This record offers the research findings of the position of knowledge management system in helping students. The research is based on a sample of 480 college students from a college in Victoria between year 2 and yr 12 of study.

The following table represents the descriptive facts for three variables in the study.

Mark

Mark

GradeID

raisedhands

VisITedResources

AnnouncementsView

Discussion

Mean

76.73333

5.6

46.775

54.79792

37.91875

43.28333

Standard Error

0.800807

0.129657

1.404873

1.509889

1.214632

1.261484

Median

80

7

50

65

33

39

Mode

92

2

10

80

12

70

Standard Deviation

17.5448

2.840653

30.77922

33.08001

26.61124

27.63774

Sample Variance

307.8202

8.069311

0.1296571.4048731.5098891.2146321.261484Median80750653339Mode92210801270Standard Deviation17.54482.84065330.7792233.0800126.6112427.63774Sample Variance307.82028.069311947.36051094.287708.1583763.8444Kurtosis-0.52198-1.04107-1.497-1.4852-1.00224-1.12656Skewness-0.704730.0269770.026962-0.342440.3992430.362594Range6510100999898Minimum3520001Maximum10012100999899Sum36832268822452263031820120776Count480480480480480480The mean mark is 76.73 with a standard deviation of 17.54. The kurtosis for marks is -0.52 and the skewness is -0.70. The mean grade ID is 5.6 with a standard deviation of 2.84. The Kurtosis for the distribution of GradeID is -1.04 while the skewness is 0.02. The average value for the variable “raisedhands” is 46.78 with a standard deviation of 30.78. The kurtosis for “raisedhands” is -1.50 while the skewness is 0.03. The variables “mark” and “GradeID” are skewed to the left while “raisedhands” is skewed to the right. The three variables have a lower peak than the standard normal distribution. However, the peakedness and the skewness of the three variables is not significant and therefore, the three variables are normally distributed.The above descriptive statistics would not be appropriate for the categorical variables like female and math. However, the statistics would show the composition of the dummy variables. For example the variable, Female is coded for 1 to represent female and 0 to represent male. Therefore, the mean value for the variable would lie between 0 and 1. The median value would be 0.5. However, if the females are greater than the number of males, the mean value would be greater than 0.5. On the other hand, if the males are greater than the number of females, the mean value would be less than 0.5.The histogram below shows the distribution of students’ marks.From the above histogram, we can conclude that students' marks are negatively skewed. The distribution of the variable seems to follow the normal distributionThe figure below represents the boxplot for the distribution of the number of times that students raised hands.The above boxplot shows that the data for the number of times that students raised hands is negatively distributed. However, the data do not have outliers.D. The contingency table below represents the distribution of students’ marks between the mathematics class and those who did not take mathematics.Row Labels01Grand TotalGreater28271353Less8245127Grand Total364116480Greater represents the students whose marks were at least 70 while less represents the number of students whose marks were less than 70.Therefore, the probability of selecting a student whose mark is at least 70 randomly is shown below.Probability = 353/480 = 0.7354 or 73.54%Further, we can determine the likelihood of a student enrolled in maths having a mark of at least 70 as shown below.Probability = 71/116 = 0.6121 or 61.21%The probability of a random student scoring at least 70 is different from the probability of a random student enrolled in math scoring at least 70. Therefore, a student mark is not independent of whether they are enrolled in math class.E. We wish to construct a 95% confidence interval for the mean number of times that male and female students raised hands. The table below shows the descriptive statistics for male and female students. maleFemaleMean43.2819752.86286Standard Error1.7522752.28427Median3960Mode1070Standard Deviation30.6021730.21805Sample Variance936.4926913.1305Kurtosis-1.50072-1.35243Skewness0.177319-0.23274Range100100Minimum00Maximum100100Sum132019251Count305175Confidence Level(95.0%)3.4481234.508444The mean number of times that male students raised hands is 43.28 with a standard deviation of 30.60. The mean number of times that female students raised hands is 52.86 with a standard deviation of 30.22. The confidence interval for the male mean is 3.45 while the confidence interval for the female mean is 4.51. The following confidence intervals can, therefore, be constructed for male and female students. Confidence intervalsMale: 43.2820 + 3.4481 = 39.8339 < Mean < 46.7301Female: 52.8629 + 4.5084 = 48.3545 < Mean < 57.3713The confidence interval for the male mean does not intersect with the confidence interval for the female