Uvtize the appropriate ttest to Utilize the appropriate t test to answer the fol

Question

Uvtize the appropriate ttest to Utilize the appropriate t test to answer the following questions for each experiment described below. Does the measured data suggest that a significant difference exists between the means of the samples for a values to 0.05, 0.10 and 0.15. If not, identify which two sets of sample data have significant differences between their respective means for each value. (Sample One versus Sample Two at values-to 0.05, 0.10 and 0.15, Sample Two versus Sample Three at values = to 0.05, 0.10 and 0.15, etc.) For each comparison, identify the hypothesis you are testing and provide the probability of them being significantly different and the actual value of the test statistic that you chose to utilize for each of the three levels. Organize your results in a table for each experiment and attach to this sheet.

Explanation / Answer

we go for t-test for testing of difference of two sample means

t=(mean1-mean2)/(sp/n1/2) and sp2=((n1-1)s12+(n2-1)s22)/n and n=n1+n2-2

sample 1 Vs sample 2

null hypothesis: there is no difference in mean value of sample 2 and Sample 3

alternative hypothesis : theis is difference in mean value of sample 2 and Sample 3

sp2=(11*0.05633+11*0.0407)/22=0.0485

sp=0.2203

t=(1.6825-1.69)/(0.0485/sqrt(22))=-0.7253

t(.05,22)=2.07 , t(0.1,22)=1.72, t(0.15,22)=1.49

since calculate t is less than the critical t-value at 22 df for each of alpha=0.05,0.1,0.15

so significant difference doesnot exizt between the sample 1 and sample 2

sample 2 Vs Sample 3

null hypothesis: there is no difference in mean value of sample 2 and Sample 3

alternative hypothesis : theis is difference in mean value of sample 2 and Sample 3

sp2=(11*0.0407+12*0.2235)/23=0.1361

sp=0.3689

t= (1.69-2.18)/(0.3689*sqrt(23))=-6.39

critical t at different value of alpha

t(.05,23)=2.07 , t(0.1,23)=1.71, t(0.15,23)=1.49

so difference of sample2 and sample 3 is significant at alpha=0.05.0.1,0.15 as the absolute value of calculated t is more than critical t.

age 3=sample 1 age 7 =sample 2 age 28=sample 3 1.69 1.82 2.76 1.69 1.86 2.6 1.97 1.72 2.38 2.09 1.73 2.06 1.81 1.7 1.81 1.53 1.44 2.76 1.63 2 2.41 1.7 1.78 2.29 1.73 1.47 2 1.72 1.32 2.15 1.48 1.87 0.97 1.15 1.57 1.91 2.26 n= 12 12 13 mean= 1.6825 1.69 2.181538462 s= 0.237338462 0.201629724 0.472772781 s2= 0.056329545 0.040654545 0.223514103

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.