The correlation coefficient, *r*, tells us about the strength and direction of the linear relationship between *x* and *y*. However, the reliability of the linear model also depends on how many observed data points are in the sample. We need to look at both the correlation coefficient *r* and the sample size *n*, together.

We perform a hypothesis test of the *significance of the correlation coefficient* to decide whether the linear relationship in the sample data is strong enough to use to model the relationship in the population.

The sample data are used to compute *r*, the correlation coefficient for the sample. If we had data for the entire population, we could find the population correlation coefficient. But, because we have only sample data, we cannot calculate the population correlation coefficient. The sample correlation coefficient, *r*, is our estimate of the unknown population correlation coefficient.

- The symbol for the population correlation coefficient is
*ρ*, the Greek letter rho. *ρ*= population correlation coefficient (unknown).*r*= sample correlation coefficient (known; calculated from sample data).

The hypothesis test lets us decide whether the value of the population correlation coefficient *ρ* 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 the correlation coefficient is significantly different from zero, we say the correlation coefficient is *significant*.

- Conclusion: There is sufficient evidence to conclude there is a significant linear relationship between
*x*and*y*because the correlation coefficient is significantly different from zero. - What the conclusion means: There is a significant linear relationship between
*x*and*y*. We can use the regression line to model the linear relationship between*x*and*y*in the population.

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

- Conclusion: There is insufficient evidence to conclude there is a significant linear relationship between
*x*and*y*because the correlation coefficient is not significantly different from zero. - What the conclusion means: There is not a significant linear relationship between
*x*and*y*. Therefore, we*cannot*use the regression line to model a linear relationship between*x*and*y*in the population.

## Note

- If
*r*is significant and the scatter plot shows a linear trend, the line can be used to predict the value of*y*for values of*x*that are within the domain of observed*x*values. - If
*r*is not significant*or*if the scatter plot does not show a linear trend, the line should not be used for prediction. - If
*r*is significant and the scatter plot shows a linear trend, the line may*not*be appropriate or reliable for prediction*outside*the domain of observed*x*values in the data.

## Performing the Hypothesis Test

- Null hypothesis:
*H*_{0}:*ρ*= 0. - Alternate hypothesis:
*H*:_{a}*ρ*≠ 0.

What the Hypothesis Means in Words:

**Null hypothesis**The population correlation coefficient*H*_{0}:*is not*significantly different from zero. There*is not*a significant linear relationship (correlation) between*x*and*y*in the population.**Alternate hypothesis**The population correlation coefficient*H*:_{a}*is*significantly different from zero. There*is*a significant linear relationship (correlation) between*x*and*y*in the population.

Drawing a Conclusion:There are two methods to make a conclusion. The two methods are equivalent and give the same result.

**Method 1: Use the***p*-value.**Method 2: Use a table of critical values.**

In this chapter, we will always use a significance level of 5 percent, *α* = 0.05.

## Note

Using the *p*-value method, you could choose any appropriate significance level you want; you are not limited to using *α* = 0.05. But, the table of critical values provided in this textbook assumes we are using a significance level of 5 percent, *α* = 0.05. If we wanted to use a significance level different from 5 percent 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*-value to Make a Decision

## Using the TI-83, 83+, 84, 84+ Calculator

To calculate the *p*-value using `LinRegTTEST`

:

- Complete the same steps as the
`LinRegTTest`

performed previously in this chapter, making sure on the line prompt for

is highlighted.*β*or*σ*, ≠ 0 - When looking at the output screen, the
*p*-value is on the line that reads`p =`

.

*p*-value is less than the significance level (

*α*= 0.05):

- Decision: Reject the null hypothesis.
- Conclusion: There is sufficient evidence to conclude there is a significant linear relationship between
*x*and*y*because the correlation coefficient is significantly different from zero.

*p*-value is

*not*less than the significance level (

*α*= 0.05):

- Decision: Do not reject the null hypothesis.
- Conclusion: There is insufficient evidence to conclude there is a significant linear relationship between
*x*and*y*because the correlation coefficient is not significantly different from zero.

You will use technology to calculate the *p*-value, but it is useful to know that the *p*-value is calculated using a *t* distribution with *n* – 2 degrees of freedom and that the *p*-value is the combined area in both tails.

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

*p*-value Method

- Consider the third exam/final exam example.
- The line of best fit is
*ŷ*= –173.51 + 4.83*x*, 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)?

