12.5: Testing the Significance of the Correlation Coefficient

The hypothesis test lets us decide whether the value of the population correlation coefficient \(\rho\) is "close to zero" or "significantly different from zero". We decide this based on the sample correlation coefficient \(r\) and the sample size \(n\).

If the test concludes that the correlation coefficient is significantly different from zero, we say that the correlation coefficient is "significant."

If the test concludes that the correlation coefficient is not significantly different from zero (it is close to zero), we say that correlation coefficient is "not significant".

NOTE

PERFORMING THE HYPOTHESIS TEST

WHAT THE HYPOTHESES MEAN IN WORDS:

DRAWING A CONCLUSION:There are two methods of making the decision. The two methods are equivalent and give the same result.

In this chapter of this textbook, we will always use a significance level of 5%, \(\alpha = 0.05\)

NOTE

Using the \(p\text\) method, you could choose any appropriate significance level you want; you are not limited to using \(\alpha = 0.05\). But the table of critical values provided in this textbook assumes that we are using a significance level of 5%, \(\alpha = 0.05\). (If we wanted to use a different significance level than 5% with the critical value method, we would need different tables of critical values that are not provided in this textbook.)

METHOD 1: Using a \(p\text\) to make a decision

Using the TI83, 83+, 84, 84+ CALCULATOR

To calculate the \(p\text\) using LinRegTTEST:

On the LinRegTTEST input screen, on the line prompt for \(\beta\) or \(\rho\), highlight "\(\neq 0\)"

The output screen shows the \(p\text\) on the line that reads "\(p =\)".

(Most computer statistical software can calculate the \(p\text\).)

If the \(p\text\) is less than the significance level (\(\alpha = 0.05\)):

If the \(p\text\) is NOT less than the significance level (\(\alpha = 0.05\))

Calculation Notes:

An alternative way to calculate the \(p\text\) (\(p\)) given by LinRegTTest is the command 2*tcdf(abs(t),10^99, n-2) in 2nd DISTR.

THIRD-EXAM vs FINAL-EXAM EXAMPLE: \(p\text\) method

Because \(r\) is significant and the scatter plot shows a linear trend, the regression line can be used to predict final exam scores.

METHOD 2: Using a table of Critical Values to make a decision

The 95% Critical Values of the Sample Correlation Coefficient Table can be used to give you a good idea of whether the computed value of \(r\) is significant or not. Compare \(r\) to the appropriate critical value in the table. If \(r\) is not between the positive and negative critical values, then the correlation coefficient is significant. If \(r\) is significant, then you may want to use the line for prediction.

Example \(\PageIndex\)

Suppose you computed \(r = 0.801\) using \(n = 10\) data points. \(df = n - 2 = 10 - 2 = 8\). The critical values associated with \(df = 8\) are \(-0.632\) and \(+0.632\). If \(r \) positive critical value, then \(r\) is significant. Since \(r = 0.801\) and \(0.801 > 0.632\), \(r\) is significant and the line may be used for prediction. If you view this example on a number line, it will help you.

Horizontal number line with values of -1, -0.632, 0, 0.632, 0.801, and 1. A dashed line above values -0.632, 0, and 0.632 indicates not significant values.

Exercise \(\PageIndex\)

For a given line of best fit, you computed that \(r = 0.6501\) using \(n = 12\) data points and the critical value is 0.576. Can the line be used for prediction? Why or why not?

Answer

If the scatter plot looks linear then, yes, the line can be used for prediction, because \(r >\) the positive critical value.

Example \(\PageIndex\)

Suppose you computed \(r = –0.624\) with 14 data points. \(df = 14 – 2 = 12\). The critical values are \(-0.532\) and \(0.532\). Since \(-0.624 < -0.532\), \(r\) is significant and the line can be used for prediction

Horizontal number line with values of -0.624, -0.532, and 0.532.

Exercise \(\PageIndex\)

For a given line of best fit, you compute that \(r = 0.5204\) using \(n = 9\) data points, and the critical value is \(0.666\). Can the line be used for prediction? Why or why not?

Answer

No, the line cannot be used for prediction, because \(r

Example \(\PageIndex\)

Suppose you computed \(r = 0.776\) and \(n = 6\). \(df = 6 - 2 = 4\). The critical values are \(-0.811\) and \(0.811\). Since \(-0.811 < 0.776 < 0.811\), \(r\) is not significant, and the line should not be used for prediction.

