Mathematics > QUESTIONS & ANSWERS > University of the People MATH 1281 WRITTEN ASSIGNMENT UNIT 6. (All)

University of the People MATH 1281 WRITTEN ASSIGNMENT UNIT 6.

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WRITTEN ASSIGNMENT UNIT 6 SUBMISSION 1. The distribution is More symmetric. > transfusion plot(frequency~march2007,data=transfusion) > plot(log(frequency)~march2007,data=transfusion) 2. First case: The computed p-val ...

WRITTEN ASSIGNMENT UNIT 6 SUBMISSION 1. The distribution is More symmetric. > transfusion<- read.csv(file.choose(), header=T) > plot(frequency~march2007,data=transfusion) > plot(log(frequency)~march2007,data=transfusion) 2. First case: The computed p-value is = 4.174e-06.Consequently the null hypothesis is rejected at significance level > t.test(frequency~march2007,data=transfusion) Welch Two Sample t-test data: frequency by march2007 t = -4.7229, df = 216.86, p-value = 4.174e-06 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -4.246285 -1.745712 sample estimates: mean in group no mean in group yes 4.801754 7.797753 > t.test(log(frequency)~march2007,data=transfusion) Welch Two Sample t-test data: log(frequency) by march2007 t = -6.1571, df = 289.04, p-value = 2.472e-09 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -0.6316889 -0.3256582 sample estimates: mean in group no mean in group yes 1.178089 1.656762 3. 3. Mark the null hypotheses that you reject with a significance level of 5% and those that you do not reject: (Reject/Don't Reject) H0: The variance of "frequency" is the same in the two subsets, (Reject/Don't Reject) H0: The variance of "log(frequency)" is the same in the two subsets. Explain your answer: The function “var.test” is applied in order to test the null hypothesis. First case: The computed p-value is = 2.2e-16. Consequently the null hypothesis is rejected at significance level of 5%. > var.test(frequency~march2007,data=transfusion) F test to compare two variances data: frequency by march2007 F = 0.34883, num df = 569, denom df = 177, p-value < 2.2e-16 alternative hypothesis: true ratio of variances is not equal to 1 95 percent confidence interval: 0.2725525 0.4397267 sample estimates: ratio of variances 0.3488348 Second case: The computed p-value is: 0.6249. Consequently the null hypothesis is not rejected at significance level of 5%. > var.test(log(frequency)~march2007,data=transfusion) F test to compare two variances data: log(frequency) by march2007 F = 0.9449, num df = 569, denom df = 177, p-value = 0.6249 alternative hypothesis: true ratio of variances is not equal to 1 95 percent confidence in

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