Principles of statistics - Essay Example

PRINCIPLES OF STATISTICS QUESTION ONE In this question we sought to find out the relationship between the dependent variable (FACTOR) and the explanatory variables (AGE, GENDER, YRSLIC and MTSLIC). Clearly from the table below we can observe that all the explanatory variables except gender have a negative correlation with the FACTOR. This means that, any unit increase in the three variables (AGE, YRSLIC and MTSLIC) results to a decrease in the value of the FACTOR by the factors indicated under the column coef.

For instance, a unit increase in age results to a decrease in FACTOR by -0.00232. However, unit increase in GENDER i.e. moving from female to male results to an increase in the FACTOR by 0.3132.Regression Analysis: FACTOR versus AGE, GENDER, YRSLIC, MTSLIC Table 1: Regression AnalysisThe regression equation isFACTOR = 2.89 - 0.0023 AGE + 0.313 GENDER - 0.0007 YRSLIC - 0.0476 MTSLICPredictor Coef SE Coef T PConstant 2.8925 0.5453 5.30 0.000AGE -0.00232 0.02738 -0.08 0.933GENDER 0.3132 0.1988 1.58 0.116YRSLIC -0.00068 0.02925 -0.02 0.981MTSLIC -0.04758 0.05200 -0.91 0.361S = 1.

69969 R-Sq = 1.1% R-Sq(adj) = 0.0%Analysis of VarianceSource DF SS MS F PRegression 4 9.287 2.322 0.80 0.524Residual Error 295 852.243 2.889Total 299 861.530In figure 1 below, we present a line graph factor against age for both male and female. The line plot shows a variation in FACTOR for the different ages based on gender.QUESTION TWOIn this question we seek to determine the distribution of the given data, from the table below, the very first line shows our data are definitely normally distributed, because the p-value for Normal is greater than 0.005.Table 2: Distribution ID Plot for no_of_weeks Descriptive StatisticsN N* Mean StDev Median Minimum Maximum Skewness Kurtosis5 0 22.4 13.8311 29 7 36 -0.

464644 -3.08671Box-Cox transformation: Lambda = 0.5Goodness of Fit TestDistribution AD P LRT PNormal 0.480 0.120Box-Cox Transformation 0.531 0.086Lognormal 0.573 0.0643-Parameter Lognormal 0.794 * 0.889Exponential 0.627 0.2762-Parameter Exponential 0.556 0.152 0.218Weibull 0.684 0.0553-Parameter Weibull 0.588 0.069 0.755Smallest Extreme Value 0.514 0.171Largest Extreme Value 0.621 0.

084Gamma 0.657 0.094The graph below (Figure 2) suggests that the given data follows a normal distributionThe Chi-Square Goodness-of-Fit Test can tell you whether the proportions of all the groups are equal or whether the proportions of each category are equal to specific values. For example:Table 3: Goodness-of-Fit Test for Poisson DistributionData column: no_of_weeksFrequency column: no_of_risesPoisson mean for no_of_weeks = 22.0667 Poisson Contributionno_of_weeks Observed Probability Expected to Chi-Sq=32 5 0.

027476 0.4121 51.0714 N N* DF Chi-Sq P-Value15 0 1 52.7668 0.000Probability plotsQUESTION THREEIn this question we sought to find out if there exists a significant difference e in the strength between a new brand and an old brand of concrete. From the table 3 below, p-value=0.018