Question

Using your calculator, run a regression analysis on the following bivariate set of data with y as the response variable.

x y

71.2 23.1

90.8 122.9

88.8 82.9

57.7 18.3

71.4 8.6

60.4 25.9

60.2 -43.1

88.5 77.4

68.2 -5.9

41.1 -83

61.4 -2.5

87.1 39.4

Find the correlation coefficient and report it accurate to three decimal places. r =

What proportion of the variation in y can be explained by the variation in the values of x? Report answer as a percentage accurate to one decimal place. (If the answer is 0.84471, then it would be 84.5%...you would enter 84.5 without the percent symbol.) r² = %

Based on the data, calculate the regression line (each value to three decimal places) y = x +

Predict what value (on average) for the response variable will be obtained from a value of 82.2 as the explanatory variable. Use a significance level of α = 0.05 to assess the strength of the linear correlation.

What is the predicted response value? (Report answer accurate to one decimal place.) y =

Answer 1

533.61 5069.44 23.1 1644.72 71.2 1 8244.64 15104.41 11159.32 122.9 90.8 6872.41 7885.44 7361.52 82.9 88.8 334.89 1055.91 3329.29 18.3 57.7 73.96 5097.96 71.4 8.6 670.81 1564.36 25.9 60.4 1857.61 3624.04 43.1 2594.62 60.2 7832.25 5990.76 6849.9 77.4 88.5 34.81 402.38 4651.24 5.9 68.2 1689.21 6889 -83 41.1 3769.96 2.5 153.5 61.4 1552.36 7586.41 3431.74 39.4 87.1 39920.88 62428.04 264 27119.71 846.8 Sum =

The correlation coefficient r is computed using the following expression SSxy where Ху. 7l 12 In this case, based on the data provided, we get that SSXY 271 19.71 12 (846.8 x 264) 8490.11 SSxx 62428.048.8) 2672.187 12 SSyy 39920.88(264)2 34112.88 12

Therefore, based on this information, the sample correlation coefficient is computed as follows 81011 V2672.187 x 34112.88 SSxY 0.889 VSSxxSSyy

Proportion of the variation in y can be explained by the variation in the values of x = r² = 0.889² = 79.0%

Now,

Therefore, based on the above calculations, the regression coefficients (the slope m, and the y- intercept n) are obtained as follows: 8490.11 SSxy SSxx 2672.1866666667 Y-X . m 22-70.566666666667 × 3.1772 -202.2054 n Therefore, we find that the regression equation is: Y202.2054 3.1772X

The regression line is : - 202.205 + 3.177 X

Predicted response value = - 202.205 + 3.177*82.2 = 58.9

Significance of correlation coefficient :

The following needs to be tested:

Ho : ρ=0

$H_1: \rho\neq0$

where ρ corresponds to the population correlation 12, so then the number of degrees of freedom is df = n-2-12-2-10 The sample size is n 0.05, for a two-tailed The corresponding critical correlation value re for a significance level of α test is: た= 0.576 Observe that in this case, the null hypothesis Ho: p 0 is rejected if 0.576 Based on the sample correlation provided, we have that lr -0.889 > rc-0.576, from which is concluded that the null hypothesis is rejected.

Hence, Linear correlation is significant at 5% level of significance.