Question

A study was conducted that measured the total brain volume (TBV) (in mm3 ) of
patients that had schizophrenia and patients that are considered normal. Table
#9.3.5 contains the TBV of the normal patients and table #9.3.6 contains the TBV
of schizophrenia patients ("SOCR data oct2009," 2013). Compute a 90%
confidence interval for the difference in TBV of normal patients and patients with
Schizophrenia.

Table #9.3.5: Total Brain Volume (in mm3) of Normal Patients

1663407 1583940 1299470 1535137 1431890 1578698

1453510 1650348 1288971 1366346 1326402 1503005

1474790 1317156 1441045 1463498 1650207 1523045

1441636 1432033 1420416 1480171 1360810 1410213

1574808 1502702 1203344 1319737 1688990 1292641

1512571 1635918

Table #9.3.6: Total Brain Volume (in mm3) of Schizophrenia Patients

1331777 1487886 1066075 1297327 1499983 1861991

1368378 1476891 1443775 1337827 1658258 1588132

1690182 1569413 1177002 1387893 1483763 1688950

1563593 1317885 1420249 1363859 1238979 1286638

1325525 1588573 1476254 1648209 1354054 1354649

Answer 1

SE( $\bar{x}$ - $\bar{y}$ )=sqrt(σ1²/n1+σ2²/n2)=31718

(1-alpha)*100% confidence interval for population mean difference=

=sample mean difference ±z(alpha/2)*SE( $\bar{x}$ - $\bar{y}$ )

90% confidence =(1463339-1445132)±z(0.05/2)*31718=18207 ±1.645*31718=18207 ±52176

=(-33969, 70383)

following infirnation is generated using ms-excel

	x	y
n=	32	30
mean=	1463339	1445132
var= $\sigma^2$ =	15247910906	28384048331
va/n= $\sigma^2$ /n=	476497216	946134944