A study used X-ray computed tomography to collect data on brain volumes for a gr

Question

A study used X-ray computed tomography to collect data on brain volumes for a group of patients with obsessive-compulsive disorders and a control group of healthy persons. Sample results (in mL) are given below for total brain volumes. Normality requirement is satisfied.

Obsessive-compulsive patients: n = 12, x = 1590.03, s = 166.84

Control group: n = 12, x = 1368.41, s = 135.97

a) Construct a 99% confidence interval for the difference between the mean brain volume of obsessive-compulsive patients and the mean brain volume of healthy persons. Assume that the two populations have unequal variances.

b) Assuming that the population variances are unequal, use a 0.05 significance level to test the claim that there is no difference between the mean of obsessive compulsive patients and the mean for healthy persons.

c) Based on the results from parts a) and b), does it appear that the total brain volume can be used as an indicator of obsessive compulsive disorders?

d) With 0.1 significance level, test the claim that the populations of total brain volumes for obsessive-compulsive patients and the control group have different amounts of variation.

e) Imagine you prepare these data for a publication. Present your data in a graphical form and indicate if your results are statistically significant. Create a legend for your figure.

Explanation / Answer

Obsessive-compulsive patients: n = 12, x1 = 1590.03, s1 = 166.84

Control group: n = 12, x2 = 1368.41, s2 = 135.97

(a) Here confidence interval = 99%

difference between the mean brain volume of obsessive-compulsive patients and the mean brain volume of healthy persons. = x1 -  x2 = 1590.03 - 1368.41 = 221.62

Pooled standerd deviation x1 -  x2= sqrt [ s12/n1 + s22/n2]= sqrt [ 166.842/12 + 135.972/12 ]

= 62.1311

Here Degree of Freedom here Df = [ s12/n1 + s22/n2]2/ [(s12/n1)2/(n1 - 1) + (s22/n2)2/(n2 - 1) = 21

t-stastic for DF =21 is 2.819 from t - table

so 99 % confidence interval is ( x1 -  x2) +- 2.831 * x1 -  x2 so the confidence interval values are

221.62 +- 2.831 * 62.1311 = (45.7268, 397.5131)

(b) Hypothesis testing

H0 :  x1 = x2

H1 :  x1 x2

Here alpha = 0.05

so by t - statstics

t value = ( x1 -  x2 )/ sqrt [ s12/n1 + s22/n2] = 221.62/62.1311 = 3.567

Then, the critical value approach tells us to reject the null hypothesis in favor of the alternative hypothesis if:

t > t0.025,21=2.080

We reject the null hypothesis because the test statistic (t = 3.57) falls in the rejection region:

(c) Based on the observations of a) and b), that the total brain volume can be used as an indicator of obsessive compulsive disorderd as it is perfectly visible that mean of brain volume for obsessive comppulsive disorders children is high than normal childre.

d) Here the signification level is 0.1, so here we have to check the variance level.

Hypothesis testing

H0 :  s1 = s2

H1 :  s1 s2

Here alpha = 0.1

t = (166.84/1590.03 - 135.97/1368.41)/sqrt [ (166.84/1590.03)2/12 + (135.97/1368.41)2/12]

t = 0.0055/0.0041 = 0.1357

Then, the critical value approach tells us to reject the null hypothesis in favor of the alternative hypothesis if:

t > t0.05,21=1.721

We accept the null hypothesis because the test statistic (t = 0.1357) falls in the significant region

so We can say there is no difference in variation between total brain volumes of OCP and the CP group children.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.