- CFA Exams
- 2025 Level I
- Topic 1. Quantitative Methods
- Learning Module 8. Hypothesis Testing
- Subject 10. Testing Concerning Tests of Variances (Chi-Square Test)
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Subject 10. Testing Concerning Tests of Variances (Chi-Square Test) PDF Download
Suppose an analyst is interested in testing whether the variance from a single population is statistically equal to some hypothesized value. Let σ2 represent the variance and let σ02 represent the hypothesized value. The null and alternative hypotheses would be expressed as:
Also note that directional hypotheses could be made instead:
The test statistic to be used is a chi-square (χ2) statistic with n-1 degrees of freedom.
The formula for the test statistic is:
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n = sample size
s2 = sample variance
σ02 = the hypothesized value for σ under H0
Unlike t-graphs and z-graphs, a chi-square graph is positively skewed. It is also truncated at zero, and thus is not defined for negative values.
Like the family of t-graphs, the shape of the graph varies; the graph becomes more symmetrical as the degrees of freedom increase.
The graph looks like this:
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Critical values for this distribution are found in the chi-square table. The table is titled "Probability in Right Tail," so be aware that this is the region the table gives and modify your calculation accordingly.
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It is possible to use both tails in this distribution, despite the fact that it is non-symmetrical.
In contrast to the t-test, the chi-square test is sensitive to violations of its assumptions. If the sample is not actually random or if it does not come from a normally distributed population, inferences based on a chi-square test are likely to be faulty.
Example
You are an institutional investor evaluating a hedge fund that seeks to deliver a return comparable to the domestic broad market equity index while keeping the monthly standard deviation in asset value under 5%. According to data in the prospectus, during the last 30 months the monthly standard deviation in asset value was 4.5%. You wish to test this claim statistically, using a significance level of 0.10. Assume that returns are normally distributed and monthly asset values are independent observations.
The hypotheses are:
H0: σ2 ≤ 0.0025, versus H1: σ2 > 0.0025. Note that by squaring the standard deviation, 5% = 0.05, you get 0.0025.
The test statistic will be chi-square with 30 - 1 = 29 degrees of freedom.
The critical value will be found in the table for a chi-square distribution, 29 degrees of freedom, alpha = 0.10. Note that this is a one-tailed test. The critical value is 39.087.
The test statistic will be: chi-square = [(29) x 0.0452] / (0.05)2 = 23.49.
The test statistic is not greater than the critical value, so you do not reject the null hypothesis.
Note: If the null hypothesis had a >= symbol in it, then you would reject if the test statistic was less than or equal to the lower alpha point.
The chi-square distribution is asymmetrical. Like the t-distribution, the chi-square distribution is a family of distributions; a different distribution exists for each possible value of degrees of freedom, n - 1 (n is sample size). Unlike the t-distribution, the chi-square distribution is bounded below by 0 (no negative values).
User Contributed Comments 11
User | Comment |
---|---|
vincenthuang | Chi-square=(n-1)*S2/sigma2 |
fiderikumo | Why is Chi square the most appropriate test here? |
tanyak | because it is the test of variance |
achu | If we had made the H1 hypothesis [variance < .025 ] instead of [var> than .025] : Then H0 would have been (var) >= .025; Chi-squared value from table is 19.768 (since left tail = 100%- 90% in right tail), and H1 is var < .025. Because test stat is 23.49, Wouldn't that also be a 'do not reject' at 10% signficance level? |
chamad | Back to the example. I Understood from thr previous readings that Alternative Hyp Ha is what we try to proof. Shouldn't be Ha:var<0.025 and H0 : var>=0.025 ??? someone please? |
SuperKnight | I don't understand I would have thought that the null hypothesis would be opposite of what they show here? You are trying to prove the alternate hypothesis so I would figure you'd have to set the null hypothesis at var >= 0.0025, and then if you find that the 4.5% std dev is significantly lower than 5%, you could reject the null hypothesis and conclude something? Perhaps this difference has something to do with the fact that Ch-square graph is skewed to the right? |
Capital | No SuperKnight. You can reject null hypothesis. That's the purpose of hypothesis testing. However, you cannot accept therefore the alternative hypothesis. |
SaeedAlam | SuperKnight I'm sure you can do it your way, with reference to the third pargraph from the end. Just switch everything |
gill15 | I got all this stuff but I`m confused about setting the null and alternative hypothesis as well...dont get it....it`s the opposite of how we`ve been doing it |
gill15 | It doesnt seem to matter actually....your rejection or acceptance would just be opposite then ---- what i`m trying to say it looks like it doesnt matter what you choose your null to be... |
Shaan23 | I did the opposite way. Ho: o^2 > .0025 but I'm missing something. Get the same test statistic = 23.49. My critical value is at the 90th percentile because I'm doing the Ha the other end of the chi graph which at 29 df is 19.768. So I do not reject the null hypothesis but I should I right? to be compatible with the example. |
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