Once again we've already done the bulk of the theoretical work in developing a hypothesis test for the slope parameter \(\beta\) of a simple linear regression model when we developed a \((1-\alpha)100\%\) confidence interval for \(\beta\). We had shown then that:

\(T=\dfrac{\hat{\beta}-\beta}{\sqrt{\frac{MSE}{\sum(x_i-\bar{x})^2}}}\)

follows a \(t_{n-2}\) distribution. Therefore, if we're interested in testing the null hypothesis:

\(H_0:\beta=\beta_0\)

against any of the alternative hypotheses:

\(H_A:\beta \neq \beta_0\), \(H_A:\beta < \beta_0\), \(H_A:\beta > \beta_0\)

we can use the test statistic:

\(t=\dfrac{\hat{\beta}-\beta_0}{\sqrt{\frac{MSE}{\sum(x_i-\bar{x})^2}}}\)

and follow the standard hypothesis testing procedures. Let's take a look at an example.

The hypothesis test for the slope \(\beta\) that we developed on the previous page was developed under the assumption that a response \(Y\) is a linear function of a nonrandom predictor \(x\). This situation occurs when the researcher has complete control of the values of the variable \(x\). For example, a researcher might be interested in modeling the linear relationship between the temperature \(x\) of an oven and the moistness \(y\) of chocolate chip muffins. In this case, the researcher sets the oven temperatures (in degrees Fahrenheit) to 350, 360, 370, and so on, and then observes the values of the random variable \(Y\), that is, the moistness of the baked muffins. In this case, the linear model:

\(Y_i=\alpha+\beta x_i+\epsilon_i\)

implies that the average moistness:

\(E(Y)=\alpha+\beta x\)

is a linear function of the temperature setting.

There are other situations, however, in which the variable \(x\) is not nonrandom (yes, that's a double negative!), but rather an observed value of a random variable \(X\). For example, a fisheries researcher may want to relate the age \(Y\) of a sardine to its length \(X\). If a linear relationship could be established, then in the future fisheries researchers could predict the age of a sardine simply by measuring its length. In this case, the linear model:

\(Y_i=\alpha+\beta x_i+\epsilon_i\)

implies that the average age of a sardine, given its length is \(X=x\):

\(E(Y|X=x)=\alpha+\beta x\)

is a linear function of the length. That is, the conditional mean of \(Y\) given \(X=x\) is a linear function. Now, in this second situation, in which both \(X\) and \(Y\) are deemed random, we typically assume that the pairs \((X_1, Y_1), (X_2, Y_2), \ldots, (X_n, Y_n)\) are a random sample from a bivariate normal distribution with means \(\mu_X\) and \(\mu_Y\), variances \(\sigma^2_X\) and \(\sigma^2_Y\), and correlation coefficient \(\rho\). If that's the case, it can be shown that the conditional mean:

\(E(Y|X=x)=\alpha+\beta x\)

must be of the form:

\(E(Y|X=x)=\left(\mu_Y-\rho \dfrac{\sigma_Y}{\sigma_X} \mu_X\right)+\left(\rho \dfrac{\sigma_Y}{\sigma_X}\right)x\)

That is:

\(\beta=\rho \dfrac{\sigma_Y}{\sigma_X}\)

Now, for the case where \((X_i, Y_i)\) has a bivariate distribution, the researcher may not necessarily be interested in estimating the linear function:

\(E(Y|X=x)=\alpha+\beta x\)

but rather simply knowing whether \(X\) and \(Y\) are independent. In STAT 414, we've learned that if \((X_i, Y_i)\) follows a bivariate normal distribution, then testing for the independence of \(X\) and \(Y\) is equivalent to testing whether the correlation coefficient \(\rho\) equals 0. We'll now work on developing three different hypothesis tests for testing \(H_0:\rho=0\) assuming \((X_i, Y_i)\) follows a bivariate normal distribution.

Given our wordy prelude above, this test may be the simplest of all of the tests to develop. That's because we argued above that if \((X_i, Y_i)\) follows a bivariate normal distribution, and the conditional mean is a linear function:

\(E(Y|X=x)=\alpha+\beta x\)

then:

\(\beta=\rho \dfrac{\sigma_Y}{\sigma_X}\)

That suggests, therefore, that testing for \(H_0:\rho=0\) against any of the alternative hypotheses \(H_A:\rho\neq 0\), \(H_A:\rho> 0\) and \(H_A:\rho< 0\) is equivalent to testing \(H_0:\beta=0\) against the corresponding alternative hypothesis \(H_A:\beta\neq 0\), \(H_A:\beta<0\) and \(H_A:\beta>0\). That is, we can simply compare the test statistic:

