8: Key, Odd

Review Questions

1. Data points in independent samples are unrelated. You should think of these as SRSs from separate populations. In contrast, each data point in a paired sample is uniquely matched to a data point in the other paired sample.
2. Use one variable to store data for the response (outcome) variable and a separate variable to store group information. 
3. Stemplots on a common stem; side-by-side boxplots. [Techniques not covered: mean ± standard deviation plot; mean ± standard error plot; dotplots.]
4. ₁ - ₂
5. df = df₁ + df₂  =  (n₁ - 1) + (n₂ - 1) = n₁ + n₂ - 2
6. There is no difference in the population  means. Or, the mean in population 1 is equal to the mean in population 2.
7. H₀: µ₁ - µ₂ =  0. Or, H₀: µ₁ = µ₂ 
8. (a) the standard deviation of the variable, (2) difference worth detectiing, (3) desired alpha, and (4) desired power.
9. independent samples, Normality, equal variance
10. True. Population do not need to be Normal. The sampling distribution of the mean (SDM), however, need to be Normal.
11. The 95% confidence of the interval refers to the procedure used to draw the confidence interval, and not to the confidence interval once it is drawn. Once the confidence interval is drawn, there is either a 100% or 0% is contains the true mean difference. 

Exercises

8.1 Sampling designs. (A) Autism. Independent samples (B) Husbands and wives. Paired samples (C) Nutritional knowledge. A single sample 

8.3 Facetious study
(A) Group 1 mean = 98. Group 2 mean = 108. Independent samples.
(B) Mean change in group 1 = 4. Paired samples. 
(C) Mean change within group 2 is equal to 1. This is a paired difference. 
(D) Group 1 had a greater mean change (4 vs. 1). This is an independent comparison of the two groups.

8.5 Large t statistic. When the sample is more than just several observations, the t distribution is almost identical to a a z distribution. Based on the 68-95-99.7 rule, we know that we would almost never get a statistic this far from 0 on a z distribution. Therefore, this t statistic will be in the far extent of the right-hand tail and the P value will be very very small (surely less than 0.01). This indicates a highly significant result. 

8.7 Bone density in newborns. Data are bone mineral density measures in grams / centimeter³. The 95% confidence interval for mean difference is based on:

• df₁ = 77 - 1 = 76; df₂ = 161 - 1 = 160; df = 76 + 160 = 236. With this many df, we can use the z  value of  1.96 for 95% confidence.
• s²_p = (76)(.026²) + (160)(.025²) / 236 = 0.000641
• SE_xbar1-xbar2 = sqrt [.000641 × (1/77 + 1/161)] = 0.00351
• 95% confidence interval for µ₁ - µ₂ = (0.098 - 0.095) ±  (1.96)(0.00351) = 0.003  ±  0.007 = (-0.004, 0.010)
• Interpretation: We can be 95% confidence that the mean difference in the population is between  -0.004 and 0.010. Our confidence is in the method used to calculate the interval, and not in this particular result as such.

8.9 Cytomegalovirus and coronary stenosis. Data represent luminal diameter reduction (mm) of coronary arteries over 6 months of follow-up.

• Step A (Hypotheses): H₀: µ₁ = µ₂ vs. H₁: µ₁   µ₂  
• Step B (Statistics): df₁ = 49 - 1 = 48, df₂ = 26 - 1 = 25, df = 48 + 25 = 73; s²_p = [(48)(0.83²) + (25)(0.69²)] / 73 = 0.616; SE_{mean dif} = sqrt[0.616 × (1/49 + 1/26)] = 0.1904 
  t_stat = (1.24 - 0.68) / (0.1904) = 2.94
• Step C (P value): Two-sided P is between 0.01 and 0.002
• Step D ("Significance conclusion"): , The difference is highly significant (reject H₀). Data support the theory that CMV plays a role in coronary restenosis. 

8.11 Pregnancy-induced hypertension and aspirin 

• H₀: µ₁ = µ₂ vs. H₁: µ₁   µ₂  
• df₁ = 22, df₂ = 23, df = 45
• s²_pooled = [(22)(8²) + (23)(8²)] / 45 = 64 [The pooled estimate of variance is a weighted average of the two sample variances. Both samples have a variance of 8², so the pooled variance is also 64.]
• se_meandif = sqrt[64 × (1/23 + 1/24)] = 2.334
• t_stat = (111 - 109) / 2.334 = 0.86
• The two-tailed P-value is between 0.30 and 0.40. 
• The difference is not significant (do not  reject H₀)

8.13 Risk Taking 

Group 1 (Girls): 5-point summary: 72, 86, 95, 97, 125 IQR = 97 - 86 = 11 FU = 97 + (1.5)(11) = 113.5; 125 is outside; upper inside value is 99 FL = 86 - (1.5)(11) = 69.5; no lower outside values Group 2 (Boys): 5-point summary: 89, 93, 105.5, 126, 130 IQR = 129 - 93 = 33 FU = 126 + (1.5)(33) = 175.5; no upper outside values FL = 93 - (1.5)(33) = 43.5; no lower outside values Interpretation:  Location: Girls have lower scores on average.  Shape: The small sample sizes make statements about shape tenuous.  Group 1 has an upper outside value.  Spread: Girls tend to have less variability (except for the outlier, which may need further scrutiny).

8.15 Efficacy of echinacea in treating upper respiratory infections (severity of symptoms). H₀: µ₁ = µ₂: Pooled variance = 5.536; t_stat = -0.5644 ( 705 df) P-value = 0.5727. The test reveal no significant difference between the herbal remedy (echinacea) and placebo. 

Sample size questions

SS.1. Vegetarians and non-vegetarians.  n = [(16)(40²) / 10²] + 1 = 257 per group