Horizontal number line with values -0.924, -0.532, and 0.532.

Exercise \(\PageIndex\)

For a given line of best fit, you compute that \(r = -0.7204\) using \(n = 8\) data points, and the critical value is \(= 0.707\). Can the line be used for prediction? Why or why not?

Answer

Yes, the line can be used for prediction, because \(r

THIRD-EXAM vs FINAL-EXAM EXAMPLE: critical value method

Consider the third exam/final exam example. The line of best fit is: \(\hat = -173.51 + 4.83x\) with \(r = 0.6631\) and there are \(n = 11\) data points. Can the regression line be used for prediction? Given a third-exam score (\(x\) value), can we use the line to predict the final exam score (predicted \(y\) value)?

Because \(r\) is significant and the scatter plot shows a linear trend, the regression line can be used to predict final exam scores.

Example \(\PageIndex\)

Suppose you computed the following correlation coefficients. Using the table at the end of the chapter, determine if \(r\) is significant and the line of best fit associated with each r can be used to predict a \(y\) value. If it helps, draw a number line.

  1. \(r = –0.567\) and the sample size, \(n\), is \(19\). The \(df = n - 2 = 17\). The critical value is \(-0.456\). \(-0.567 < -0.456\) so \(r\) is significant.
  2. \(r = 0.708\) and the sample size, \(n\), is \(9\). The \(df = n - 2 = 7\). The critical value is \(0.666\). \(0.708 > 0.666\) so \(r\) is significant.
  3. \(r = 0.134\) and the sample size, \(n\), is \(14\). The \(df = 14 - 2 = 12\). The critical value is \(0.532\). \(0.134\) is between \(-0.532\) and \(0.532\) so \(r\) is not significant.
  4. \(r = 0\) and the sample size, \(n\), is five. No matter what the \(dfs\) are, \(r = 0\) is between the two critical values so \(r\) is not significant.
Exercise \(\PageIndex\)

For a given line of best fit, you compute that \(r = 0\) using \(n = 100\) data points. Can the line be used for prediction? Why or why not?

Answer

No, the line cannot be used for prediction no matter what the sample size is.

Assumptions in Testing the Significance of the Correlation Coefficient

Testing the significance of the correlation coefficient requires that certain assumptions about the data are satisfied. The premise of this test is that the data are a sample of observed points taken from a larger population. We have not examined the entire population because it is not possible or feasible to do so. We are examining the sample to draw a conclusion about whether the linear relationship that we see between \(x\) and \(y\) in the sample data provides strong enough evidence so that we can conclude that there is a linear relationship between \(x\) and \(y\) in the population.

The regression line equation that we calculate from the sample data gives the best-fit line for our particular sample. We want to use this best-fit line for the sample as an estimate of the best-fit line for the population. Examining the scatter plot and testing the significance of the correlation coefficient helps us determine if it is appropriate to do this.

The assumptions underlying the test of significance are:

The left graph shows three sets of points. Each set falls in a vertical line. The points in each set are normally distributed along the line — they are densely packed in the middle and more spread out at the top and bottom. A downward sloping regression line passes through the mean of each set. The right graph shows the same regression line plotted. A vertical normal curve is shown for each line.

Summary

Linear regression is a procedure for fitting a straight line of the form \(\hat = a + bx\) to data. The conditions for regression are:

The slope \(b\) and intercept \(a\) of the least-squares line estimate the slope \(\beta\) and intercept \(\alpha\) of the population (true) regression line. To estimate the population standard deviation of \(y\), \(\sigma\), use the standard deviation of the residuals, \(s\). \(s = \sqrt>\). The variable \(\rho\) (rho) is the population correlation coefficient. To test the null hypothesis \(H_: \rho =\) hypothesized value, use a linear regression t-test. The most common null hypothesis is \(H_: \rho = 0\) which indicates there is no linear relationship between \(x\) and \(y\) in the population. The TI-83, 83+, 84, 84+ calculator function LinRegTTest can perform this test (STATS TESTS LinRegTTest).

Formula Review

Least Squares Line or Line of Best Fit:

Standard deviation of the residuals: