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英国paper代写范文-Business statistics coursework assignment

2017-07-17 17:21:30 | 日記
本篇英国paper代写范文-Business statistics coursework assignment将利用课堂学习的统计知识分析两个不同网站上的数据。第一部分包含英国土地注册处网站关于林肯郡房价的数据。直方图将用于显示数据的分布和分散。那么平均值,标准差,最大值和最小值也将被提出。第二部分分析电影发行商协会网站的票房数据。运用回归来探索变量“周末总计”和“发布周”之间的关系。本篇paper代写由51due英国代写平台整理,供大家参考阅读。

Introduction
This report will utilize the statistical knowledge learned in class to analyse the data on two different websites. The first section contains data from the UK Land Registry Website about house prices in Lincolnshire. A histogram will be used to show the distribution and dispersion of the data. Then the average, standard deviation, maximum and minimum values will also be presented. The second section analyses the box-office data from the Film Distributor Associations website. A regression is run to explore the relationship between variables “Weekend total” and “Weeks on release”.
Section A: Land Registry
A1 Cost benefit analysis of open data
Open data can bring more transparency for how the government is performing. If anyone has any doubts on what the government did, they can simply go online to find open data. Second, it is easier for students and professors to find data for academic research. Third, open data can help firms such as McKinsey consulting to utilize and analyse these data. However, there are also costs associated with free data because those people who worked hard to gather these data may not be paid as much as when the data is accessed with a fee.
A2 Price paid data analysis
Figure 1
As shown in Figure 1, the mean is found using the “average” function in Excel. The median is found using the “Median” function in Excel. The variance is found using “VAR.S.” in Excel because not all of the data is used and only the sample from Lincolnshire is included. The standard deviation is the square root of variance. The kurtosis is found using the “Kurtosis” function in Excel.
Figure 2
Figure 3
From Figure 2, we can conclude that the data is mostly centred on the 22,000 to 648,000 range. The data is skewed to the right. The mean is 182,354 while the median is 154,000. When the mean is greater than the median, it means that the data is skewed to the right. Figure 3 is a partial one and only included data from 22,000 to 648,000 since that is the area where the data is most concentrated and I broke it down to further analyse the data and see the distribution. From the second histogram, we can see that the data is also skewed to the right. The excess kurtosis is positive and the data is leptokurtic.
A3 Highest and lowest prices
The table with the five lowest prices in Lincolnshire is shown in Figure 4:
Figure 4
The table with the five highest prices in Lincolnshire is shown in Figure 5:
Figure 5
Section B: Film Box-office data analysis
B1 Description of the data
The time series chart for the four years is shown in the line graph in Figure 6.
Figure 6

The percentage changes from year to year is shown in the table below. The data is calculated using this year’s data dividing by last year’s data with the corresponding month. A number greater than 100% indicates that the data increased from last year and a number lower than 100% suggests that the data decreased from last year.
Figure 7
From Figure 6 and 7, we can see that the admissions data for 2015 was generally higher than the 2014 data for the same month. The 2014 data was generally lower than that of 2013’s. There is no apparent increasing or decreasing trend for the data from 2012 to 2013 if one looks the data by month. However, the total showed that the admissions decreased from 2013 to 2012, and from 2014 to 2013. Only in 2015 to 2014 did the admissions data increase. There is seasonality because the data is lowest for September in all four years. August, December, and January are months with the higher admissions. This pattern is fairly consistent over the past four years.
B2 Regression analysis
The scatter plot, Figure 9, shows that the two variables have a negative relationship, though not very linear.
Regression analysis
1.Linear
The first regression has “weekends on release” on the y axis and “Weekend total” on the x axis. The result of the linear model is presented in Figure 10.
Weekends on release=b + weekend total
Figure 10
The fact that the R squared is only 0.015 and that the p values are not significant means that the linear model is not a good fit for this data.
2.X variable Log model
The second model is the log transformation of “weekend total”, shown in Figure 11:
Weekends on release= b + ln (weekend total)
Figure 11
The result is better than the linear model because R square included to 0.09 and the p value lowered. However, since the p value is still not significant, this is still not the appropriate model.
3.Y variable log model
The best fitting model found is by transforming the dependent variables into log forms and since one of them is negative and only a positive number has a corresponding log form, that negative number was deleted. The result is presented in Figure 12.
Ln (Weekends on release) = b + weekend total
The R squared is 0.908, a very high one, and the p value for both the intercept and the x variable are significant at the 1% level. The scatter plot for this regression is shown in Figure 13:

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