12.6 Testing the significance of the correlation coefficient  (Page 3/3)

• $H_o$ : $\rho$ = 0
• $H_a$ : $\rho$ ≠ 0
• $\alpha$ = 0.05
• The p-value is 0.026 (from LinRegTTest on your calculator or from computer software)
• The p-value, 0.026, is less than the significance level of $\alpha$ = 0.05
• Decision: Reject the Null Hypothesis $H_o$
• Conclusion: There is sufficient evidence to conclude that there is a significant linear relationship between $x$ and $y$ because the correlation coefficient is significantly different from 0.
• 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

Suppose you computed $r=0.801$ using $n=10$ data points. $\text{df}=n-2=10-2=8$ . The critical values associated with $\text{df}=8$ are -0.632 and + 0.632. If $r$ $\text{negative critical value}$ or $r>\text{positive critical value}$ , 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.

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Suppose you computed $r=-0.624$ with 14 data points. $\text{df}=14-2=12$ . The critical values are -0.532 and 0.532. Since $-0.624$ $$ $-0.532$ , $r$ is significant and the line may be used for prediction

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Suppose you computed $r=0.776$ and $n=6$ . $\text{df}=6-2=4$ . 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.

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Third exam vs final exam example: critical value method

• Consider the third exam/final exam example .
• The line of best fit is: $\hat{y}=-173.51+\text{4.83x}$ with $r=0.6631$ and there are $\mathrm{n\; =\; 11}$ 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)?

• $H_o$ : $\rho$ = 0
• $H_a$ : $\rho$ ≠ 0
• $\alpha$ = 0.05
• Use the "95% Critical Value" table for $r$ with $\text{df}=\text{n}-2=11-2=9$
• The critical values are -0.602 and +0.602
• Since $0.6631>0.602$ , $r$ is significant.
• Decision: Reject $H_o$ :
• Conclusion:There is sufficient evidence to conclude that there is a significant linear relationship between $x$ and $y$ because the correlation coefficient is significantly different from 0.
• Because $r$ is significant and the scatter plot shows a linear trend, the regression line can be used to predict final exam scores.

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 scatterplot 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:

• There is a linear relationship in the population that models the average value of $y$ for varying values of $x$ . In other words, the expected value of $y$ for each particular $$ value lies on a straight line in the population. (We do not know the equation for the line for the population. 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 that there are more $y$ values scattered closer to the line than are scattered farther away. Assumption (1) above implies that these normal distributions are centered on the line: the means of these normal distributions of $y$ values lie on the line.
• The standard deviations of the population $y$ values about the line are equal for each value of $x$ . In other words, each of these normal distributions of $y$ values has the same shape and spread about the line.
• The residual errors are mutually independent (no pattern).

**With contributions from Roberta Bloom