Input Data Analysis - Assignment Example

Input Data Analysis [Student Name] [Course Title] [Instructor Name] [Date] Table of Contents Introduction 4 Matlab implementation 4 Conclusion 10 Works cited 11 Appendix 12 List of figures Figure 1: q-q plot 5 Figure 2: normal distribution curve 6 Figure 3: scatter diagram 7 Figure 4: comparison of theoretical and function 9 Introduction Matlab can be used to perform many functions in mathematics and engineering. It can be used to calculate the mean, standard deviation as well as carry out tests of normality. This paper is going to carry out a number of functions that average of data, standard deviation and chi-square test. Various plots will be made while carrying out those functions. Matlab implementation In order to open the file saved as data.csv matlab function the csvread (`filename`) is opened and saved in the file called Implementation. The data file was in the desktop so it was opened as m=csvread('D:\Documents and Settings\Administrator\desktop\Data.csv'); When the data was saved it was used to calculate the mean and standard deviation using the following functions k1=sum(m)/50 s1 = 9.8365 This was verified using the inbuilt functions s=std(m); s = 0.8143 >> k=mean(m) k = 9.8365 From the results obtained using the own custom functions, it could be suggested that the standard deviations and mean are close to the mean value and standard deviation obtained functions built into MATLAB. The reason for using functions built into MATLAB is that it improves and verify the findings of the statistical values. The standard deviations are 0.8143 on either side of the mean and mean is 9.8365. Furthermore, by observing the values of the variable provided it could be concluded that the data is normally distributed as majority of the values lie within two standard deviations of the mean value. This is further tested by plotting q-q plot which a diagram, data quartiles and standard normal quartiles are two variables to be displayed in the form of a collection of numerous points, where each point is comprised of the values of two variables. A q-q plot actually suggests the type of correlation between two variables. This is done using using the following function qqplot(m). the output is as follows; Figure 1; qqplot(m) Figure 1: q-q plot From the plot above it can be concluded that the data is normally distributed that is normal probability distribution. The results show that the histograms of data are bell-shaped normal distributions whereas that of data are not symetric as shown below; Figure 2: normal distribution curve Therefore, this probability distribution will be used to approximately the same number of elements of another data. This is done using the following Matlab command R=k+s.*randn(50,1) data generated is as follows 10.7143, 10.0874, 10.0805, 9.6759, 9.7172, 9.7525, 7.5572, 10.1567, 10.6428, 8.7798, 8.5971, 10.3422, 8.6089, 8.4689, 10.4789, 9.3037, 10.8502, 8.7842, 9.3362, 10.0333, 10.2838, 10.2173, 9.9924, 9.6493, 9.3648, 10.2277, 9.5215, 10.1798, 10.7222, 8.0048 , 11.3059, 9.3218, 10.9085, 11.1000, 8.6403, 9.9806, 12.6591, 9.6617, 10.2325, 10.1059, 10.8689, 10.7245, 9.0658, 9.4794, 10.1160, 9.7889, 11.9007, 10.1937, 10.1927, 9.1543 Bearing in mind the two sets of data that is data provided and data generated we can can carry out complete a regression analysis as well as determine the equation of the line using own functions; format long b1=x/y; this will provide x coefficient that is required in the equation. ycal=b1*x; figure(2) scatter(x,y) hold on plot(x,ycal) xlabel('data'); ylabel('generated data') y= -0.15x + 11.42 Comparing this value (1) with the value provide generated by inbuilt matlab function, it can be observed that there is a difference between both values. From the matlab function, the value of y=0.843x+9.8365 less than the value obtained using the regression equation created. This difference can be explained on the basis of the ability of the regression to predict the variations assessed by the value of coefficient of determination to predict the total variations in the values of variables. The plot is shown below; Figure 3: scatter diagram In sample testing has suggested that the regressions pertaining to data variability results in attractive outcomes in terms of the downbeat relationships amongst mean and unexpected standard deviation. The predictions relating to presentation of such provisional and changing variation models have been related to the more recent asymmetrical models The findings reveal that the regression model that is characteristic of fat tailed density tends to improve over all estimation in the measurement of conditional variation. Additional objectives Chi-Square function in MATLAB The function in Matlab is as follows; In built function which is h=chi2gof(x) produces the same results as the function created. The answer is 0. This means the value calculated is the sa as the value from the other function at 95% level of confidence. This is clear that the two functions arrive at the same answer. Figure 4: comparison of theoretical and function Chi-Square analysis of the data This will look at the best fit line, comparison of our best fit line and theoretical model and the goodness of that fit, and In our case we will use the chi-square goodness-of-fit test to test data whether it is it is normally distributed. Therefore the data null hypotheses are H0: the variable is normally distributed. H1: the variable is not normally distributed. Then at this stage we set . This is significant level which we will use to test the data. Reject H0 if Fcalc < F crit Do not reject H0 if Fcalc ≥ F crit In analysis data we first determine Degree of Freedom (df) = n-1, where n= the number of classifications of data, therefore, n=1. The Chi Square demonstrates the expected for the data is .0. Critical value at 0.05 or 95% level of significance Reject H0 if Fcalc < F crit (0) Do not reject H0 if Fcalc ≥ F crit (0) Comparing Since F calc =F crit (0 =0), we accept H0. Then we calculate the value. The calculated value is 0 while from the table the value is 0. The C.V. with .f. = 0 and is 0, so the null hypothesis is not rejected. There is sufficient evidence from the sample, which suggest that the data is normally distributed. It means that we accept null hypothesis and reject alternative hypothesis at the 5% level of significance. Conclusion It is felt that the test used in this paper proves to be an efficient determinant of for the normality of the data. Firstly, it is well recognized that the estimate pertaining to the standard normal distribution numbers is more specific with the enhancement of sampling frequencies. The findings of the data reveal mean is 9.8365 and standard deviation is 0.365. The study also made use of regression to determine skewness, and variance data. This reflects that for all these variables the normality has been violated Works cited Mathworks. Linear Regression. 30 Jan. 2010. web 12April 2016 (http://www.mathworks.com/help/matlab/data_analysis/linear-regression.html) Gonzalez RC & Woods, RE, 2008, ‘Digital Image Processing, Instructor's Manual’ New York: Prentice Hall. Gonzalez, RC, Woods, RE and Eddins, SL, 2009, Digital Image Processing Using MatLab’, New York: Prentice Hall. Appendix Read More