Question

Can someone please explain this to me in details. thank you

2) The volume of the right cordate section of the brain was estimated from CT scans for patients with obsessive-compulsive disorder, and a control group of healthy persons. The following data were obtained (volumes in mL). OCD patients 0.34 0.08 10 Control 0.45 0.09 12 Mean 03 a. Conduct a hypothesis tests to compare the variance between the groups. b. Conduct a hypothesis test to examine if there is a difference in mean volume of this part of the brain between the two groups. C. Calculate a 95% confidence interval for the difference in the mean volume of this part of the brain between the two groups d. Describe the type l and type Il errors that could occur in this hypothesis-testing situation.

Answer 1

Solution

Part (a)

Let

X = Volume of the right cordate section of the brain OCD patients,

Y = Volume of the right cordate section of the brain of healthy persons (control group)

Then, X ~ N(µ₁, σ₁²) and Y ~ N(µ₂, σ₂²)

Hypotheses:

Null H₀: σ₁² = σ²₂   Vs Alternative H_A: σ₂² > σ²₁ [since SD of control group > SD of OCD]

Test statistic:

F= s₂/s₁²where s₁and s₂ are standard deviations based on n₁ observations on X and n₂ observations on Y.

Calculations

n1	10
n2	12
s1	0.08
s2	0.09
Fcal	1.265625
α	0.05*
Fcrit	3.102485
n1 - 1	9
n2 - 1	11
p-value	0.367619

* α = 0.05 is assumed

Distribution, Significance Level, α Critical Value and p-value

Under H₀, F~ F_{n1 – 1, n2 – 1}

Critical value = upper α% point of F_{n1 – 1, n2 – 1}

p-value = P(F_{n1 – 1, n2 – 1} > F_(cal))

Using Excel Function: Statistical FINV and FDIST, these are found as shown in the above table.

Decision:

Since Fcal < Fcrit, or equivalently, since p-value > α. H₀ is accepted.

Conclusion:

There is not sufficient evidence to suggest the two variances are unequal. Answer 1

Part (b)

Hypotheses:

Null: H₀: µ1 = µ2 Vs Alternative: H_A: µ1 ≠ µ2

Test Statistic:

t = (Xbar - Ybar)/[s√{(1/n₁) + (1/n₂)}] where

s² = {(n₁ – 1)s₁² + (n₂ – 1)s₂²}/(n₁ + n₂ – 2);

Xbar and Ybar are sample averages and s₁,s₂ are sample standard deviations based on n₁ observations on X and n₂ observations on Y respectively.

Calculations

Summary of Excel calculations is given below:

n1	10
n2	12
Xbar	0.34
Ybar	0.45
s1	0.08
s2	0.09
s^2	0.007335
s	0.085645
tcal	2.999659
α	0.05*
tcrit	2.085963
p-value	0.007081

* α = 0.05 is assumed

Distribution, Significance Level, α Critical Value and p-value:

Under H₀, t ~ t_{n1 + n2 - 2}. Hence, for level of significance α%, Critical Value = upper (α/2)% point of t_{n1 + n2 - 2} and p-value = P(t_{n1 + n2 - 2} > | tcal |).

Using Excel Function: Statistical TINV and TDIST, these are found to be as shown in the above table.

Decision:

Since | tcal | > tcrit, or equivalently since p-value < α, H₀ is rejected.

Conclusion:

There is sufficient evidence to suggest that there is difference in the mean volume of this part of the brain between the two groups. Answer 2

Part (c)

100(1 - α) % Confidence Interval for (μ₁ - μ₂) is: (Xbar – Ybar) ± {(t_{2n – 2, α/2})(s)√{(1/n₁) + (1/n₂)}

substituting the values from Parts (a) and (b),

95% confidence interval for the difference in the mean volume of this part of the brain between the two groups is: [0.034, 0.186] Answer 3

Part (d)

Under Part (a), Type I error would occur when we conclude that the variances are different when in reality they are equal and Type II error would occur when we conclude that the variances are the same when in reality they are not.

Under Part (b), Type I error would occur when we conclude that the means are different between the two groups when in reality they are equal and Type II error would occur when we conclude that the means are equal between the two groups when in reality they are not. Answer

DONE