Answering Mathematics Exam Questions - Essay Example

Answers to the Test Here you need to go to excel, click on the office button, go to excel options, click add-ins, and patiently wait for the Analysis ToolPack to install after you prompt it. Once you have that set, you can now proceed to the Data tab on the excel menu, and select the Data Analysis option (am using 2007 version, 2003 will vary slightly but 2010 has almost similar qualities to 2007). Once you select the Data Analysis tab, you will get various options to scroll down; for the analysis here, select ANOVA Single Factor Analysis. The output will be large, but you don’t really need the frequency bit, concentrate on the more refined bit, like I have copied here. Take interest in the F-stat and P-value. They are the only bits you need to validate any investigation (hypothesis). If asked to calculate physically, F=Mean sum of squares (between groups)/ MS (within groups). You realize that for this question we treat the Variation within groups as the ‘error group’, thus Error Sum of Squares (SSE) where we have SS (within groups) and Mean Error sum of squares (MSE).Now when you are required to use F-Stat to validate your test, you got to have two values of it. The first one you compute as in the case shown in the table below (the ANOVA). It is called the computed value of F or F-Computed (you went the long ANOVA way to compute it). To get the second one it is really easy, especially when you’ve got the table values. Take the degrees of freedom in the row you want to test AGAINST THE TOTAL, very important, and bear in mind the level of significance. For example here we shall have F (3, 79, 0.05). They flow in the order Numerator, Denominator and Level of Significance (LoS). You call this the F-tabulated (derived from the tables). If F-computed>F-tabulated, reject the H0. The P-Value works slightly different. If the P-Value is much less than the Level of Significance, you reject the null hypothesis. PLEASE NOTE THE DIFFERENCE WELL. YOU MAY WRITE IT DOWN FOR NOW, THEN AFTER A FEW APPLICATIONS YOU WILL BE ABLE TO USE THEM SUMMARY Groups Count Sum Average Variance Column 1 20 4875 243.75 145.7763 Column 2 20 4499 224.95 181.7342 Column 3 20 5094 254.7 164.2211 Column 4 20 5057 252.85 159.8184 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 11115.24 3 3705.079 22.74625 1.31E-10 2.724944 Within Groups 12379.45 76 162.8875 Total 23494.69 79         a) F-Statistic= 22.74625 P-Value= 1.31?10-10 b) Let ?i denote the average service time for outlet i. H0: ?1= ?2= ?3= ?4 In order to test the hypothesis on means, the null hypothesis becomes the proposition that all means are equal. Thus by default, anything different will form the alternative hypothesis. c) Yes I would reject the null hypothesis at 5% Level of Significance. This is because the P-Value is smaller than the Level of significance. In order to answer this question, you either refer to the P-value or the F-statistic, when you have both it is your choice. Your Professor was good enough to facilitate your having both. Remember that if the P-value is smaller than the LoS, you reject H0; and if F-computed is greater than F-tabulated, you reject. THAT IMPLIES THAT P-VALUEF-TABULATED. IN BOTH CASES YOU REJECT THE NULL HYPOTHESIS. 1.31?10-10 Read More