Weare presents with the following information in which we are presumed to answer.We should use chi square test program to come up with numerous hypotheticalconclusions.TheNortheast region is operating the software sold 165 and the group with nosoftware sold 100, The Southeast region with software sold 200 and the groupwith no software sold 125. The Central regionwith software sold 175 and the group with no software sold 125.
The West groupwith software sold 180 and the group with no software sold 130.Softwareprogram is categorical variable which can have 2 possibilities either Yes, andor No. this consider the cases whether the software is assigned or notassigned. Performance is again a categorical variable which can have 2possibilities either Improved or not improved. Contingency Table is shown below. Further calculations With Software Without Software Row Totals Northeast Region 165 (159.00) 0.
23 100 (106.00) 0.34 265 Southeast Region 200 (195.00) 0.
13 125 (130.00) 0.19 325 Central Region 175 (180.
00) 0.14 125 (120.00) 0.21 300 West Region 180 (186.00) 0.19 130 (124.00) 0.
29 310 Column Total 720 480 1200 (Grand Total) In my Chi-SquareCalculation, the compiled the data set in contingency table. Using theChi-Square method, the numbers calculate as expected, differences in all theregions of the sales group sold. The significance level of operation is: 0.05. Thecalculations gives the chi-square statistical figure of is 1.
7176, the degreeof freedom =3, the probability, p-value is 0.633018. Accordingly, the final developmentis not significant at p < 0 .05. This variables are independent. (G F,2014).VPof Sales at WidgeCorp, would like a trial of whether these two variables areindependent or not. A nonparametric test using Chi square test is used to testthe following hypothesis.
NullHypothesis Ho: the two variables which are the Software Program andPerformance are independent.AlternativeHypothesis Ha: the two variables which are the Software Program andPerformance are not independent.Whenthe conclusion is statistically significant (p-value less than 0.05) it isshown that the alternative hypothesis is; otherwise (p-value greater than0.05), the null is chosen.Therelationship is not statistically significant as a result of the p-value,0.633, is greater than 0.05.
Accordingly, it is apparent that it is in favor ofthe null hypothesis – the two variables which are Software Program andPerformance are independent.The VP of Sales atWidgeCorp, knows quite a bit about statistics, would like to know feasible nulland alternative hypotheses for a nonparametric test on this data using thechi-square distribution. A nonparametric test is utilized on information thatare qualitative or categorical. This is predominately is utilized when theinformation is does not give the right perception when observed at the mean ofsuch variables.When looking problem, thehypothesis shows that the sales estimates are generated from the four regionsare basically random and this is known as the null hypothesis. This also showsthat the null hypothesis is accurate if the observed data sales for everyregion did not change from what was originally expected.
To know this is onlyexpected by chance. What usually correlates the null hypothesis is known as thealternative hypothesis. The null hypothesis is shown as H0 and the alternativehypothesis as H1. (C. Z.
2013).The significance level isusually the acceptable level of type I error, it is usually given as ?. Most ofthe time, a significance level of ? = 0 .05 is used. The significance levelform part one is: less than 0.05. The P-value (that is the probability value)is the statistic value used to test the null hypothesis.
If p < ? then we reject the nullhypothesis. Here it is not. 0.633018 is greater than 0.05. Therefore; H0: p =<0.
5, Ha: p <> 0.5Since P is not less than?; hence it is reasonable the state to the VP of Sales at WidgeCorp that thenull hypothesis is accurate and should be accepted. And hence state that weshould take the other correlation of the Alternative Hypothesis.