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
The
95% Critical Values of the Sample Correlation Coefficient Table at the end of
this chapter (before the
Summary ) may 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.
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$$\text{negative critical value}$ or
$r>\text{positive critical value}$ , then
$r$ is
significant. Since
$r=0.801$ and
$0.801>0.632$ ,
$r$ is significant and the line may be used
for prediction. If you view this example on a number line, it will help you.
Suppose you computed
$r=-0.624$ with 14 data points.
$\text{df}=14-2=12$ . The critical values are -0.532 and 0.532. Since
$-0.624$$$$-0.532$ ,
$r$ is significant and
the line may be used for prediction
Suppose you computed
$r=0.776$ and
$n=6$ .
$\text{df}=6-2=4$ . The
critical values are -0.811 and 0.811. Since
$-0.811$$$$0.776$$$$0.811$ ,
$r$ is not significant
and the line should not be used for prediction.
The line of best fit is:
$\hat{y}=-173.51+\text{4.83x}$ with
$r=0.6631$ and there are
$\mathrm{n\; =\; 11}$ data points.
Can the regression line be used for prediction?
Given a third exam score (
$x$ value), can we
use the line to predict the final exam score (predicted
$y$ value)?
${H}_{o}$ :
$\rho $ = 0
${H}_{a}$ :
$\rho $ ≠ 0
$\alpha $ = 0.05
Use the "95% Critical Value" table for
$r$ with
$\text{df}=\text{n}-2=11-2=9$
The critical values are -0.602 and +0.602
Since
$0.6631>0.602$ ,
$r$ is significant.
Decision: Reject
${H}_{o}$ :
Conclusion:There is sufficient evidence to conclude that there is a significant linear relationship between
$x$ and
$y$ because the correlation coefficient is significantly different from 0.
Because
$r$ is significant and the scatter plot shows a linear trend, the regression line can be used to predict final exam scores.
Additional practice examples using critical values
Suppose you computed the following correlation coefficients. Using the
table at the end of the chapter, determine if
$r$ is significant and the line of best fit associated
with each
$r$ can be used to predict a
$y$ value. If it helps, draw a number line.
$r=-0.567$ and the sample size,
$n$ , is 19. The
$\text{df}=n-2=17$ . The critical value is -0.456.
$-0.567$$$$-0.456$ so
$r$ is significant.
$r=0.708$ and the sample size,
$n$ , is 9. The
$\text{df}=n-2=7$ . The critical value is 0.666.
$0.708>0.666$ so
$r$ is significant.
$r=0.134$ and the sample size,
$n$ , is 14. The
$\text{df}=\mathrm{14}-2=12$ . The critical value is 0.532. 0.134 is between -0.532 and 0.532 so
$r$ is not significant.
$r=0$ and the sample size,
$n$ , is 5. No matter what the dfs are,
$r=0$ is between the two critical values so
$r$ is not significant.
Assumptions in testing the significance of the correlation coefficient
Testing the significance of the correlation coefficient requires that certain assumptions about the data are satisfied. The premise of this test is that the data are a sample of observed points taken from a larger population. We have not examined the entire population because it is not possible or feasible to do so. We are examining the sample to draw a conclusion about whether the linear relationship that we see between
$x$ and
$y$ in the sample data provides strong enough evidence so that we can conclude that there is a linear relationship between
$x$ and
$y$ in the population.
The regression line equation that we calculate from the sample data gives the best fit line for our particular sample. We want to use this best fit line for the sample as an estimate of the best fit line for the population. Examining the scatterplot and testing the significance of the correlation coefficient helps us determine if it is appropriate to do this.
The assumptions underlying the test of significance are:
There is a linear relationship in the population that models the average value of
$y$ for varying values of
$x$ . In other words, the expected value of
$y$ for each particular
$$ 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).
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
1 It is estimated that 30% of all drivers have some kind of medical aid in South Africa. What is the probability that in a sample of 10 drivers: 3.1.1 Exactly 4 will have a medical aid. (8) 3.1.2 At least 2 will have a medical aid. (8) 3.1.3 More than 9 will have a medical aid.