Chapter 8 Key (Odd)

(8.1) REVIEW QUESTIONS
(A) Independent samples represent random observations from separate unrelated populations. Paired samples represent matched or natural couples.
(B) (a) independent (b) paired (husband/wife pairing) (c) independent (d) independent (e) paired 
(C) You would use one variable to store the outcome and a separate variable to store the group indicator.
(D) One variable (let's call it ANXIETY) would contain information about anxiety score and a binary variable (STU_TYPE) would contain information about whether the student was an undergrad or grad student (1 = undergrad, 2 = grad). 
(E) Boxplots are plotted side-by-side. You then make a visual comparison of hinge-spreads and whiskers-spreads. If these spread differ greatly, you have evidence of unequal variance within groups. This technique is reliable only when samples are large. 
(F) The sample mean difference (₁ - ₂).
(G) d 
(H) b
(I)  c
(J) df = df₁ + df₂  =  (n₁ - 1) + (n₂ - 1)
(K) We are 95% confident that �₁ - �₂ (the true population mean difference) lies between -3.3 and 1.7.
(L) The mean in population 1 is equal to the mean in population 2 (or their difference equals 0).
(M) H₀: �₁ = �₂  (or H₀: �₁ - �₂ =  0)
(N) Independent samples; Normality; Equal variance
(O) True
(P) You conduct sample size calculations before a study is begun so you study will have sufficient power or be sufficiently precise.
(Q) You need to specify (a) the standard deviation of the variable, (b) the size of the difference you want to detect, (c) the alpha level of the test, and (d) the required power. 

(8.3) TWOGRPS

(A) Side-by-side boxplot

Calculations for group 1: IQR = 97 - 86 = 11 FL = 86 - (1.5)(11) = 69; no lower outside values FU = 97 + (1.5)(11) = 113.5; one upper outside value of 125, upper inside value is 99 Calculations for group 2: IQR = 129 - 93 = 33 FL = 93 - (1.5)(33) = 43.5; no lower outside values FU = 126 + (1.5)(33) = 175.5; no upper outside values Interpretation: Group 1 has a lower average (median) value. In addition, the boxes hardly overlap. This provides evidence that distribution 1 is located to the left of distribution 2. Group 1 has a smaller IQR but a greater range that group 2.

(B) Hypothesis Test
s²_p = [(10 - 1)(14.86²) + (10 - 1)(15.37²)] / [(10 - 1) + (10 - 1)] = 227.935
se = [227.935 * (1/10 + 1/10)] = 6.76
H₀: �₁ =  �₂ = 0 vs. H₁: �₁   �₂  
Let a = .05
t_stat = (93.3 - 107.6) - 0 / 6.76 = -2.115, df = 18
.01 < p < .05
Therefore, reject H₀. 

(8.5) JOGGERS
(A) 95% CI for �₁ - �₂

• s²_pooled = [4.8²(24) + 5.1²(25)] / [24 + 25] = 24.60
• se = sqrt[24.60 (1/25 + 1/26)] = 1.389
• t_49,.975 @ t_40,.975 = 2.02
• 95% CI for �₁ - �₂ = (47.5 - 37.5) � (2.02)(1.389) = 10.0 � 2.8 = (7.2, 12.8)

Interpretation:  With 95% confidence we can say the mean difference in V0₂ uptake in joggers and non-jogger populations is between 7.2 and 12.8. 

(B) 99% CI for �₁ - �₂

• t_49,.995 @ t_40,.995 = 2.70
• 99% CI for �₁ - �₂ = (47.5 - 37.5) � (2.70)(1.389) = 10.0 � 3.8 = (6.2, 13.8)

Interpretation: This confidence locates the population difference with 99% confidence. 

(8.7) CMV. Luminal diameter reduction (mm) of coronary arteries. 
s²_p = 0.616
se _{mean dif} = .1904
H₀: �₁ = �₂ vs. H₁: �₁ not = �₂  
a = .05
t_stat = (1.24 - 0.68) / (.1904) = 2.94; df = (49 - 1) + (26 - 1) = 73; [draw curve, locate t_stat in right tail, and then determine using t table that the two-sided  p < .01)
Reject H₀  

Interpretation: The study shows that the the CMV+ group had significantly greater luminal reduction than the CMV- negative group. 

(8.9) ASPIRIN
H₀: H₁: �₁ = �₂ vs. H₁: �₁ not = �₂  
a = .05
t_stat = 0.86; df = 45; [draw curve and locate the t_stat .86 standard error units to right of center]; p > .2
retain H₀ is 

(8.11) INDTSIZE
n = [(16)(40²) / 10²] + 1 = 257