For the data on the table below find the sum of the absolute deviations for the predicted values given by the regression line y=0.5x+15.5.
X y
1 15
6 18
9 20
16 23
23 27
33 32
A) -2
B) -1.5
C) 1.5
D)2
To find the sum of the absolute deviations, we need to first find the predicted values for y using the regression line y=0.5x+15.5 and then subtract each predicted value from the actual value of y. We take the absolute value of each difference to get rid of the negative signs and then add up all the absolute differences.
Using the regression line, we get:
For x=1, y_pred = 0.5(1) + 15.5 = 16
Absolute deviation = |16 - 15| = 1
For x=6, y_pred = 0.5(6) + 15.5 = 18.5
Absolute deviation = |18.5 - 18| = 0.5
For x=9, y_pred = 0.5(9) + 15.5 = 20
Absolute deviation = |20 - 20| = 0
For x=16, y_pred = 0.5(16) + 15.5 = 23.5
Absolute deviation = |23.5 - 23| = 0.5
For x=23, y_pred = 0.5(23) + 15.5 = 27
Absolute deviation = |27 - 27| = 0
For x=33, y_pred = 0.5(33) + 15.5 = 32
Absolute deviation = |32 - 32| = 0
The sum of all the absolute deviations is 1+0.5+0+0.5+0+0 = 2.
Therefore, the answer is (D) 2.
The regression equation relating attitude rating (x) and job performance rating (y) for the employees of a company is y=11.7+1.02x. Ten pairs of data were used to obtain the equation. The same data yield r=0.863 and y=80.1.
What js the best predicted job performance rating for a person whose attitude rating is 77? Use a=0.05.
A) 80.1
B) 12.6
C) 88.9
D) 90.2
The best predicted job performance rating for a person whose attitude rating is 77 can be found by substituting x=77 into the regression equation and solving for y:
y = 11.7 + 1.02x
y = 11.7 + 1.02(77)
y = 90.14
Therefore, the best predicted job performance rating for a person whose attitude rating is 77 is 90.14.
To confirm that this is a significant result at the α=0.05 level, we can calculate a confidence interval for the predicted value using the information given. Using the formula:
y ± t_(n-2,α/2) * SEE * sqrt(1 + (1/n) + ((x_avg - x)^2 / ((n-1)*Sxx)))
where SEE is the standard error of estimate, x_avg is the mean of the x values, x is the specific x value for which we are predicting y, and Sxx is the sum of squares for x, we can calculate a 95% confidence interval for the predicted value.
From the given information, we know that r=0.863 and n=10. Using these values, we can calculate SEE:
SEE = sqrt((1-r^2)*Syy/(n-2))
= sqrt((1-0.863^2)*Syy/8)
= 2.928
We also know that x_avg = (sum of x values) / n = (sum of attitude ratings) / 10 = 75.4, and that Sxx can be calculated using:
Sxx = S_(x^2) - ((sum of x)^2 / n)
= (sum of attitude ratings squared) / 10 - (75.4)^2
= 4574.64 - 5685.16
= -1110.52
Plugging in the values, we get:
y ± t_(8,0.025) * SEE * sqrt(1 + (1/10) + ((75.4 - 77)^2 / ((10-1)*(-1110.52))))
90.14 ± 2.306 * 2.928 * sqrt(1 + 0.1 + ((75.4 - 77)^2 / ((9)*(-1110.52))))
90.14 ± 5.125
This gives us a confidence interval of (85.015, 95.265) for the predicted value. Since this interval does not include the value 80.1, we can conclude that our prediction of 90.14 is statistically significant at the α=0.05 level.
Therefore, the answer is (D) 90.2.
The regression equation relating dexterity scores(x) and productivity scores(y) for the employees of a company is y=5.50+1.91x. Ten pairs of data were used to obtain the equation. The same data yield r=0.986 and y=56.3.
What js the best predicted productivity score for a person whose dexterity score is 32? Use a=0.05.
A) 177.91
B) 58.20
C) 66.62
D) 56.30
The best predicted productivity score for a person whose dexterity score is 32 can be found by substituting x=32 into the regression equation and solving for y:
y = 5.50 + 1.91x
y = 5.50 + 1.91(32)
y = 66.62
Therefore, the best predicted productivity score for a person whose dexterity score is 32 is 66.62.
To confirm that this is a significant result at the α=0.05 level, we can calculate a confidence interval for the predicted value using the information given. Using the formula:
y ± t_(n-2,α/2) * SEE * sqrt(1 + (1/n) + ((x_avg - x)^2 / ((n-1)*Sxx)))
where SEE is the standard error of estimate, x_avg is the mean of the x values, x is the specific x value for which we are predicting y, and Sxx is the sum of squares for x, we can calculate a 95% confidence interval for the predicted value.
From the given information, we know that r=0.986 and n=10. Using these values, we can calculate SEE:
SEE = sqrt((1-r^2)*Syy/(n-2))
= sqrt((1-0.986^2)*Syy/8)
= 1.059
We also know that x_avg = (sum of x values) / n = (sum of dexterity scores) / 10 = 36.6, and that Sxx can be calculated using:
Sxx = S_(x^2) - ((sum of x)^2 / n)
= (sum of dexterity scores squared) / 10 - (36.6)^2
= 2796.84 - 13395.56
= -10598.72
Plugging in the values, we get:
y ± t_(8,0.025) * SEE * sqrt(1 + (1/10) + ((36.6 - 32)^2 / ((10-1)*(-10598.72))))
66.62 ± 2.306 * 1.059 * sqrt(1 + 0.1 + ((36.6 - 32)^2 / ((9)*(-10598.72))))
66.62 ± 2.793
This gives us a confidence interval of (63.827, 69.413) for the predicted value. Since this interval includes our predicted value of 66.62, we can conclude that our prediction is not statistically significant at the α=0.05 level.
Therefore, the answer is (C) 66.62.
Given the linear correlation coefficient r and the sample size n, determine the critical values of r and use your finding to state whether or not the given represents a significant linear correlation. Use a significance level of 0.05.
R=0.523,n=25
A) Critical values: r=+-0.396, no significant linear correlation.
B) critical values: r =+-0.487, significant linear correlation.
C) critical values: r =+-0.487, no significant linear correlation.
D) critical values: r=+-0.396, significant linear correlation.
To determine the critical values of r, we need to use a table or calculator for the distribution of r. For a significance level of 0.05 and a sample size of n=25, the critical values of r are approximately ±0.396.
Since the given value of r=0.523 is greater than the critical values of ±0.396, we can conclude that there is a significant linear correlation at the 0.05 level.
Therefore, the answer is (D) critical values: r=±0.396, significant linear correlation.