\(t=\dfrac{\hat{\beta}-0}{\sqrt{MSE/\sum(x_i-\bar{x})^2}}\)

to a \(t\) distribution with \(n-2\) degrees of freedom. It should be noted, though, that the test statistic can be instead written as a function of the sample correlation coefficient:

\(R=\dfrac{\dfrac{1}{n-1} \sum\limits_{i=1}^n (X_i-\bar{X}) (Y_i-\bar{Y})}{\sqrt{\dfrac{1}{n-1} \sum\limits_{i=1}^n (X_i-\bar{X})^2} \sqrt{\dfrac{1}{n-1} \sum\limits_{i=1}^n (Y_i-\bar{Y})^2}}=\dfrac{S_{xy}}{S_x S_y}\)

That is, the test statistic can be alternatively written as:

\(t=\dfrac{r\sqrt{n-2}}{\sqrt{1-r^2}}\)

and because of its algebraic equivalence to the first test statistic, it too follows a \(t\) distribution with \(n-2\) degrees of freedom. Huh? How are the two test statistics algebraically equivalent? Well, if the following two statements are true:

1. \(\hat{\beta}=\dfrac{\dfrac{1}{n-1} \sum\limits_{i=1}^n (X_i-\bar{X}) (Y_i-\bar{Y})}{\dfrac{1}{n-1} \sum\limits_{i=1}^n (X_i-\bar{X})^2}=\dfrac{S_{xy}}{S_x^2}=R\dfrac{S_y}{S_x}\)

2. \(MSE=\dfrac{\sum\limits_{i=1}^n(Y_i-\hat{Y}_i)^2}{n-2}=\dfrac{\sum\limits_{i=1}^n\left[Y_i-\left(\bar{Y}+\dfrac{S_{xy}}{S_x^2} (X_i-\bar{X})\right) \right]^2}{n-2}=\dfrac{(n-1)S_Y^2 (1-R^2)}{n-2}\)

then simple algebra illustrates that the two test statistics are indeed algebraically equivalent:

\(\displaystyle{t=\frac{\hat{\beta}}{\sqrt{\frac{MSE}{\sum (x_i-\bar{x})^2}}} =\frac{r\left(\frac{S_y}{S_x}\right)}{\sqrt{\frac{(n-1)S^2_y(1-r^2)}{(n-2)(n-1)S^2_x}}}=\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}} \)

Now, for the veracity of those two statements? Well, they are indeed true. The first one requires just some simple algebra. The second one requires a bit of trickier algebra that you'll soon be asked to work through for homework.

It would be nice to use the sample correlation coefficient \(R\) as a test statistic to test more general hypotheses about the population correlation coefficient:

\(H_0:\rho=\rho_0\)

but the probability distribution of \(R\) is difficult to obtain. It turns out though that we can derive a hypothesis test using just \(R\) provided that we are interested in testing the more specific null hypothesis that \(X\) and \(Y\) are independent, that is, for testing \(H_0:\rho=0\).

Now that we know the p.d.f. of \(R\), testing \(H_0:\rho=0\) against any of the possible alternative hypotheses just involves integrating \(g(r)\) to find the critical value(s) to ensure that \(\alpha\), the probability of a Type I error is small. For example, to test \(H_0:\rho=0\) against the alternative \(H_A:\rho>0\), we find the value \(r_\alpha(n-2)\) such that:

\(P(R \geq r_\alpha(n-2))=\int_{r_\alpha(n-2)}^1 \dfrac{\Gamma[(n-1)/2]}{\Gamma(1/2)\Gamma[(n-2)/2]}(1-r^2)^{\frac{(n-4)}{2}}dr=\alpha\)

Yikes! Do you have any interest in integrating that function? Well, me neither! That's why we'll instead use an \(R\) Table, such as the one we have in Table IX at the back of our textbook.

Okay, the derivation for this hypothesis test is going to be MUCH easier than the derivation for that last one. That's because we aren't going to derive it at all! We are going to simply state, without proof, the following theorem.

The theorem, therefore, allows us to test the general null hypothesis \(H_0:\rho=\rho_0\) against any of the possible alternative hypotheses comparing the test statistic:

\(Z=\dfrac{\dfrac{1}{2}ln\dfrac{1+R}{1-R}-\dfrac{1}{2}ln\dfrac{1+\rho_0}{1-\rho_0}}{\sqrt{\dfrac{1}{n-3}}}\)

to a standard normal \(N(0,1)\) distribution.

What? We've looked at no examples yet on this page? Let's take care of that by closing with an example that utilizes each of the three hypothesis tests we derived above.

To develop an approximate \((1-\alpha)100\%\) confidence interval for \(\rho\), we'll use the normal approximation for the statistic \(Z\) that we used on the previous page for testing \(H_0:\rho=\rho_0\).