*H*_{0}: *ρ* = 0

*H _{a}*:

*ρ*≠ 0

*α* = 0.05

- The
*p*-value is 0.026 (from LinRegTTest on a calculator or from computer software). - The
*p*-value, 0.026, is less than the significance level of*α*= 0.05. - Decision: Reject the null hypothesis
*H*_{0}. - Conclusion: There is sufficient evidence to conclude there is a significant linear relationship between the third exam score (
*x*) and the final exam score (*y*) because the correlation coefficient is significantly different from zero.

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 Percent Critical Values of the Sample Correlation Coefficient Table** (Table 12.9) can be used to give you a good idea of whether the computed value of *r* is significant. Use it to find the critical values using the degrees of freedom, *df* = *n* – 2. The table has already been calculated with *α* = 0.05. The table tells you the positive critical value, but you should also make that number negative to have two critical values. If *r* is not between the positive and negative critical values, then the correlation coefficient is significant. If *r* is significant, then you may use the line for prediction. If *r* is not significant (between the critical values), you should not use the line to make predictions.

## Example 12.6

Suppose you computed *r* = 0.801 using *n* = 10 data points. The degrees of freedom would be 8 (*df* = *n* – 2 = 10 – 2 = 8). Using Table 12.9 with df = 8, we find that the critical value is 0.632. This means the critical values are really ±0.632. 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 to see that *r* is not between the two critical values.

## Try It 12.6

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

## Example 12.7

Suppose you computed *r* = –0.624 with 14 data points, where *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.

## Try It 12.7

For a given line of best fit, you compute that *r* = 0.5204 using *n* = 9 data points, and the critical values are ±0.666. Can the line be used for prediction? Why or why not?

## Example 12.8

Suppose you computed *r* = 0.776 and *n* = 6, with *df* = 6 -– 2 = 4. The critical values are – 0.811 and 0.811. Since 0.776 is between the two critical values, *r* is not significant. The line should not be used for prediction.

## Try It 12.8

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?

## Third Exam vs. Final Exam Example: Critical Value Method

Consider the third exam/final exam example.
The line of best fit is: *ŷ* = –173.51 + 4.83*x*, with *r* = .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)?

*H*_{0}:*ρ*= 0-
*H*:_{a}*ρ*≠ 0 *α*= 0.05

- Use the 95 Percent Critical Values table for
*r*with*df*=*n*– 2 = 11 – 2 = 9. - Using the table with
*df*= 9, we find that the critical value listed is 0.602. Therefore, the critical values are ±0.602. - Since 0.6631 > 0.602,
*r*is significant. - Decision: Reject the null hypothesis.
- Conclusion: There is sufficient evidence to conclude there is a significant linear relationship between the third exam score (
*x*) and the final exam score (*y*) because the correlation coefficient is significantly different from zero.

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

## Example 12.9

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

*r*= –0.567 and the sample size,*n*, is 19.

To solve this problem, first find the degrees of freedom.*df*=*n*- 2 = 17.

Then, using the table, the critical values are ±0.456.

–0.567 < –0.456, or you may say that –0.567 is not between the two critical values.

*r*is significant and may be used for predictions.

*r*= 0.708 and the sample size,*n*, is 9.

*df*=*n*- 2 = 7

The critical values are ±0.666.

0.708 > 0.666.

*r*is significant and may be used for predictions.

*r*= 0.134 and the sample size,*n*, is 14.

*df*= 14 –- 2 = 12.

The critical values are ±0.532.

0.134 is between –0.532 and 0.532.

*r*is not significant and may not be used for predictions.*r*= 0 and the sample size,*n*, is 5.

It doesn’'t matter what the degrees of freedom are because*r*= 0 will always be between the two critical values, so*r*is not significant and may not be used for predictions.

## Try It 12.9

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?

## Assumptions in Testing the Significance of the Correlation Coefficient

Testing the significance of the correlation coefficient requires that certain assumptions about the data be 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 that we can conclude 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 whether it is appropriate to do this.

- There is a linear relationship in the population that models the sample data. Our regression line from the sample is our best estimate of this line in the population.
- The
*y*values for any particular*x*value are normally distributed about the line. This implies there are more*y*values scattered closer to the line than are scattered farther away. Assumption 1 implies that these normal distributions are centered on the line; the means of these normal distributions of*y*values lie on the line. - Normal distributions of all the
*y*values have the same shape and spread about the line. - The residual errors are mutually independent (no pattern).
- The data are produced from a well-designed, random sample or randomized experiment.