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  • H o : ρ = 0
  • H a : ρ ≠ 0
  • α = 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 α = 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. df = n - 2 = 10 - 2 = 8 . The critical values associated with df = 8 are -0.632 and + 0.632. If r negative critical value or 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.
r is not significant between -0.632 and +0.632. r = 0.801 > +0.632 . Therefore, r is significant.

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 may be used for prediction

Horizontal number line with values of -0.624, -0.532, and 0.532.
r = -0.624 -0.532 . Therefore, r is significant.

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.
-0.811 r = 0.776 0.811 . Therefore, r is not significant.

    Third exam vs final exam example: critical value method

  • Consider the third exam/final exam example .
  • The line of best fit is: y ^ = -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)?
  • H o : ρ = 0
  • H a : ρ ≠ 0
  • α = 0.05
  • Use the "95% Critical Value" table for r with df = 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.

  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 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).

A downward sloping regression line is shown with the y values normally distributed about the line with equal standard deviations for each x value. For each x value, the mean of the y values lies on the regression line. More y values lie near the line than are scattered further away from the line.
The y values for each x value are normally distributed about the line with the same standard deviation. For each x value, the mean of the y values lies on the regression line. More y values lie near the line than are scattered further away from the line.

**With contributions from Roberta Bloom

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
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Sherica
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Sherica
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Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
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abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
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Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Quantitative information analysis iii. OpenStax CNX. Dec 25, 2009 Download for free at http://cnx.org/content/col11155/1.1
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