UsableStats: 1-Sample Z-test Examples

Log-in | Contact Jeff | Email Updates

The 1-Sample Z-test: You'll probably never use it

View All Tutorials

Fundamentals of Statistics 3: Sampling :: 1-Sample Z-test Examples

While you're unlikely to use the 1-sample Z test in real life (since we almost never know the population standard deviation) it is still good to understand how it works since we'll use slight variations on the formula when we encounter the 1-sample t-test. Below are three examples to illustrate the concept of finding the probability of a sample mean using the normal distribution given you know the population standard deviation.

1. A teacher wanted to know how well the gifted students in here class perform relative to her other classes. She administers a standardized test with a mean of 50 and standard deviation of 10. Her class of 31 students has an average score of 55, what percent of classes is their average score higher than?

First find the standard error of the mean for a sample size of 31 =10/SQRT(31) = 1.796
How many standard errors are there between the observed sample mean and the population mean: (55-50)/1.796 = 2.783 standard errors.
The percent of area under the normal curve from negative infinity to 2.783 = 99.7%, meaning their class score is higher than 99.7% of all other samples of 31.

2. A researcher wanted to study the effects of mentoring on intelligence scores. He wanted to know as a baseline what the average intelligence of his students were relative to the general population. He used a standardized IQ test which has a mean of 100 and standard deviation of 16. The 50 students in his study scored an average of 102 on the IQ test. At what percentile of intelligences is the average IQ of his students relative to his population ?

First find the standard error of the mean for a sample size of 50 =16/SQRT(50) = 2.26
How many standard errors are there between the observed sample mean and the population mean: (102-100)/2.26 = .8849 standard errors.
The percent of area under the normal curve from negative infinity to .8849 =81.2%, meaning the average intelligence of this set of students is higher than 81.2% of all other samples of 50 students. It is also 31.2% above an average IQ given this sample size.

Finally, we can use the 1-sample z-test to test hypotheses. The general framework of Hypothesis testing will be covered in future lessons.

3. A rental car company claims the mean time to rent a car on their website is 60 seconds with a standard deviation of 30 seconds. A random sample of 36 customers attempted to rent a car on the website. The mean time to rent was 75 seconds. Is this enough evidence to contradict the company's claim?

First find the standard error of the mean for a sample size of 36 =30/SQRT(36) = 5
How many standard errors are there between the observed sample mean and the population mean: (75-60)/5 =3 standard errors.
The percent of area under the normal curve from negative infinity to 3 =99.865%. That means we would only expect to see a sample mean this extreme less than 1% of the time. We conclude that this is enough evidence to question their claim.

If you need more practice understanding z-scores including these 1-sample z questions, I put together a Crash Course in Z-scores which provides a visual and easy-to-follow guide with plenty of practice understanding the normal curve and z-scores. It can also be downloaded along with an Excel calculator which will show you how the calculations are done so you can check your work. You can see a demonstration of the calculator below.

View All Tutorials

Home | Contact Jeff | Sign up For Newsletter

Not at all					Neutral					Extremely
0	1	2	3	4	5	6	7	8	9	10

Name
Email	Not Published
Comment
	To prevent comment spam, please answer the following question before submitting (tags not permitted) : What is 2 + 4: (enter the number)

How well did you understand this lesson?

What didn't make sense?

Copyright © 2004-2022 Measuring Usability LLC