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A study was conducted that measured the total brain volume (TBV) (in mm) of patients that...

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

Solutions

Expert Solution

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.


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