![e Computethe t static for these data [2 marks] Based on your t statistic, what would the researcher conclude about whether de](http://img.homeworklib.com/images/fc0b0391-2edd-4c75-86c7-75633b1a0d85.png?x-oss-process=image/resize,w_560)
(e) the t-statistic is =10.0275
(f) yes, deep brain stimulation is more beneficial
as the one-tailed p-value is less than the typical level of significance alpha=0.05, so we reject null hypothesis and concolcude that deep brain stimulation is more beneficial
following information has been generated using ms-excel
| n | 9 | 9 |
| sum(x) | 45 | 54 |
| sum(x2) | 249 | 358 |
| sum((x-mean(x))2) | 24 | 34 |
sum((x-mean(x))2)=sum(x2)-sum(x)*sum(x)/n
here we use t-test with
null hypothesis H0:mean1=mean2 and alternate hypothesis H1:mean1< mean2 ( this is left-one-tailed test)
statistic t=|(mean1-mean2)|/((sp*(1/n1 +1/n2)1/2) with df is n=n1+n2-2 and sp2=((n1-1)s12+(n2-1)s22)/n
| t-test | ||||||
| sample | mean | n | (n-1)s2 | |||
| landmark | 45.0000 | 9 | 24.0000 | |||
| absolute | 54.0000 | 9 | 34.0000 | |||
| difference= | 9.0000 | 18 | 58.0000 | |||
| sp2= | 3.6250 | |||||
| sp= | 1.9039 | |||||
| SE= | 0.8975 | |||||
| t= | 10.0275 | |||||
| one tailed | p-value= | 0.0000 | ||||
| two tailed | p-value= | 0.0000 | ||||
| two tialed critical | t(0.05) | 2.1199 | ||||
| one tailed critical | t(0.05) | 1.7459 |
g and h are for problem #1 Problem Set Chapters 10 & 12 [44 MARKS] Please print out and complete this problem set on your own and bring it to the next exam to be handed in at the start of the...