# The Chi-Square Distribution

## Test of a Single Variance

A test of a single variance assumes that the underlying distribution is normal. The null and alternative hypotheses are stated in terms of the population variance or population standard deviation. The test statistic is
where
1.  n = the total number of data,
2. s2 = sample variance, and
3. σ2 = population variance.
You may think of s as the random variable in this test. The number of degrees of freedom is df = n – 1. A test of a single variance may be right-tailed, left-tailed, or two-tailed.  will show you how to set up the null and alternative hypotheses. The null and alternative hypotheses contain statements about the population variance.

### References

Apple Insider. (n.d.). Retrieved from http://appleinsider.com/mac_price_guide
World Bank. (n.d.). Retrieved from http://www.worldbank.org/

### Chapter Review

To test variability, use the chi-square test of a single variance. The test may be left-, right-, or two-tailed, and its hypotheses are always expressed in terms of the variance or standard deviation.

### Formula Review


 

  Test of a single variance statistic where
   n: sample size
   s: sample standard deviation
   σ: population standard deviation
df = n – 1 degrees of freedom
2. Use the test to determine variation.
3. The degrees of freedom is the number of samples – 1.
4. The test statistic is 
 , where n = the total number of data, s2 = sample variance, and σ2 = population variance.

5. The test may be left-, right-, or two-tailed.
Use the following information to answer the next three exercises. An archer’s standard deviation for his hits is six, where the data are measured in distance from the center of the target. An observer claims the standard deviation is less than six.
Use the following information to answer the next three exercises. The standard deviation of heights for students in a school is 0.81. A random sample of 50 students is taken, and the standard deviation of heights of the sample is 0.96. A researcher in charge of the study believes the standard deviation of heights for the school is greater than 0.81.
Use the following information to answer the next four exercises: The average waiting time in a doctor’s office varies. The standard deviation of waiting times in a doctor’s office is 3.4 minutes. A random sample of 30 patients in the doctor’s office has a standard deviation of waiting times of 4.1 minutes. One doctor believes the variance of waiting times is greater than originally thought.

### Homework

Use the following information to answer the next 12 exercises. Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes.

For each word problem, use a solution sheet to solve the hypothesis test problem. Go to 