mean. Therefore, female students raised their hands more times than male students.F. It is claimed that students who take religion have marks that are greater than 70. We wish to test whether the mean mark for the students who took religion is significantly greater than 70 at 5% significance level. An independent sample t-test will be carried out to determine the significance of the difference. The table below shows the results of the sample t test.t-Test: Two-Sample Assuming Unequal Variances Marktest valueMean76.9090970Variance313.61040Observations2222Hypothesized Mean Difference0df21t Stat1.829941P(T<=t) one-tail0.040746t Critical one-tail1.720743P(T<=t) two-tail0.081492t Critical two-tail2.079614 The mean mark for the students who took religion is 76.91 with a variance of 313.61. The test statistic, t = 1.8299 is significant because the p-value, 0.04 is less than the level of significance. Therefore, the mean mark for the students who took religion is significantly greater than 70. This confirms the school administrator’s claim that students choose religion because they believe it is a subject that they can obtain a mark greater than 70.G. We are to carry out a multiple linear regression analysis using “mark” as the dependent variable and the other variables as the independent variables. The table below shows the results of the multiple linear regression analysis. CoefficientsStandard Errort StatP-valueLower 95%Upper 95%Intercept44.83863.61175412.414638.5E-3137.7413751.93582GradeID0.0612270.2182150.2805810.779156-0.367570.490026raisedhands0.138720.0284364.878351.47E-060.0828430.194598VisITedResources0.2108150.025588.2412321.72E-150.1605480.261081AnnouncementsView0.0917140.0319262.87270.0042540.0289780.15445Discussion0.0367920.0236181.5577550.119966-0.009620.083203Female4.9567041.2236954.0506055.98E-052.5521017.361307Math8.95913.2453082.7606310.0059952.58195615.33624Science4.0570733.0841591.3154550.189-2.0034110.11755Language8.0650483.0808582.6177920.0091362.01105314.11904Religion2.5894623.9455470.65630.511953-5.1636810.3426The following regression equation can be obtained from the above results.Marks = 44.8386 + 0.0612(GradeID) + 0.1387(raisedhands) + 0.2108(VisiteDResources) + 0.0917(AnnouncementsView) + 0.0368(Discussion) + 4.9567(Female) + 8.9591(math) + 4.0571(Science) + 8.0650(language) + 2.5895(Religion)H. We are to determine the significance of Raised Hands in predicting Marks in the model. We shall use both the critical value method and the p-value method to determine the significance. The following hypothesis will be used to analyze the significance of the predictor.Null hypothesis: The coefficient for raised hands is not different from zero. Alternative hypothesis: The coefficient for raised hands is different from zero.Critical value methodWe first determine the critical t-value at 5% significance level and 469 degrees of freedom.Critical value = 1.965Calculated t-statistic = 4.87835We reject the null hypothesis because the calculated t-statistic is greater than the critical value of t. Therefore, the coefficient for raised hands is different from zero.P-value methodUsing the p-value method, we shall compare the p-value to the significance level. The significance level is 0.05 while the p-value is 1.47E-06. We reject the null hypothesis because the p-value is less than the level of significance. Therefore, the coefficient for raised hands is different from zero.The coefficient for raised hands is 0.13872. This implies that for every ten times that a student raised hands in class, the mean mark increased by 1.3872.We are to interpret the slopes of the independent variables.Grade ID: The coefficient for grade ID implies that a unit increase in the grade of a student in primary or secondary would lead to an increase in the mark by 0.0612 units.VisiTedResources: The coefficient for visited resources implies that a unit increase in the number of times that a student visited resources in blackboard increased the student mark by 0.2108 units.Announcements view: The coefficient for announcements view indicates that a unit increase in the number of announcements viewed on blackboard increased the student mark by 0.0917.Discussion: The coefficient for discussion implies that a unit increase in comments on the discussion board increases the student marks by 0.0368.Female: The coefficient for female implies that female students obtained 4.9567 marks higher than male students.Math: The coefficient for math implies that students who took mathematics obtained 8.9591 marks higher than those who did not take mathematics.Science: The coefficient for science implies that students who took science obtained 4.0571 marks higher than those who did not take science.Language: The coefficient for language implies that students who took language subject obtained 8.0650 marks higher than those who did not take a language subject. Religion: The coefficient for religion implies that students