if New Jersey receives significantly more funding for public school education (per student per year).
NOTE: This entire example works the same way if you have a dataset. Using the dataset, you would need to first calculate the sample mean. To run a z-test, it is generally expected that you have a larger sample size (30 or more) and that you have information about the population mean and standard deviation. If you do not have this information, it is sometimes best to use the t-test.
Step 1: Set up your hypothesis
Hypothesis: The mean per student per year funding in New Jersey is significantly greater than the average per student per year funding over the entire USA.
Step 2: Create Ho and Ha
NOTE: There are many ways to write out Ho.
Ho: mean per student per year funding for New Jersey = mean per student per year funding for the USA
This can also be written as the following. Ho: New Jersey mean – Population mean = 0
Ha: mean per student per year funding for New Jersey > mean per student per year funding for the USA
NOTICE1: The Ha in this example is ONE-TAILED because we are interested in seeing if New Jersey is significantly GREATER than the population mean. In a two-tailed test, the Ha contains a NOT EQUAL and the test will see if there is a significant difference (greater or smaller).
NOTICE2: The Ho is the null hypothesis and so always contains the equal sign as it is the case for which there is no significant difference between the two groups.
Step 3: Calculate the z-test statistic
Now, calculate the test statistic. In this example, we are using the z-test and are doing this by hand. However, there are many applications that run such tests. This Site has several examples under the Stats Apps link.
z = (sample mean – population mean) / [population standard deviation/sqrt(n)]
z = (8801 – 6300) / [400/sqrt(100)]
z = 2501 / [400/10]
z = 2501 / [40]
z = 62.5
So, the z-test result, also called the test statistic is 62.5.
Step 4: Using the z-table, determine the rejection regions for you z-test.
To do this, you must first select an alpha value. The alpha value is the percentage chance that you will reject the null (choose to go with your Ha research hypothesis as you conclusion) when in fact the Ho really true (and your research Ha should not be selected). This is also called a Type I error (choosing Ha when Ho is actually correct).
The smaller the alpha, the smaller the percentage of error, BUT the smaller the rejection regions and more difficult to reject Ho.
Most research uses alpha at .05, which creates only a 5% chance of Type I error. However, in cases such as medical research, the alpha is set much smaller.
In our case, we will use alpha = .05
This is ONE-TAILED test, therefore the rejection region is any z-test value greater than the critical z value for a one-tailed test with alpha = .05.
The critical value for one-tailed z-test at alpha = .05 is 1.645.
HOW TO Find Critical Values and Rejection Regions
Therefore, the rejection region is any value GREATER than 1.645.
Step 5: Create a conclusion
Our z-test result is 62.5.
This is very large!
62.5 is MUCH LARGER than 1.645 and so the result of the z test is INSIDE the rejection region.
(The rejection region is all portions of the curve to the right of 1.645. Why to the right? Because this is a one-tailed test for which we are looking at whether the sample is GREATER than the population).
Note: The p –value in this case would be the probability of getting a result of 62.5 randomly given that the rejection region starts at 1.645. The probability of getting 62.5 is nearly zero and so the p –value = 0. The actual calculation of the p value is mathematically complicated and is often done using a software like Excel, SPSS, SAS, R, etc.
This z-test has a significant result.
Conclusion:
The funding for New Jersey public schools (per student per year) is significantly GREATER than the average funding per student per year for the USA.
http://www.ascd.org/publications/educational-leadership/may02/vol59/num08/Unequal-School-Funding-in-the-United-States.aspx