<blockquote>
  <p>Do you agree with the way I have set up this hypothesis test? If the estimate were less than 1, I would have stated <span class="math-container">$\mathrm H_0: \beta_{\ln m} \geq 1$</span> against the alternative <span class="math-container">$\mathrm H_1: \beta_{\ln m} &lt; 1$</span>.</p>
</blockquote>

<p>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>

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

CC BY-SA 4.0
2019-04-30T04:18:57.153Z
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
CC BY-SA 4.0
2019-04-30T04:48:18.593Z
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
CC BY-SA 4.0
2019-05-30T07:33:18.387Z
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
CC BY-SA 4.0
2019-05-30T07:35:50.317Z
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: https://i.sstatic.net/9zYG8.png .
ahorn
CC BY-SA 4.0
2019-05-30T08:06:48.083Z
@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
CC BY-SA 4.0
2019-05-30T08:07:57.377Z
Sorry, I mean the alternative hypotheses of elasticity and inelasticity.
ahorn
CC BY-SA 4.0
2019-04-29T20:21:42.613Z
2019-04-29T20:21:42.613Z
AlexK
<blockquote>
  <p>Given the regression output
  <span class="math-container">$$\widehat{\ln cons} = \underset{(0.6018)}{0.4054} + \underset{(0.0744)}{1.2739}\, \ln m - \underset{(0.1902)}{0.6666}\, \ln p_1 -\underset{(0.2645)}{1.6146}\, \ln p_2$$</span>
  where </p>
  
  <ul>
  <li><span class="math-container">$\ln cons$</span> is the log of chocolate consumption,</li>
  <li><span class="math-container">$\ln m$</span> is the log of income,</li>
  <li><span class="math-container">$\ln p_1$</span> is the log of the price of chocolate, and</li>
  <li><span class="math-container">$\ln p_2$</span> is the log of the price of sweets,</li>
  </ul>
  
  <p>test whether chocolate is a luxury good.</p>
</blockquote>

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

<p><span class="math-container">$\mathrm H_0: \beta_{\ln m} \leq 1$</span></p>

<p><span class="math-container">$\mathrm H_1: \beta_{\ln m} &gt; 1$</span></p>

<p><span class="math-container">$\displaystyle t = \frac{1.2739-1}{0.0744}\approx 3.681$</span></p>

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

<p>Do you agree with the way I have set up this hypothesis test? If the estimate were less than 1, I would have stated <span class="math-container">$\mathrm H_0: \beta_{\ln m} \geq 1$</span> against the alternative <span class="math-container">$\mathrm H_1: \beta_{\ln m} &lt; 1$</span>.</p>

CC BY-SA 4.0
2019-04-29T17:30:58.417Z
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
CC BY-SA 4.0
2019-04-29T18:03:29.313Z
@NuclearWang sorry! I mangled my interpretation. But that wasn't the point of the question, so please accept my edit.
ahorn
CC BY-SA 4.0
2019-08-03T12:12:48.667Z
what is indicated by values e.g.  (0.0744 ) in the regression output and how do you interpret ) ?
Subhash C. Davar
CC BY-SA 4.0
2019-08-03T12:44:59.143Z
"test whether chocolate is a luxury good." This needs a clear version and goal of your study.
Subhash C. Davar
CC BY-SA 4.0
2019-08-04T10:38:43.737Z
@SubhashC.Davar the values in brackets are the standard errors of the estimates they are below.
ahorn
CC BY-SA 4.0
2019-08-04T14:17:59.977Z
Q: hypothesis test for elasticity? I can not make out whether it is correct  to use term -elasticity while the test is  checking  the  significance of regression coefficients produced by your model.d
Subhash C. Davar
CC BY-SA 4.0
2019-08-04T15:33:48.793Z
@SubhashC.Davar the explained variable and explanatory variable are both in logarithmic form. Therefore, the coefficient is an elasticity. To see this, exponentiate the model to become $cons = m^{\beta_m}p_1^{\beta_1}p_2^{\beta_2}\varepsilon$ then note that $\frac{\mathrm d cons}{\mathrm d m}\cdot\frac m{cons} = \left(\beta_m m^{\beta_m - 1}p_1^{\beta_1}p_2^{\beta_2}\varepsilon\cdot m\right) / \left(m^{\beta_m}p_1^{\beta_1}p_2^{\beta_2}\varepsilon\right) = \beta_m$.
ahorn
<econometrics><elasticity><statistics><hypothesis-testing>
How does one do a hypothesis test for elasticity?
ahorn
# How does one do a hypothesis test for elasticity?

(Asked by: username_0 on 2019-04-29)

> 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 $\mathrm H\_1: \beta\_{\ln m} < 1$.

**Tags:** econometrics, elasticity, statistics, hypothesis-testing

> **Comments:**
> **username_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.  
> **username_0**: @NuclearWang sorry! I mangled my interpretation. But that wasn't the point of the question, so please accept my edit.  
> **username_2**: what is indicated by values e.g.  (0.0744 ) in the regression output and how do you interpret ) ?  
> **username_2**: "test whether chocolate is a luxury good." This needs a clear version and goal of your study.  
> **username_0**: @SubhashC.Davar the values in brackets are the standard errors of the estimates they are below.  
> **username_2**: Q: hypothesis test for elasticity? I can not make out whether it is correct  to use term -elasticity while the test is  checking  the  significance of regression coefficients produced by your model.d  
> **username_0**: @SubhashC.Davar the explained variable and explanatory variable are both in logarithmic form. Therefore, the coefficient is an elasticity. To see this, exponentiate the model to become $cons = m^{\beta_m}p_1^{\beta_1}p_2^{\beta_2}\varepsilon$ then note that $\frac{\mathrm d cons}{\mathrm d m}\cdot\frac m{cons} = \left(\beta_m m^{\beta_m - 1}p_1^{\beta_1}p_2^{\beta_2}\varepsilon\cdot m\right) / \left(m^{\beta_m}p_1^{\beta_1}p_2^{\beta_2}\varepsilon\right) = \beta_m$.  

---

## 1 answers

### Answer by username_3 on 2019-04-29. Score: 1 | Accepted 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.

> **Comments:**
> **username_0**: 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 $0.7<1$?  
> **username_3**: 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.  
> **username_0**: 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.  
> **username_0**: 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: https://i.sstatic.net/9zYG8.png .  
> **username_3**: @username_0 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.  
> **username_0**: Sorry, I mean the alternative hypotheses of elasticity and inelasticity.  

---

