#1. Severe anxiety often occurs in patients who must undergo chronic hemodialysis. A set of progressive...

Solution

Let X = Post-test score of Experimental Group , Y = Post-test score of Experimental Group

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

Part (a)

To test equality of population standard deviations.

Hypotheses:

Null H₀: σ₁² = σ²₂   Vs Alternative H_A: σ₁² ≠ σ²₂

Test statistic:

F= s₁²/s₂²where s₁and s₂ are standard deviations based on n1 observations on X and n2 observations on Y.

= 1.2351

[Note: F is defined as above when s₁² > s₂². If s₂² > s₁², F is defined as F = s₂²/s₁²]

Calculations

n1	38
n2	23
s1	10.18
s2	9.16
Fcal	1.235107
Assuming,
α =	0.05
Fcrit	1.949457
n1 - 1	37
n2 - 1	22
p-value	0.304313

Distribution, Critical Value and p-value

Under H₀, F~ F_{n1 – 1, n2 – 1} or F_{n2 – 1, n1 – 1} depending on whether s₁² > s₂² or s₂² > s₁²

Critical value = upper α% point of F_{n1 – 1, n2 – 1} or F_{n2 – 1, n1 – 1} depending on whether s₁² > s₂² or s₂² > s₁²

p-value = P(F_{n1 – 1, n2 – 1} > F_(cal)) or P(F_{n2 – 1, n1 – 1} > F_(cal)), depending on whether s₁² > s₂² or s₂² > s₁²

Fcrit and and p-value are found to be as shown in the above table [using Excel Function: Statistical   FINV and FDIST respectively]

Decision:

Since p-value > α. H₀ is accepted.

Conclusion:

There is not sufficient evidence to suggest that the population variances and hence the population standard deviations are different for the two groups. Answer

Part (b)

To test equality of population means where σ₁² = σ₂² = σ², say but σ² is unknown [vide Answer 1].

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.

= 2.427

Calculations

Summary of Excel calculations is given below:

n1	38
n2	23
Xbar	33.42
Ybar	39.71
s1	10.18
s2	9.16
s^2	96.27664
s	9.812066
tcal	- 2.426504
Assuming α = 0.05
tcrit	2.000995
p-value	0.018319

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 |) = 2P(t_{n1 + n2 - 2} > tcal) if tcal > 0 and 2P(t_{n1 + n2 - 2} < tcal) if tcal < 0

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 the two groups differ with respect to the means. i.e., we conclude that the population means of post-test scores differ from group to group. Answer

DONE