Statistics Final Semester Exam

Statistics Final Semester Exam Do the questions provided below! 1. Name 3 differences between Descriptive Statistics and Inferential Statistics (make it in a matrix table) Answer : No. 1. Descriptive Statistics Aims to describe data characteristics Inferential Statistics Aims to conclude the population by analyzing the sample 2. 3. Usually presented in table form and Presented in the form of possibilities- chart possibility. The test statistics used are simple like The test statistics used are quite mean, variance, etc. complicated, such as regression analysis tools, T-Test, Hypothesis Testing, etc. 2. Solve the Case below! A student conducts research with the aim of finding out the relationship between length of work, frequency of truancy, and work productivity. For this purpose, students took a sample of 12 employees at the artificial eyelash company PT Indolentik. Here is the data it has collected: Employee 1 2 3 4 5 6 7 8 Work Period 7 6 7 8 7 6 8 8 Frequency Of Skipping 3 7 5 1 3 2 1 2 Work Productivity 75 55 65 84 75 82 95 84 9 10 11 12 9 9 10 8 4 1 1 1 95 105 110 90 Question : What is the correlational relationship between length of service and frequency of truancy, then between length of service and work productivity, finally between frequency of truancy and work productivity? (show the calculation manually, then enter the calculation results in the following matrix table). Correlational Relationship Table Work Period Frequency Of Work Frequency Of Work Productivity Period Skipping - -0,552 0,872 -0,552 - -0,791 0,872 -0,791 - Skipping Work Productivity Answer : 1. Correlational relationship between length of service and frequency of truancy Answer : The magnitude of the relationship between the length of service variable and the frequency of truancy variable is -0.552, which shows that the correlation relationship is quite strong and in the opposite direction (negative). 2. Correlational relationship between length of service and work productivity Answer : The magnitude of the relationship between the length of service variable and the work productivity variable is 0.872, which shows that the correlation relationship is very strong and in the same direction (positive). 3. Correlational relationship between truant frequency and work productivity Answer : The magnitude of the relationship between the variable frequency of truancy and the variable work productivity is -0.791, which shows a strong correlation and is in the opposite direction (negative). Step 1 Resolving the case of a correlational relationship between length of service and frequency of truancy Employee Work Period (N) (X) Frequency Of Skipping (Y) 3 1 7 2 6 7 3 7 5 4 5 8 7 1 3 6 6 2 7 8 8 8 1 2 9 9 4 10 11 9 10 1 1 12 Total 8 93 1 31 X Y 2 1 4 2 3 5 8 2 1 1 2 8 1 6 3 6 9 1 0 8 2 2 6 X2 Y2 49 9 36 49 49 25 64 49 1 9 36 4 64 64 1 4 81 16 81 100 1 1 64 737 1 121 The Pearson Product Moment correlation coefficient formula is as follows: 𝑅𝑥𝑦 = 𝑅𝑥𝑦 = 𝑅𝑥𝑦 = 12 (226) − (93)(31) √[12 (737) − (93)2][12 (121) − (31)2] 2.712 − 2.883 √[8.844 − 8.649][1.452 − 961] −171 √[195][491] 𝑅𝑥𝑦 = 𝑅𝑥𝑦 = −171 √95.745 −171 309,42 = -0,552 Step 2 Resolving the case of the correlational relationship between length of service and work productivity Employee Work Period (N) (X) 1 2 3 4 5 6 7 8 9 10 11 12 Total 7 6 7 8 7 6 8 8 9 9 10 8 93 Work Productivity (Y) 75 55 65 84 75 82 95 84 95 105 110 90 1.015 XY X2 Y2 525 330 455 672 525 492 760 672 855 945 1.100 720 8.051 49 36 49 64 49 36 64 64 81 81 100 64 737 5.625 3.025 4.225 7.056 5.625 6.724 9.025 7.056 9.025 11.025 12.100 8.100 88.611 The Pearson Product Moment correlation coefficient