who took religion obtained 2.5895 marks higher than those who did not take religion.All coefficients except for female, math, and science had positive signs according to my expectation. I expected math and science to have a negative sign because most students score lower marks in these subjects. I expected the sign for female to be positive or negative. However, I expected the coefficient to be closer to zero to indicate low differences in male and female marks.J. We are to interpret the value of adjusted R2The value of R2 is 0.50843 while the value of adjusted R2 is 0.497948. This implies that 49.79% of the variation in the dependent variable is explained by the variations in the independent variables.K. We will test for the significance of the regression model using ANOVA. The table below shows the results of the ANOVA analysis.ANOVA dfSSMSFSignificance FRegression1074965.857496.58548.508524.24E-66Residual46972480.02154.5416Total479147445.9 The test statistic, F = 48.5085 is significant because the p-value, 4.24E-66is less than the level of significance. Therefore, the regression model is a significant predictor of the dependent variable.L. We will determine the significant predictors of student marks. Other than raised hands, the other significant predictors of the dependent variable are: visited resources, announcements view, female, math and language. Visited resources and announcements view have positive relationships with the dependent variable. An increase in the predictors results in an increase in students' marks. The coefficient for female, math, and language are positive. This implies that female students, as well as students who took mathematics and language subjects, obtained higher marks than male students and students who did not take mathematics and language subjects.M. We are required to predict the average marks of a student in Year 6 who has raised their hand 35 times, visited 40 resources, looked at 75 announcements, participated in 5 discussions, is a Female in a Math course. We will use the regression equation and assume that the other dummy variables are zero.Marks = Marks = 44.8386 + 0.0612(6) + 0.1387(35) + 0.2108(40) + 0.0917(75) + 0.0368(5) + 4.9567(1) + 8.9591(1) + 4.0571(0) + 8.0650(0) + 2.5895(0)Marks = 44.8386 + 0.3672 + 4.8545 + 8.432 + 6.8775 + 0.184 + 4.9567 + 8.9591 + 0 = 79.4696It is appropriate to predict the students’ average mark using the regression equation because the regression model is significant. Hence, the independent variables significantly predict the dependent variable.N. We wish to test for the linearity and normality of the distribution of the quantitative variables. The figure below shows the relationship between the dependent variable and the independent variables.The table above shows a linear relationship between the quantitative independent variables and the dependent variable. Therefore, the data satisfies the assumption of linearity.The table below shows the distribution of the dependent variable.The P-P plot above shows that the dependent variable is normally distributed. The figure below shows the distribution of residual values.The distance between the plots and the line through the plots is fairly consistent for the graph. This shows that the data has equal error variances and therefore homoscedastic.O. The data provides sufficient information about the true distribution of the population of students because it is sufficient to reject the null hypothesis. Therefore, the sample size is sufficient to identify significant predictors in the data collected.P. We are to determine the binomial distribution of female students’ marks. The marks are divided into two; less than 90 (less) and more or equal to 90 (more). The table below shows the distribution of the female students’ marks.MathsRow Labels01Grand TotalMale22679305less17167238more551267Female13837175less7228100more66975Grand Total364116480There were 175 female students in the sample. 37 female students took mathematics subject. 28 scored less than 90 while 9 scored at least 90.The probability that none of the female students scored at least 90 = 28/37 = 0.7568Probability of female students scoring at least 90 = 9/37 = 0.2432The probability of all girls scoring at least 90 = px. qn-x = 0.24325*0.75680 = 0.00085 or 8.51*10-4We will compare the above probabilities with those for male students. There are 305 male students in the sample. 79 male students took mathematics. 67 of the students scored less than 90 in mathematics and at 12 scored at least 90 in mathematics.The probability that none of the female students scored at least 90 = 67/79 = 0.8481Probability of female students scoring at least 90 = 12/79 = 0.1519Probability of all boys scoring at least 90 = px. qn-x = 0.15195*0.75680 = 0.000081 or 8.09*10-5Female students have a higher probability of scoring at least 90 in mathematics than male students.

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