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

Answer 1

Using the t-distribution, it is found that since the test statistic is less than the critical value for the left tailed test, there is not enough evidence to show that the patients with schizophrenia have less TBV on average than a  patient that is considered normal.

At the null hypothesis, we test if patients with schizophrenia do not have less TBV on average than a  patient that is considered normal, that is, the subtraction is not negative, hence:

$H_0: \mu_S - \mu_N \geq 0$

At the alternative hypothesis, we test if they have less, that is, if the subtraction is negative, hence:

$H_1: \mu_S - \mu_N < 0$

Regarding each sample, the mean and standard deviation are found using a calculator.

$\mu_N = 1463339.2, s_N = 125458.28 , n_N = 32$

$\mu_S = 1445132.3, s_S = 171355.8, n_S = 30$

We have the standard deviation for each sample, hence, the t-distribution will be used.

The standard errors are given by:

$Se_N = \frac{125458.28}{\sqrt{32}} = 22178.1$

$Se_S = \frac{171355.8}{\sqrt{30}} = 31285.1$

The distribution of the difference has mean and standard error given by:

$\overline{x} = \mu_S - \mu_N = 1445132.3 - 1463339.2 = -18206.9$

$Se = \sqrt{Se_N^2 + Se_S^2} = \sqrt{22178.1^2 + 31285.1^2} = 38348.7$

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{Se}$

In which $\mu$ is the value tested at the hypothesis.

Hence:

$t = \frac{\overline{x} - \mu}{Se}$

$t = \frac{-18206.9 - 0}{38348.7}$

$t = -0.47$

The critical value for a left-tailed test, as we are testing if the mean is less than a value, with a significance level of 0.1 and 32 + 30 - 2 = 60 df is $t^{\ast} = 2$

Since the test statistic is less than the critical value for the left tailed test, there is not enough evidence to show that the patients with schizophrenia have less TBV on average than a  patient that is considered normal.

A similar problem is given at https://brainly.com/question/13873630