formula is as follows: 𝑅𝑥𝑦 = 𝑅𝑥𝑦 = 𝑅𝑥𝑦 = 𝑅𝑥𝑦 = 𝑅𝑥𝑦 = 12 (8.051) − (93)(1.015) √[12 (737) − (93)2][12 (88.611) − (1.015)2] 96.612 − 94.395 √[8.844 − 8.649][1.063.332 − 1.030.225] 2.217 √[195][33.107] 2.217 √6.455.865 2.217 2.540,83 = 0,872 Step 3 Resolving the correlational relationship between Freq. Skipping classes with work productivity Employee Frequency (N) Of Skipping (X) 1 3 2 7 3 5 4 1 5 3 6 2 7 1 8 2 9 4 10 1 11 1 12 1 Total 31 Work Productivity (Y) 75 55 65 84 75 82 95 84 95 105 110 90 1.01 5 XY X2 Y2 225 385 325 84 225 164 95 168 380 105 110 90 2.356 9 49 25 1 9 4 1 4 16 1 1 1 121 5.625 3.025 4.225 7.056 5.625 6.724 9.025 7.056 9.025 11.025 12.100 8.100 88.611 The Pearson Product Moment correlation coefficient formula is as follows: 𝑅𝑥𝑦 = 𝑅𝑥𝑦 = 𝑅𝑥𝑦 = 12 (2.356) − (31)(1.015) √[12 (121) − (31)2][12 (88.611) − (1.015)2] 28.272 − 31.465 √[1.452 − 961][1.063.332 − 1.030.225] −3.193 √[491][33.107] 𝑅𝑥𝑦 = 𝑅𝑥𝑦 = −3.193 √16.255.537 −3.193 4.031,81 = -0,791 1. Write a functional relationship regression model for this case? The functional relationship regression model used in this case is a multiple linear regression model, namely a regression analysis involving more than two variables, one dependent (dependent) variable and more than one independent or independent variable. 𝑌̂ = 𝑎 + 𝑏1 + 𝑏2 1. What is meant by dependent variable and independent variable? Answer : Dependent variable (dependent variable) is a variable that is influenced, as a result of the existence of an independent variable. It is said to be a dependent variable because the dependent variable is influenced by the independent variable (independent variable). Independent variables are variables that influence or cause changes in the emergence of the dependent variable. 2. From the variables above, determine which is the dependent variable and which is the independent variable? Answer : The dependent variable is work productivity. The independent variables are frequency of truancy and length of service 3. Find the coefficients and constants of this multiple linear regression model (show the manual calculations) Answer : No. 1 2 3 4 5 6 7 8 9 10 11 Y 75 55 65 84 75 82 95 84 95 105 110 X1 7 6 7 8 7 6 8 8 9 9 10 X2 3 7 5 1 3 2 1 2 4 1 1 X12 Y2 5.625 49 3.025 36 4.225 49 7.056 64 5.625 49 6.724 36 9.025 64 7.056 64 9.025 81 11.025 81 12.100 100 X22 9 49 25 1 9 4 1 4 16 1 1 X1 . Y 525 330 455 672 525 492 760 672 855 945 1.100 X2 . Y 225 385 325 84 225 164 95 168 380 105 110 X1. X2 21 42 35 8 21 12 8 16 36 9 10 12 Total 90 1.015 8 93 1 31 8.100 88.611 64 737 A = Constant value B = Regretion coefficient value Using elimination method: b0n + b1ΣX1+ b2ΣX2 =ΣY ....................................... (1) b0ΣX1 + b1ΣX12 + b2ΣX1X2 = ΣX1Y. .................... (2) b0ΣX2 + b1ΣX2X1 + b2ΣX22= ΣX2Y. ................... (3) For example : b0 = x b1 = y b2 = z 12x + 93y + 31z = 1.015. ................................ (1) 93x + 737y + 226z = 8.051 ............................. (2) 31x + 226y + 121z = 2.356 ................ (3) Completion 1 12x + 93y + 31z = 1.015 |x93 93x + 737y + 226z = 8.051 |x12 1.116x + 8.649y + 2.883z = 94.395 1.116x + 8.844y + 2.712z = 96.612 _ -195y + 171z = -2.217 ........................ (4) Completion 2 93x + 737y + 226z = 8.051 |x31 31x + 226y + 121z = 2.356 |x93 2.883x + 22.847y + 7.006z = 249.581 2.883x + 21.018y + 11.253z = 219.108 1.829 y -4.247z = 30.473 ....... (5) 1 121 720 8.051 90 2.356 8 226 Completion 3 -195y + 171z = -2.217 |x1.829 1.829 y - 4.247z = 30.473 |x195 -356.655y + 312.759z = - 4.054.893 356.655y – 828.165z = 5.942.235 + - 515.406z = 1.887.342 z (b2) = -3,661 Completion 4 -195y + 171z = -2.217 -195y + 171(-3,661) = -2.217 -195y - 626,031 = -2.217 -195y = -2.217 + 626,031 y (b1) = 8,158 Completion 5 12x + 93y + 31z = 1.015 12x + 93(8,158) + 31(-3,661) = 1.015 12x + 758,694 – 113,491 = 1.015 12x + 645,203 = 1.015 x = 30,818 1. Formulate a functional relationship hypothesis for this case! Answer : Hypothesis: Ho: There is no significant influence between the frequency of truancy and the length of work together on work productivity. Ha: There is a significant influence between the frequency of truancy and the length of work together on work productivity. 2. Perform a partial coefficient test (t test) at α = 0.05 Answer : Determining the level of significance. The significance level uses 5%; 0.05 1. Determine t count 𝑏𝑖 𝑠𝑒(𝑏𝑖) Based on calculations using regression analysis in Ms. Excel 𝑡= SUMMARY OUTPUT Regression Statistics Multiple R 0.948404 R Square 0.899469 AdjustedR Square 0.877129 Standard Error 5.551331 Observations 12 ANOVA df SS Regression Residual Total Intercept X Variable 1 X Variable 2 𝑡𝑥1 = 8,1581 𝑡𝑥2 = −3,6619 1,6524 MS F 2 2481.561 1240.781 9 11 277.3555 30.81727 2758.917 Standard Coefficients Error 30.8181 14.5557 8.1581 1.6524 -3.6619 1.0413 40.2625 t Stat 2.1173 4.9372 -3.5166 = 4.9372 = - 3,5166 1,0413 2. Determine t table Ttable (𝛼2 , 𝑛) = 𝑡(0,25;12) = 2,179 3. Significance F Testing Criteria Ho accepted if -t table < t count < t table Ho rejected if -t count < -t table atau t count > t table P-value 0.0633 0.0008 0.0066 3.24E-05 Lower 95% -2.1091 4.4202 -6.0175 Upper 95% 63.7453 11.8960 -1.3062 4. Compare tcount with t table Based on the calculation results above, with a significance level of 5% it can be concluded that: X1 : |tcount| (4.9372) > t table (2.179), then Ho is rejected. This means that the length of service affects work productivity. X2 : |tcount| (3.1566) > t table (2.179), then Ho is rejected. This means that the frequency of truancy affects work productivity. 2. Carry out the overall model test (F test) α = 0.05 Answer : Determine the hypothesis Ho : b1 = b2 = 0 (the independent variable has no effect on the dependent variable) Ha : at least one bi ≠ 0 (the independent variable has an effect on the dependent variable) Significance Level 5% Hypothesis Testing Rules: If Fcount > F table, then Ho is rejected Test Statistics: F Test R2/(k − 1) F= (1 − R2)/(n − k) K: the number of independent variables, including the intercept n : lots of data 𝐹= = 40.2625 . 1. Conclusion: F table (k-1,n-k) = 𝐹(0,05;2;9) = 4,256 Based on the F test above, using a significance level of 5%, the calculated F value (40.2625) > F table (4.256) is obtained, so Ho is rejected. This means that the independent variable has an effect on the dependent variable. 2. Carry out a correlational test for these three variables together (R2 test) Answer : 𝑹𝟐 = (𝒃𝟏𝚺𝒙𝟏𝒚) + (𝒃𝟐𝚺𝒙𝟐𝒚) 𝚺𝒚𝟐 𝑹𝟐 = (𝟖, 𝟏𝟓𝟖𝟏)(𝟏𝟖𝟒, 𝟕𝟓) + [(−𝟑, 𝟔𝟔𝟏𝟗)(−𝟐𝟔𝟔, 𝟎𝟖𝟑𝟑) 𝟐𝟕𝟓𝟖, 𝟗𝟏𝟔𝟕 = 0,8995 3. Arrange the most appropriate regression equation (based on the calculation results) Answer : Y = 30,818 + 8,158 X1 -3,662 X2 4. Interpret the regression equation above Answer : If X1=X2=0 then Y = 30.8181 This means that if an employee has no work experience and has never been absent (X1=X2=0), then his work productivity is estimated to be worth 30.8181. If X= 1/increase by one; then Y = 35.3143 This means that when an employee has a work period of 1 and a frequency of absenteeism of 1, then his work productivity is estimated to be worth 35.3143. For each additional period of work, it is estimated that an employee's work productivity will increase by 8.1581. For each additional frequency of truancy, it is estimated that an employee's work productivity will decrease by 3.6619.