. Homework Wife Husband X 5 7 5 7 7 3 3 9 9 5 9 5 3 3 9 Sum 5 7 Mean 7.5.375 SS.5 37.75 r = ( )( 7) - 5.5 ( )( 37.75) = 55.5 7.7 =.9 With r Crit () =.77, we would reject H : r =. Thus, it would make sense to compute the regression equation to allow us to make predictions. ˆ =.73X + 3. We would also want to compute the standard error of estimate in order to have a sense of the accuracy of predictions made using the regression equation. SEE = 3.7 =.99. Anxiety Exam Score X 5 7 7 5 7 79 553 3 5 5 5 Sum 3 9 5 Mean 5 3 SS 7 ( )( 9) 5-3 r = ( )( 7) = -3 3 = -.9
With r Crit () =., we would reject H : r =. Thus, we should construct the regression equation in order to make predictions. ˆ = -.7X + 9.9 To get a sense of the accuracy of predictions, we should compute the standard error of estimate: SEE =.97 =.93. The Spearman correlation is exactly the same (computationally) as the Pearson correlation. The only difference is that the computation of r is based on the ranked data, rather than the actual scores. X ank X ank X 7 9 3 3 5 3 3 5 5 5 Sum 7 9 5 5 5 SS 55. 59. ( )( 5) 5-5 r = 5 ( )( ) = 9 =.9 With r Crit (5) =., we would actually retain H in this case.. a. = $X + $ (Company A) and = $5X + $ (Company B) b. $7 from Company A and $7 from Company B, so the costs would be equal for rats. c. $3 from Company A and $ from Company B, so Company B is the better deal.. X X ˆ 7 9-3 5 5-3 - 5 7 3 7 Sum 3 3 Mean 3 SS - ˆ
( )( 3) 3 - r = ( )( ) = 3 3. =.9 With r Crit () =., we would reject H : r =. Thus, we should construct the regression equation in order to make predictions. ˆ = 3X - 3 If we sum the differences between the observed and predicted values, we will always get, as seen above. However, if we square the differences before adding, we get, which would be the SS Error. Note that if you compute r =. and then compute -r =.7, we can multiply the coefficient of alienation by SS and also get SS Error. (We actually get.3, due to rounding.). egression Summary vs. X Squared vs. X egression esidual.9.3.7.59 DF Sum of Squar Mean Square F-Value P-Value.33.33.9. 9.5.55 7 5.75 9 7 5 3 egression Plot 3 5 7 9 X =.5 -.7 * X; ^ =.3 egression Coefficients vs. X X.5..5.39 <. -.7.5 -.9-5.. As you can see, there is a significant correlation between X and, with p <.5.
egression Summary vs. X+5 Squared.9.3.7.59 9 7 5 3 egression Plot 7 9 3 5 X+5 =.55 -.7 * X; ^ =.3 vs. X+5 egression esidual.33.33.9. 9.5.55 7 5.75 egression Coefficients vs. X+5 X+5.55.75.55.7. -.7.5 -.9-5.. As you can see, adding 5 to each of the X values has no impact on the correlation coefficient. If you think of r as the mean product of z scores, that should make sense to you. That is, the addition of 5 to each of the X values would have no effect on the z scores, so r should remain the same. It s also the case that adding a constant to a set of scores would leave the SS intact, so the SS X would stay the same after the addition of a constant of 5. As you can see in the output below, multiplying each value of X by a constant (3) also has no impact on the correlation coefficient. The SS X goes from 3.5 to 57.5 (3 3 or 9 times larger), while SS would remain unchanged. The SP would be 3 times larger after multiplying each X by 3.
egression Summary vs. X*3 Squared vs. X*3 egression esidual.9.3.7.59.33.33.9. 9.5.55 7 5.75 9 7 5 3 egression Plot 5 5 5 3 X*3 =.5 -.9 * X; ^ =.3 egression Coefficients vs. X*3 X*3.5..5.39 <. -.9.53 -.9-5..
. egression Summary Errors vs. eaction Time Squared.773.597.53.5 Errors egression Plot Errors vs. eaction Time egression esidual 53.7 53.7.97.5 3.99.33 7 9.75 9 3 eaction Time = 35.37 -.33 * X; ^ =.597 egression Coefficients Errors vs. eaction Time eaction Time 35.37 9.3 35.37 3.7.9 -.33.5 -.773 -.93.5 As you can see, there is a significant negative linear relationship (r = -.773) between speed and accuracy. As speed increases, errors decrease and vice versa. ou could describe the relationship precisely by using the regression equation.
.7 egression Summary Doctor Visits vs. LCU Squared.77.77.7.5 Doctor Visits egression Plot 5 5 5 3 LCU =.39 +.3 * X; ^ =.77 Doctor Visits vs. LCU egression esidual 7.3 7.3 3.. 9.3.37 9.77 egression Coefficients Doctor Visits vs. LCU LCU.39.739.39 3.33.3.3..77 5.3. As you can see, there is a significant positive linear relationship between Doctor Visits and LCUs. Placing the same data in the Spearman Correlation analysis (under Nonparametric analyses) yields the following analysis: Spearman ank Correlation for LCU, Doctor Visits Sum of Squared Differences.5 ho. Z-Value.5 P-Value.99 ho corrected for ties.5 Tied Z-Value.577 Tied P-Value. # Ties, LCU # Ties, Doctor Visits