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.