## Question

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

## Answers

Solution

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

Then, X ~ N(µ

_{1}, σ_{1}^{2}) and Y ~ N(µ_{2}, σ_{2}^{2})Part (a)

To test equality of population standard deviations.

Hypotheses:

Null H

_{0}: σ_{1}^{2}= σ^{2}_{2 }Vs Alternative H_{A}: σ_{1}^{2}≠ σ^{2}_{2}Test statistic:

F= s

_{1}^{2}/s_{2}^{2}where s_{1}and s_{2}are standard deviations based on n1 observations on X and n2 observations on Y.= 1.2351

[

Note:F is defined as above when s_{1}^{2}> s_{2}^{2}. If s_{2}^{2}> s_{1}^{2}, F is defined as F = s_{2}^{2}/s_{1}^{2}]Calculations

n1

38

n2

23

s1

10.18

s2

9.16

Fcal

1.235107

Assuming,

α =

0.05

Fcrit

1.949457n1 - 1

37

n2 - 1

22

p-value

0.304313Distribution, Critical Value and p-value

Under H

_{0}, F~ F_{n1 – 1, n2 – 1}or F_{n2 – 1, n1 – 1}depending on whether s_{1}^{2}> s_{2}^{2}or s_{2}^{2}> s_{1}^{2}Critical value = upper α% point of F

_{n1 – 1, n2 – 1}or F_{n2 – 1, n1 – 1}depending on whether s_{1}^{2}> s_{2}^{2}or s_{2}^{2}> s_{1}^{2}p-value = P(F

_{n1 – 1, n2 – 1}> F_{(cal)}) or P(F_{n2 – 1, n1 – 1}> F_{(cal)}), depending on whether s_{1}^{2}> s_{2}^{2}or s_{2}^{2}> s_{1}^{2}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. H

_{0}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.

AnswerPart (b)

To test equality of population means where σ

_{1}^{2}= σ_{2}^{2}= σ^{2}, say but σ^{2}is unknown [vide Answer 1].Hypotheses:

Null: H

_{0}: µ1 = µ2 Vs Alternative: H_{A}: µ1≠µ2Test Statistic:

t = (Xbar - Ybar)/[s√{(1/n

_{1}) + (1/n_{2})}] wheres

^{2}= {(n_{1}– 1)s_{1}^{2}+ (n_{2}– 1)s_{2}^{2}}/(n_{1}+ n_{2}– 2);Xbar and Ybar are sample averages and s

_{1},s_{2}are sample standard deviations based on n_{1}observations on X and n_{2}observations on Y respectively.=

2.427Calculations

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

0.018319Distribution, Significance Level, α Critical Value and p-value:

Under H

_{0}, 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 < 0Using 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, H

_{0}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