In: Statistics and Probability
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). 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 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 |
1636119 |
In order to solve this question I used Excel.
Step.1 Enter data in excel sheet.
Step.2 Go to 'Data' menu ---> 'Data Analysis' ---> Select 't Test :Assuming unequal variances'.
Step.3 New window will pop-up on screen. Refer following screen shot and enter information accordingly.
Excel output:
Schizophrenia | Normal | ||||
1331777 | 1663407 | Variance of schi | 29560690487 | ||
1368378 | 1453510 | Variance of normal | 15739779000 | ||
1690182 | 1474790 | ||||
1563593 | 1441636 | t-Test: Two-Sample Assuming Unequal Variances | |||
1325525 | 1574808 | ||||
1636119 | 1512571 | Schizophrenia | Normal | ||
1487886 | 1583940 | Mean | 1451293.194 | 1463339.219 | |
1476891 | 1650348 | Variance | 29560690487 | 15739779000 | |
1569413 | 1317156 | Observations | 31 | 32 | |
1317885 | 1432033 | Hypothesized Mean Difference | 0 | ||
1588573 | 1502702 | df | 55 | ||
1066075 | 1635918 | t Stat | -0.316842857 | ||
1443775 | 1299470 | P(T<=t) one-tail | 0.376281198 | ||
1177002 | 1288971 | t Critical one-tail | 1.673033966 | ||
1420249 | 1441045 | P(T<=t) two-tail | 0.752562396 | ||
1476254 | 1420416 | t Critical two-tail | 2.004044769 | ||
1297327 | 1203344 | ||||
1337827 | 1535137 | ||||
1387893 | 1366346 | ||||
1363859 | 1463498 | ||||
1648209 | 1480171 | ||||
1499983 | 1319737 | ||||
1658258 | 1431890 | ||||
1483763 | 1326402 | ||||
1238979 | 1650207 | ||||
1354054 | 1360810 | ||||
1861991 | 1688990 | ||||
1588132 | 1578698 | ||||
1688950 | 1503005 | ||||
1286638 | 1523045 | ||||
1354649 | 1410213 | ||||
1292641 | |||||
Hypothesis:
Since p-value for one tailed test is 0.3763 which is greater than 0.10,hence we accept null hypothesis and conclude that the patients with schizophrenia have approximately equal TBV on average than a patient that is considered normal.