What is the relationship between the amount of time statistics students study per week and their...

The provided data are shown in the table below х Y 3 58 10 75 15 89 5 77 12 79 0 54 15 96 Also, the following calculations are needed to compute the correlation coefficient: X Y X*Y X2 y2 3 58 174 9 3364 10 75 750 100 5625 15 89 1335 225 7921 5 77 385 25 5929 12 79 948 144 6241 0 54 0 0 2916 15 96 1440 225 9216 Sum = 60 528 5032 728 41212 The correlation coefficient r is computed using the following expression: r = SSXY VSSxx SSyy where n 1 SSxy = x;Y; ΣΥ, i=1 i=1 **!) *** 2 n n 1 SSxx = x п i=1 SSyy = y;? In this case, based on the data provided, we get that SSXY = 5032 - (60 (60 x 528) = 506.286 SSxx = 728 (60) = 213.714 SSYY = 41212 (528)2 = 1385.714 Therefore, based on this information, the sample correlation coefficient is computed as follows r = SSXY VSSxXSSyy 506.286 213.714 x 1385.714 = 0.93

Ho: p=0

H:P70

The following needs to be tested: Ho : p=0 HA :p=0 where p corresponds to the population correlation. The sample size is n = 7, so then the number of degrees of freedom is df = n - 2= 7 – 2 = 5 The corresponding t-statistic to test for the significance of the correlation is: t=r n - - 2 p2 = 0.931 7 – 2 0.932 - = 5.658 1 1 The p-value is computed as follows: p= Pr(|t5| > 5.658) = 0.0024 Since we have that p= 0.0024 < 0.05, it is concluded that the null hypothesis Ho is rejected. Therefore, based on the sample correlation provided, it is concluded that there is enough evidence to claim that the population correlation p is different than 0, at the 0.05 significance level.

These are the data that have been provided for the dependent and independent variable: Obs. Score Time 1 3 58 2 10 75 3 15 89 4 5 77 5 12 79 O 54 7 15 96 We need to compute the coefficient of determination, which is computed by squaring the correlation coefficient, which needs to be computed first.

Now, with the provided sample data, we need to construct the following table to compute the correlation coefficient: Obs. Score Time X? Y2 X; - Y; 3 58 9 3364 174 2 10 75 100 5625 750 3 15 89 225 7921 1335 4 5 77 25 5929 385 5 12 79 144 6241 948 6 O 54 O 2916 O 7 15 96 225 9216 1440 Sum = 60 528 728 41212 5032 Based on the table above, we compute the following sum of squares that will be used in the calculation of the correlation coefficient: 2 5ης - Σκ(Σκ) 1=1 1=1 728 1 7 X 3600 213.7143 2 SSyy ΣΥ; 1 41212 Χ 278784 7 1385.7143 1 58χν - Σκι - (Σκ) (Σ) 1 5032 X 31680 7 506.2857

Now, the correlation coefficient is computed using the following expression:: SSXY r VSSxx SSyy 506.2857 V213.7143 x 1385.7143 = 0.9303 Then, the coefficient of determination, or R-Squared coefficient (R²), is computed by simply squaring the correlation coefficient that was found above. So we get: R2 0.93032 = 0.8655 Therefore, based on the sample data provided, it is found that the coefficient of determination is RP = 0.8655. This implies that approximately 86.55% of variation in the dependent variable is explained by the independent variable.

The following data are passed: Time Score 3 58 10 75 15 89 5 77 12 79 0 54 15 96 The independent variable is Time, and the dependent variable is Score.

In order to compute the regression coefficients, the following table needs to be used: Time* Sc Time Score Time? Score? ore 3 58 174 9 3364 10 75 750 100 5625 15 89 1335 225 7921 5 77 385 25 5929 12 79 948 144 6241 0 54 0 0 2916 15 96 1440 225 9216 Sum = 60 528 5032 728 41212 Based on the above table, the following is calculated: 60 X = 8.5714285714286 n 7 i=1 528 = 75.428571428571 n 7 SSxx = 728 – 602/7 = 213.71428571429 Σ 3x -: (2x) $5y = $x -; () S$x = {xx -- (Ex) () 2 = 41212 – 5282/7 = 1385.7142857143 5032 – 60 x 528/7 = 506.28571428571 Therefore, based on the above calculations, the regression coefficients (the slope m, and the y-intercept n) are obtained as follows: m = SSXY SSxx 506.28571428571 213.71428571429 = 2.369 n = 7 – Ž. m = 75.428571428571 – 8.5714285714286 x 2.369 = 55.123 Therefore, we find that the regression equation is: Score = 55.123 + 2.369Time