Question

A study was conducted that measured the total brain volume (TBV) (in ) 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 ) 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 ) 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

1. ) State the random variable and the parameter in words.

b.) If needed, state the null and alternative hypotheses and the level of significance and don't "bump" them into one line (refer to the textbook)

Ho :

HA :

a =

c.) State and check the assumptions for a hypothesis test

d.) Find the sample statistic, test statistic, and p-value, confidence interval ( whatever the task is)

e.) Conclusion

Since the p-value……..fail to reject or reject   H o or confidence intervals...

f.) Interpretation ( statistical and real world)

Answer 1

Random variable is "total brain volume". We should use t-test. H: 4242 H:44 a=0.10 Assumptions:

The data are continuous. The data is collected from a randomly selected portion of the population. The data follow the normal probability distribution. The two samples are independent. (d) EXCEL (Data Analysis) output: t-Test: Two-Sample Assuming Unequal Variances Schizophenia patients Normal patients 1446481.75 1463339.219 29347919475 15739779000 32 Mean Variance Observations Hypothesized Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail 0 57 -0.449095102 0.327532983 1.296581044 0.655065966 1.672028889

-79,638.141 confidence interval 90.% lower 45,923.204 confidence interval 90 % upper 62,780.672 margin of error Sample statistic: t Test Statistic: -0.449. p-value = 0.3275 Confidence interval: (-79638.141.45923.204). Since the p-value is 0.3275 > a{=0.10). then fail to reject the null hypothesis. Interpretation:

There is not enough evidence to show that the patients with schizophrenia have less TBV on average than a patient that is considered normal.