Question

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Problem I (40pts) Evans et al. (A-5) examined the effect of velocity on ground reaction forces (GRF) in dogs with lameness from a torn cranial cruciate ligament. The dogs were walked and trotted over a force platform and the GRF recorded (in newtons) during the stance phase. The table contains 22 measurements (see problem 1.xlsx) of force expressed as the mean of five force measurements per dog when walking and the mean of five force measurements per dog when trotting. Use the GRF value when walking to predict the GRF value when trotting. (a) Find regression equation using the formulas for A, and derived based on the least-square method (b) Plot the regression equation with the data scatter plot. (Show your codes) Make sure you can sketch similar plots by hand. (c) Test Ho:β.-O using both ANOVA and t-test at α = 0.05. show all steps GRF-Walk GRF-Trot 50.8 43.2 44.8 39.5 31.5 33.3 28.8 38.3 36.9 14.6 27 2.8 27.4 60.1 11.1 32.3 2 2 38.2 50.8 24.9 3,6 30.7 27.2 46.3 41.8 32.4 65.8 32.2 29.5 38.7 42 37.6 30.2 28.2 24.3 31.6 29,9 34.3 24.9 0

Answer 1

This is a simple problem related to formulation of the regression equation based on the traditional approach as shown below.

The independent variable is GRF - Walk, and the dependent variable is GRF - Trot. In order to compute the regression coefficients, the following table needs to be used: GRF- Walk GRF Trot GRF-Walk GRF-Walk GRF-Trot GRF-Trot 31.5 33.3 32.3 28.8 38.3 36.9 14.6 27 32.8 1606.5 1438.56 1447.04 1137.6 1685.2 2214 162.06 872.1 1354.64 1046.68 1600.2 755 1555.68 992.25 1108.89 1043.29 829.44 1466.89 1361.61 213.16 729 1075.84 750.76 992.25 625 1128.96 2601 1866.24 2007.04 1560.25 1936 3600 123.21 1043.29 1705.69 1459.24 2580.64 912.04 2143.69 43.2 44.8 39.5 60 32.3 41.3 38.2 50.8 30.2 46.3 31.5 25 33.6

30.7 1283.26 881.28 2895.2 908.04 716.85 1222.92 1260 1289.68 750 28082.49 942.49 739.84 1936 795.24 590.49 998.56 900 1176.49 625 21021.45 1747.24 1049.76 4329.64 1036.84 870.25 1497.69 1764 1413.76 900 38147.51 32.4 65.8 32.2 29.5 38.7 42 37.6 30 882.7 27.2 28.2 24.3 31.6 30 34.3 25 668.3 Sum = Based on the above table, the following is calculated: 668.3 i =-= 30.377272727273 882.7 =-= 40.122727272727 Ỹ = ssxx= ΣΧ?--(Σ X, | = 21021.45-668.32/22 = 720.31863636364 ss,Y= ΣΥ.2_n(ΣΥ) = 38147.51-882.72/22= 2731·1786363636

SSXY = Σ xx--( Σ x.) | Σ Y; 28082.49-668.3 × 882.7/22 = 1268.4713636364 Therefore, based on the above calculations, the regression coefficients (the slope m, and the y-intercept n) are obtained as follows Sxy 1268.4713636364 SSxx720.31863636364 - 1.761 n = Y-X·m 40.122727272727-30.377272727273 x 1.761 =-13.3712 Therefore, we find that the regression equation is GRF - Trot-13.3712 +1.761GRF Walk Regression equation for GRF trot Vs GRF Halk 70.16 60.16 50.16 16 16 20.16 10.16 13.72 18.72 23.72 28.72 33.72 38.7 43.7248.72 GRF-Halk -Regression equation: Y 13.3712 1.761X

Now let us try to test the significance of the regression slope.

WE shall calculate the standard error forstly

	X	Y	(X-X-bar)^2	(Y-Y-bar)^2
	GRF-Walk	GRF-Trot
	31.5	50.8	1.3	113.9
	33.3	43.2	8.6	9.4
	32.3	44.8	3.7	21.8
	28.8	39.5	2.4	0.4
	38.3	44	63.0	15.0
	36.9	60.1	42.7	398.9
	14.6	11.1	248.5	842.6
	27	32.3	11.3	61.3
	32.8	41.3	5.9	1.4
	27.4	38.2	8.8	3.7
	31.5	50.8	1.3	113.9
	24.9	30.2	29.9	98.6
	33.6	46.3	10.5	38.1
	30.7	41.8	0.1	2.8
	27.2	32.4	10.0	59.7
	44	65.8	186.0	659.1
	28.2	32.2	4.7	62.8
	24.3	29.5	36.8	112.9
	31.6	38.7	1.5	2.0
	29.9	42	0.2	3.5
	34.3	37.6	15.5	6.4
	24.9	30.2	29.9	98.6
mean	30.4	40.1	32.8	123.9
Sum	668.0	882.8	722.6	2726.8
	SE	=sqrt(2726.8/20)/sqrt(722.6)		0.4

We shall perform the following process.

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

• State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

  H_o: The slope of the regression line is equal to zero.

  H_a: The slope of the regression line is not equal to zero.

  If the relationship between home size and electric bill is significant, the slope will not equal zero.
• Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a linear regression t-test to determine whether the slope of the regression line differs significantly from zero.
• Analyze sample data. To apply the linear regression t-test to sample data, we require the standard error of the slope, the slope of the regression line, the degrees of freedom, the t statistic test statistic, and the P-value of the test statistic.

  We get the slope (b₁) and the standard error (SE) from the regression output.

  b₁ = 1.761       SE = 0.40

  We compute the degrees of freedom and the t statistic test statistic, using the following equations.

  DF = 22 - 2 = 20

  t = b₁/SE = 1.761/0.40 = 4.40

  where DF is the degrees of freedom, n is the number of observations in the sample, b₁ is the slope of the regression line, and SE is the standard error of the slope.

  Based on the t statistic test statistic and the degrees of freedom, we determine the P-value. The P-value is the probability that a t statistic having 20 degrees of freedom is more extreme than 4.40. Since this is a two-tailed test, "more extreme" means greater than 4.40 or less than -4.40. We use the t Distribution Calculator to find P(t > 4.40) = 0.000138 and P(t < -4.40) = 0.000138. Therefore, the P-value is 0.000138 + 0.000138 or 0.000276.
• Interpret results. Since the P-value (0.000276) is less than the significance level (0.05), we cannot accept the null hypothesis.
• hence we can significantly conclude that the slope is not insignificantly zero.