Question

What is the relationship between the amount of time statistics students study per week and their final exam scores? The results of the survey are shown below. Time Score 3 10 15 512 015 58 75 89 89 77 79 54 96 a. Find the correlation coefficient: r = Round to 2 decimal places. b. The null and alternative hypotheses for correlation are: Ho: ? - 0 H: 70 (Round to four The p-value is: decimal places) c. Use a level of significance of a = 0.05 to state the conclusion of the hypothesis test in the context of the study. There is statistically significant evidence to conclude that there is a correlation between the time spent studying and the score on the final exam. Thus, the regression line is useful. There is statistically insignificant evidence to conclude that a student who spends more time studying will score higher on the final exam than a student who spends less time studying. There is statistically insignificant evidence to conclude that there is a correlation between the time spent studying and the score on the final exam. Thus, the use of the regression line is not appropriate. There is statistically significant evidence to conclude that a student who spends more time studying will score higher on the final exam than a student who spends less time studying. d. m? (Round to two decimal places) e. Interpret 2 74% of all students will receive the average score on the final exam. There is a 74% chance that the regression line will be a good predictor for the final exam score based on the time spent studying. Given any group that spends a fixed amount of time studying per week, 74% of all of those students will receive the predicted score on the final exam. There is a large variation in the final exam scores that students receive, but a fixed amount of time studuino por un

Answer 1

a) Correlation coefficient r = 0.93.

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

b) The null and alternative hypotheses are

Ho: p=0

H:P70

P-Value = 0.0024

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.

c) 1st option is correct: There is statistically significant evidence to conclude that there is a correlation between the time spent studying and the score on the final examexam. Thus, the regression line is useful.

d) r^2 = 0.87.

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.

f)

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

g) Score = 55.123 + 2.369*11 = 81.182

Score = 81.

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