Hypothesis Tests on a Single Standard Deviation

Chi-square distribution

If Z is a random variable with a standardized normal distribution (mean zero and variance 1), then Z² has a chi-square distribution with 1 degree of freedom. If Z₁, Z₂, ……, Z_k, are a set of independent and identically distributed standard normal variables, then the sum of their squares has a chi-square distribution with k degrees of freedom. 

NOTE: Let X_i be independent and normally distributed random variables with mean µ and variance 1. Then (X_i - µ) has mean zero and has a chi-square distribution with k degrees of freedom.

EXAMPLES

1) Car speed is measured using a radar unit. In an urban area, the radar readings X_i are normally distributed with a mean of 25 mph and a standard deviation of 3 mph. A recent source claims that the standard deviation of his radar is 1. Imagine your manager wishes to test the hypothesis H₀: σ = 1 versus the alternative hypothesis H₁: σ ≠ 1. To carry out this test, your manager has asked you to take five speed measurements using a test car that was programmed to travel at 25 mph. This was done and the speed measurements you recorded were 25, 24, 27, 25, 26. Should the hypothesis H₀: σ = 1 be rejected at the significance level α = 10%?  

Let's calculate the variance of X: 

The variance is

and the standard deviation is

                                          

You've found out that the sum of squared deviations, Σ (X_i-µ)² equals 6. For a significance test at α = 10% and 5 degrees of freedom you will find in a chi-square distribution table:

The calculated value of 6 is inside the acceptance region. H₀: σ = 1 cannot be rejected.

2)  Your manager was not pleased since he doubts that sigma can be less than 3. As a result, he insisted you should take a new sample to test the null hypothesis H₀: σ = 1 against the alternative hypothesis H₁: σ ≠ 1. What should you do? Obtain five new speed measurements using the same test car set to 25 mph, in the same conditions. Suppose the recorded speeds were 28, 27, 30, 28, 29. The hypothesis H₀: σ = 1 should be rejected at the significance level α = 10%?  

Let's calculate the variance of X:

                 

You found out that the sum of squared deviations, Σ (X_i-µ)² equals 63. For a significance test at α = 10% and 5 degrees of freedom you will find in a chi-square distribution table:

Since Σ (X- µ)²= 63 >11.07, the hypothesis the radar has standard deviation σ =1 was rejected at the 10% significance level. You have two samples and two different conclusions.

In fact, these two examples illustrate how random sampling variability can lead to different conclusions when testing a hypothesis with samples of the same size and the same significance level, from the same population in the same conditions. This occurs because sample statistics are not perfect estimates of their corresponding population parameters. 

3) But your manager argued, pointing out that radar readings in urban areas usually follow a normal distribution with a mean speed of 25 mph and a standard deviation of 3 mph. He suggests taking another five measurements and test again the hypotheses H₀: σ = 1 against H₁: σ ≠ 1. Following this suggestion, you record five additional speeds: 29, 27, 29, 26, and 29.  Should the hypothesis H₀: σ = 1 be rejected at a significance level α = 10%?

            

The numerator of the fraction should fall between 1.145 and 11.07 to support the hypothesis H₀: σ = 1, but 53 is unquestionably out. Therefore, you must reject the null hypothesis.

Are you satisfied with your manager's sugestions? Perhaps you should look the given examples from a different perspective. To test H₀: σ =1, what if you use the sample mean instead of the hypothesized mean µ = 25 mph? Let's calculate the sample mean and the sample variance:

              

By using the sample mean, the degrees of freedom are reduced to n - 1. Consequently, the sum of squared deviations follows a chi-square distribution with 4 degrees of freedom. Refer to the chi-square table to find the critical values for α=10 with 4 degrees of freedom.

.

Given that the sum of squared deviations equals 8 is less than 9.48, it is reasonable to conclude that σ = 1 at the 10% significance level, as claimed.

If you accept σ = 1, we can look at

You’ve made the assumption that the radar readings follow a normal distribution with a mean μ = 25 and a standard deviation σ = 1. Then  follow a normal distribution with a mean μ = 25 and a standard deviation

 

The random sample of 5 observations you’ve collected has mean  28.  Hence, you are able to test the hypothesis that the mean of the radar readings is 25 against the alternative hypothesis that it is greater than 25 using a z-test.

To calculate the z-score, we use the formula

 

Given that the z-score is greater than 1.645, you must reject the null hypothesis that the mean is 25 at 10 % level of significance.

 

Therefore, based on this sample of radar readings (29, 27, 29, 26, 29), it is reasonable to conclude that the new radar is not calibrated (mean μ = 25 and a standard deviation σ = 1).

 Now, you could think: would a radar with σ = 3 produce measurements as above? If σ = 3, σ² = 9. How will the numerator of

behave? Remember: chi-square only tell what happens when E(X-µ)² =1. But here E(X-µ)² =9. What to do?

The new test statistic is:

  

where s² comes from the sample and σ² comes from H₀. The degrees of freedom associated with the test statistic (for finding the critical statistic) is (n-1). For this test to be valid, the population must be normally distributed. So, calculate

The test statistic chi square equals which is in the 95% region of acceptance: [0.4844: 11.1433]. H₀ cannot be rejected.

IMPORTANT: This is an exercise of statistics; it doesn’t tell you radar calibration procedures. It is therefore useful to have a set of accepted standards and protocols for maintaining the quality of radars.

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