In: Math
A study was conducted that measured the total brain volume (TBV) (in mm) 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). Is there enough evidence to show that the patients with schizophrenia have less TBV on average than a patient that is considered normal? Test at the 10% level.
Table #9.3.5: Total Brain Volume (in mm) 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 mm) 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  | 
| 
 1636119  | 
Solution:
Here, we have to use two sample t test for population means assuming equal population variances.
H0: µnormal = µSchizophrenia versus H0: µnormal > µSchizophrenia
WE are given α = 0.10
From given data, we have
X1bar = 1463339.219
S1 = 125458.2759
N1=32
X2bar = 1451293.194
S2= 171932.2264
N2=31
DF = N1 + N2 – 2 = 32 + 31 – 2 = 61
Test statistic formula for pooled variance t test is given as below:
t = (X1bar – X2bar) / sqrt[Sp2*((1/n1)+(1/n2))]
Where Sp2 is pooled variance
Sp2 = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)
Sp2 = [(32 – 1)* 125458.2759^2 + (31 – 1)* 171932.2264^2]/(32 + 31 – 2)
Sp2 = 22536948583.7099
t = (X1bar – X2bar) / sqrt[Sp2*((1/n1)+(1/n2))]
t = (1463339.219 – 1451293.194) / sqrt[22536948583.7099*((1/32)+(1/31))]
t = 0.3184
Critical value = 1.2956
(by using t-table)
P-value = 0.3756
(by using t-table)
P-value > α = 0.10
So, we do not reject the null hypothesis
There is insufficient evidence to show that the patients with schizophrenia have less TBV on average than a patient that is considered normal.