Question

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 #1 contains the TBV of the normal patients and Table #2 contains the TBV of schizophrenia patients ("SOCR data Oct2009," 2013).

Table #1: 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 #2: 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

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.

(i) Let ?₁= mean TBV of patients that are considered normal. Let ?₂ = mean TBV of patients that had schizophrenia. Which of the following statements correctly defines the null hypothesis H_O?

A. ?₁ + ?₂= 0

B. ?₁ – ?₂< 0 (?₁ < ?₂)

C. ?₁ ? ?₂ > 0 (?₁ > ?₂)

D. ?₁ ? ?₂ = 0 (?₁ = ?₂)

Enter letter corresponding to correct answer

(ii)   Let ?₁= mean TBV of patients that are considered normal. Let ?₂ = mean TBV of patients that had schizophrenia. Which of the following statements correctly defines the alternate hypothesis H_A?

A. ?₁ ? ?₂ = 0 (?₁ = ?₂)

B. ?₁ – ?₂< 0 (?₁ < ?₂)

C. ?₁ ? ?₂ > 0 (?₁ > ?₂)

D. ?₁ + ?₂= 0

Enter letter corresponding to correct answer

(iii) Enter the level of significance ? used for this test:

Enter in decimal form. Examples of correctly entered answers: 0.01    0.02    0.05    0.10

(iv) For sample from population with mean = ?1 : Determine sample mean x¯1 and sample standard deviation s1iv  For sample from population with mean = ?1 :          Determine sample mean x¯1 and sample standard deviation s1 {"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mstyle mathsize="14px"><mrow><mfenced><mi mathvariant="bold">iv</mi></mfenced><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">For</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">sample</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">from</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">population</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">with</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">mean</mi><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">=</mo><mo mathvariant="bold">&#xA0;</mo><msub><mi mathvariant="bold-italic">&#x3BC;</mi><mn mathvariant="bold">1</mn></msub><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">:</mo><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mspace linebreak="newline"></mspace><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">Determine</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">sample</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">mean</mi><mo mathvariant="bold">&#xA0;</mo><msub><mover><mi mathvariant="bold">x</mi><mo mathvariant="bold">&#xAF;</mo></mover><mn mathvariant="bold">1</mn></msub><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">and</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">sample</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">standard</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">deviation</mi><mo mathvariant="bold">&#xA0;</mo><msub><mi mathvariant="bold-italic">s</mi><mn mathvariant="bold">1</mn></msub></mrow></mstyle></math>"}

Enter sample mean in integer form (no decimals), then comma, then sample standard deviation in integer form. Examples of correctly entered answers:

13145,1211

-112845,13187

(v) For sample from population with mean = ?2 : Determine sample mean x¯2 and sample standard deviation s2v  For sample from population with mean = ?2 :       Determine sample mean x¯2 and sample standard deviation s2 {"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mstyle mathsize="14px"><mrow><mfenced><mi mathvariant="bold">v</mi></mfenced><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">For</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">sample</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">from</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">population</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">with</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">mean</mi><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">=</mo><mo mathvariant="bold">&#xA0;</mo><msub><mi mathvariant="bold-italic">&#x3BC;</mi><mn mathvariant="bold">2</mn></msub><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">:</mo><mspace linebreak="newline"></mspace><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">Determine</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">sample</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">mean</mi><mo mathvariant="bold">&#xA0;</mo><msub><mover><mi mathvariant="bold">x</mi><mo mathvariant="bold">&#xAF;</mo></mover><mn mathvariant="bold">2</mn></msub><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">and</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">sample</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">standard</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">deviation</mi><mo mathvariant="bold">&#xA0;</mo><msub><mi mathvariant="bold-italic">s</mi><mn mathvariant="bold">2</mn></msub></mrow></mstyle></math>"}

Enter sample mean in integer form (no decimals), then comma, then sample standard deviation in integer form. Examples of correctly entered answers:

13145,1211

-112845,13187

(vi) Determine degrees of freedom df :

Enter value in decimal form rounded down to nearest whole number. Examples of correctly entered answers:

2 3 16 110

(vii) Determine test statistic:

Enter value in decimal form rounded to nearest ten-thousandth. Examples of correctly entered answers:

0.0104      3.0370        16.5000        110.0819

(viii) Using tables, calculator, or spreadsheet: Determine and enter p-value corresponding to test statistic.

Enter value in decimal form rounded to nearest ten-thousandth. Examples of correctly entered answers:

0.0001     0.0021     0.0305     0.6004      0.8143    1.0000

(ix) Comparing p-value and ? value, which is the correct decision to make for this hypothesis test?

A. Reject H_o

B. Fail to reject H_o

C. Accept H_o

D. Accept H_A

Enter letter corresponding to correct answer.

(x) Select the statement that most correctly interprets the result of this test:

A. The result is not statistically significant at .10 level of significance. Sufficient evidence exists to support the claim that patients with schizophrenia have less TBV on average than a patient that is considered normal.

B. The result is not statistically significant at .10 level of significance. There is not enough evidence to support the claim that patients with schizophrenia have less TBV on average than a patient that is considered normal.

C. The result is statistically significant at .10 level of significance. Sufficient evidence exists to support the claim that patients with schizophrenia have less TBV on average than a patient that is considered normal.

D. The result is statistically significant at .10 level of significance. There is not enough evidence to support the claim that patients with schizophrenia have less TBV on average than a patient that is considered normal.

Enter letter corresponding to most correct answer

Answer 1

The Null Hypothesis is

The alternative hypothesis is

#### By using R command

> X1=c(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)
> X1 ####TBV of the normal patients
[1] 1663407 1583940 1299470 1535137 1431890 1578698 1453510 1650348 1288971
[10] 1366346 1326402 1503005 1474790 1317156 1441045 1463498 1650207 1523045
[19] 1441636 1432033 1420416 1480171 1360810 1410213 1574808 1502702 1203344
[28] 1319737 1688990 1292641 1512571 1635918
> X2=c(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)
> X2 ####TBV of the schizophrenia patients
[1] 1331777 1487886 1066075 1297327 1499983 1861991 1368378 1476891 1443775
[10] 1337827 1658258 1588132 1690182 1569413 1177002 1387893 1483763 1688950
[19] 1563593 1317885 1420249 1363859 1238979 1286638 1325525 1588573 1476254
[28] 1648209 1354054 1354649 1636119
> mean(X1) ### Sample mean of X1
[1] 1463339
> mean(X2) ### Sample mean of X2
[1] 1451293
> var(X1) ##sample variance of X1
[1] 15739779000
> var(X2) ##sample variance of X2
[1] 29560690487
> sd(X1) ## sample standard deviation of X1
[1] 125458.3
> sd(X2) ## sample standard deviation of X2.
[1] 171932.2

> t.test(X1,X2)

Welch Two Sample t-test

data: X1 and X2
t = 0.31684, df = 54.817, p-value = 0.7526
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-64151.34 88243.39
sample estimates:
mean of x mean of y
1463339 1451293

The degrees of freedom is =54.817

The test statistics is =0.31841

p-value = 0.7513

The Pvalue 0.7513 indicates that we are unable to reject the null hypothesis at 10% level of significance.

The result is not statistically significant at .10 level of significance. There is not enough evidence to support the claim that patients with schizophrenia have less TBV on average than a patient that is considered normal.