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I am interested in testing the following hypothesis: \begin{align} \newcommand{\var}{\rm Var} \newcommand{\cov}{\rm Cov} \newcommand{\se}{\rm se} H_0\!:\ B_2 + B_3 &= 1 \\ H_1\!:\ B_2 + B_3 &\ne 1 \end{align} How do you find the test statistic for a hypothesis about the sum of two independent variables? I'm very confused.

The model is: \begin{align} \ln(y) &= −3.33846 + 1.49877 x_1 + 0.489858 x_2 \\ \se(B_2) &= 0.539803 \\ \se(B_3) &= 0.102043 \end{align} *this is a double log

I am guessing the test statistic is:
\begin{align} \frac{B_2 + B_3 -1}{\se(B_2)+\se(B_3)} &= \frac{1.49877 + 0.489858 -1}{0.539803+0.102043} \\ &= 1.540 \end{align} This yields a p-value = 0.071, so I don't reject at the 95% significance. Apparently both are wrong.

I have seen this formula: \begin{align} H_0\!: B_2 + CB_3 &= 1 \\ H_1\!: B_2 + \ \ \ B_3 &\ne 1 \end{align} Which leads to: \begin{align} t &= \frac{B_1+CB_2-a}{se(B_2+cB_3)} \\ \ \\ \se(B_2+cB_3) &= \sqrt{\var(B_2)+\var(B_3)+2C\times \cov(B_2,B_3)} \end{align} But I have no idea what $c$ is.

Update: $C$ turned out to be irrelevant for this question.

This ended up being the answer:

$$t=\frac{1.49877+0.489858-1}{\sqrt{0.291387+0.0104129+2(-0.0384272)}=2.084}, \quad p = 0.059$$

Insufficient evidence to reject null at the 95% significance level

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  • $\begingroup$ I don't get why the hypotheses are $H_{0}: \beta_{2} + \beta_{3} = 1$ vs $H_{1}: \beta_{2} + \beta_{3} \neq 1$ $\endgroup$
    – stochazesthai
    Commented May 21, 2015 at 16:14
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    $\begingroup$ I still don't see a clear question in this post. What are you trying to ask? Do you want to know how to test a linear hypothesis in a multiple regression model? $\endgroup$
    – whuber ♦
    Commented May 21, 2015 at 17:02
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    $\begingroup$ 1 orthogonal comment: there will be a covariance between your parameter estimates. You will need to take that into account. The denominator would be something like sqrt{Var(B1)+Var(B2)-2Cov(B1,B2)}. $\endgroup$
    – gung - Reinstate Monica
    Commented May 21, 2015 at 17:37
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    $\begingroup$ Since you now appear to be able to answer your question you should answer it (or at least outline the approach) in an answer, not in your question. You should also clarify your question so it asks a question to which the resulting posted answer is an answer. $\endgroup$
    – Glen_b
    Commented May 22, 2015 at 0:02
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    $\begingroup$ Just to point this out to you: You say "Of course they independent variables!" and then you calculate the covariance to something non-zero. This is a contradiction (meaning it's impossible) and it may be useful for you to think about that. $\endgroup$
    – KOE
    Commented May 22, 2015 at 2:04

1 Answer 1

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(I am turning my comment into an answer so that this thread isn't counted as officially unanswered. I actually hadn't realized that this was the answer that was wanted, I thought my point was orthogonal to the question.)

You are performing a simultaneous test of two parameters. Often that would be done by dropping the variables and performing a nested model test. In your case, you are trying this using the standard errors, not as a nested model test. To do that, you need to take the covariance of your parameters into consideration (in addition, variances add, not SEs). Thus, the denominator would be:
$$ \sqrt{{\rm Var}(\hat\beta_2) + {\rm Var}(\hat\beta_3) - 2{\rm Cov}(\hat\beta_2, \hat\beta_3)} $$ The covariance of your parameter estimates is not typically outputted by statistical software. It comes from the variance-covariance matrix of the betas, which the software will have calculated as part of the process of fitting the model. It should be possible to get that information, but how will differ depending on the software you use; in R, for example, you would call vcov(model_fit_object).

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  • $\begingroup$ Thank you gung this was very helpful and much appreciated. $\endgroup$
    – Ivan
    Commented Sep 9, 2015 at 6:35

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