Non Parametric and Chi-Square Distribution

Individual Paper 3 Non parametric and Chi-squargon distribution apprize Summary I worked for a logistic company. My major responsibility was in charge of the storage and transportation of parts of cars betwixt two atomic design 18as, which are about 1400 miles apart. One of my jobs is collecting the goods from suppliers and arranging the trucks to deliver them. There are five truck drivers, and each of them is assigned to deliver on each weekday throughout a whole year. Before the rake, we will check the superior of the goods and Make sure that there are no damage goods.When arriving at the destination, the staff will check the goods again and record the modify goods that occurred in transit. At the end of every month, we will pay for the compensation correspond to the number of the defective goods. In order to reduce the number of the dishonored goods during the delivery, I want to identify the reasons why they are damaged. In this study, I want to find out that whether s ome drivers are more flat to make the goods damaged during their delivery.Variable to be measured Two variables are to be measured. The first variable is just the five truck drivers, and the second mavin is the quality of the goods after the delivery. Determination of Population Population in this discipline is defined as the all goods delivered from Tianjin area to Guangzhou area. Statistical method To read relationship between the two variables above which are both tokenish in terms of info type, I decide to use Chi-squared streak of a contingency table. Sample SelectionThe information about delivery is recorded in our computer system, including the delivery date, name of the driver, the number of damaged goods and so on. I take out the data about 52 weeks during the previous year and record them into the following table Quality transport Driver Passed Damaged come in Driver A Driver B Driver C Driver D Driver E Total Hypothesis The objective is to des cribe whether there is a relationship between the five drivers and the number of damaged goods.The null hypothesis will touch on that there is no relationship between the two variables H0 The two variables are independent The alternative hypothesis specifies one variable affects the other, expressed as H1 The two variables are dependent Use the formula ?2=i=1k(fi-ei)2ei ? =(r-1) (c-1) which calculates the test statistic. Or use the Excel by importing the data into the function of Data Analysis Plus, Contingency Table. Then I could bear chi-squared Stat, p-value. The number of degrees of freedom v=(r-1)(c-1)=(5-1)(2-1)=4.If I employ a 5% significance level, the rejection region is X2 X2 a, v = X2 . 05, 4 = 9. 49 Comparing the results, if the p-value is great than 9. 49, there is not enough evidence to infer that there is a relationship between the five drivers and the number of damaged goods if the p-value is not great than 9. 49, I can reject the null hypothesis in respect of the alternative, which means there is a relationship between the five drivers and the number of damaged goods. So I can reduce the number of damaged goods through improving the drivers conditions.