# How does one do a hypothesis test for elasticity?

Given the regression output $$\widehat{\ln cons} = \underset{(0.6018)}{0.4054} + \underset{(0.0744)}{1.2739}\, \ln m - \underset{(0.1902)}{0.6666}\, \ln p_1 -\underset{(0.2645)}{1.6146}\, \ln p_2$$ where

• $$\ln cons$$ is the log of chocolate consumption,
• $$\ln m$$ is the log of income,
• $$\ln p_1$$ is the log of the price of chocolate, and
• $$\ln p_2$$ is the log of the price of sweets,

test whether chocolate is a luxury good.

Since $$1.27 > 1$$, it is logical to test whether $$\beta_{\ln m}$$ could be less than $$1$$. When I test elasticity, I base the null hypothesis on what is logical, as in this case if $$\beta_{\ln m}$$ is significantly greater than $$1$$, one shouldn't reject a (illogical) null hypothesis $$\mathrm H_0: \beta_{\ln m} \geq 1$$. So,

$$\mathrm H_0: \beta_{\ln m} \leq 1$$

$$\mathrm H_1: \beta_{\ln m} > 1$$

$$\displaystyle t = \frac{1.2739-1}{0.0744}\approx 3.681$$

Therefore, I reject the null hypothesis in favour of the alternative that chocolates are a luxury good.

Do you agree with the way I have set up this hypothesis test? If the estimate were less than 1, I would have stated $$\mathrm H_0: \beta_{\ln m} \geq 1$$ against the alternative $$\mathrm H_1: \beta_{\ln m} < 1$$.

• A t-statistic that high has less than a 0.01% chance of occurring under the null hypothesis, which is very good evidence that it should be rejected. – Nuclear Hoagie Apr 29 '19 at 17:30
• @NuclearWang sorry! I mangled my interpretation. But that wasn't the point of the question, so please accept my edit. – ahorn Apr 29 '19 at 18:03
• what is indicated by values e.g. (0.0744 ) in the regression output and how do you interpret ) ? – Subhash C. Davar Aug 3 '19 at 12:12
• "test whether chocolate is a luxury good." This needs a clear version and goal of your study. – Subhash C. Davar Aug 3 '19 at 12:44
• @SubhashC.Davar the values in brackets are the standard errors of the estimates they are below. – ahorn Aug 4 '19 at 10:38

## 1 Answer

Do you agree with the way I have set up this hypothesis test? If the estimate were less than 1, I would have stated $$\mathrm H_0: \beta_{\ln m} \geq 1$$ against the alternative $$\mathrm H_1: \beta_{\ln m} < 1$$.

Setting up null and alternative hypotheses is the first step in a t-test. You should not set them up after computing your estimates. In your specific example with income elasticity of demand, you set up the null and alternative correctly, meaning that you want the alternative hypothesis to be the statement about what you are trying to prove. You should use that null and alternative regardless of what the data then tells you.

P.S. You should report the one-sided p-value from your one-sided hypothesis test, not just the test statistic.

• What if $\hat\beta_{\ln m}=0.7$? What what your hypotheses be? How could it be intelligent to reject $\mathrm H_0$ in favour of $\mathrm H_1:\beta_{\ln m} > 1$ when clearly $0.7<1$? – ahorn Apr 30 '19 at 4:18
• Why would you reject the null in that situation? Your new t statistic would be negative and not in the rejection region, so you would fail to reject the null. – AlexK Apr 30 '19 at 4:48
• I now agree that the alternative hypothesis should be what I'm trying to prove. Though, it is a bit silly to do a formal hypothesis test when the estimate lies on the correct side. – ahorn May 30 '19 at 7:33
• Here is an example of a confusing question, because the null hypothesis of elasticity and the null hypothesis of inelasticity both are accepted in those respective tests: i.stack.imgur.com/9zYG8.png . – ahorn May 30 '19 at 7:35
• @ahorn Null hypothesis is typically formulated with just the $=$ sign. Then the $<$ or $>$ is used in the alternative, if the goal is to do a one-sided test. Even if the estimate is on the correct side of the alternative, it may still be statistically insignificant. With that in mind, I'm not sure what "null hypothesis of elasticity" means, because that indicates that the null hypothesis does not have an equal sign in it, which does not make sense. The image you included shows that the null of inelasticity was rejected in favor of the alternative, as the coefficient is significant. – AlexK May 30 '19 at 8:06