(1) List four specific types of information a user would want to put into a data documentation (DD) file to help decipher the file. [4]
ANS: Variable names, variable types, variable lengths or positions, data file name and location, each variable's coding scheme or units of measure, descriptions of prior file manipulations, etc.
(2) An ANOVA procedure tests for differences between 4 groups, with each group comprising 10 individuals. How many degrees of freedom are associated with the between-group variance estimate? ____________ [1]
ANS: dfB = k - 1 = 3
(3) Why is Bonferroni's adjustment (or a similar adjustment method) necessary when testing multiple comparisons? [2]
ANS: To maintain a reasonable family-wise alpha level/ type I error rate.
(4) List ways in which paired samples can be achieved. [1]
ANS: Matching; longitudinal samples, "pre-test / post-test" intervention, cross-over trials, etc.
(5) Provide a synonym for "within-group variance." [1]
ANS: Mean Square Within, MSError, pooled estimate of variance, "variance"
(6) What is the formal name of the non-parametric equivalent of (one-way) ANOVA? [1]
ANS: The Kruskal-Wallis Test
(7) List the null and alternative hypotheses tested by ANOVA when comparing 4 groups. [2]
ANS: H0: �1 = �2 = �3 = �4 vs. H1: at least one population mean differs (or equivalents)
(8) List the assumptions required for valid analysis of variance.[3]_
ANS: Independence; Normality; Equal variance
(9) What does "heteroscedastic" mean? [1]
ANS: Unequal variance.
The purpose of an investigation by Alahuhta et al. (Daniel, 1995, p. 240, Ex. 7.4.3) was to evaluate the influence of extradural (anaesthesia) blocks for elective caesarean sections. Study subjects were eight healthy patients in gestational weeks 38 - 42 with uncomplicated singleton pregnancies undergoing elective caesarean section. Several maternal and fetal hemodynamic variables were studied simultaneously. Among the measurements taken were maternal diastolic arterial pressure (mm Hg) during two stages of the study. Data are:
PATIENT STAGE1 STAGE2
------- ------ ------
1 70 79
2 87 87
3 72 73
4 70 77
5 73 80
6 66 64
7 63 64
8 57 60
(1) Create an Epi Info data file with these data. (No output is required.) [10]
ANS: Grading will be based on accuracy of descriptive statistics, below.
(2) Compute the mean and standard deviation of the Stage 1 measurements. [2]
mean = ________ mm Hg
standard deviation = ________ mm Hg
ANS: mean = 69.750, sd = 8.746 [answers on key not rounded so that I might check unrounded answers)
(3) Compute the mean and standard deviation of the Stage 2 measurements. [2]
mean = ________ mm Hg
standard deviation = ________ mm Hg
ANS: mean = 73.00 sd = 9.472
(4) Compute the mean and standard deviation for the difference in Stage 1 and Stage 2 measurements. [4]
mean = ________ mm Hg
standard deviation = ________ mm Hg
ANS: mean = -3.25 (or +3.25, depending on how DELTA was created), sd = 4.0
(5) Draw a stem-and-leaf plot of difference variable DELTA. Label the axis appropriately. [4]
ANS: Graded on a 4 point GPA-like scale.
There are several ways one could draw this stem-and-leaf plot. Two possibilities are:
|-0|917713
|+0|02
DELTA (x 10) - negative values represent STAGE2 values > STAGE1 values
|-9|0
|-8|
|-7|00
|-6|
|-5|
|-4|
|-3|0
|-2|
|-1|00
|-0|
|+0|0
|+1|
|+2|0
DELTA (x 1) - negative values represent STAGE2 values > STAGE1 values
(6) Test whether the observed difference is significant. Report the null and alternative hypotheses, set your alpha level, report the test statistic, its degrees of freedom, and p value using an APA-like format, state your conclusion.) [7]
H0: �d = 0 [1]
H1: �d not = 0 [1]
Let alpha = .05 (or .01, or whatever, as long as it is explicit) [1]
t(7) = 2.32 [2]
p = .052 [1]
Retain H0 [1]
(6) In one or two brief sentences, interpret your results. [4]
ANS: Graded on a GPA-like scale. The answer should include references to a [non-significant difference] in which [Stage 2 readings tended to exceed Stage 1 readings]. One could also mention that the results are "almost significant" or "marginally significant," and the test probably had limited power because of small n. When we look at the stem-and-leaf plot, we note that, with one exception, Stage 2 values exceed or are equal to Stage 1 values. The mean difference is 3.3 mm Hg.
Comment: Some students will mistake the question to mean "interpret the hypothesis testing results" when, in fact, they should interpret the descriptive statistics and graph, as well.
(7) Suppose we have three independent groups, with 6 observations per group, and the following ANOVA results are computed:
Variation SS df MS
Between 423.502 2 211.751
Within 376.216 15 25.081
Total 799.718 17
Let us assume equal variances within groups and an alpha level of .05. What was the power of the above study to find a minimal detectable difference of 10? Show all work. [4]
ANS: phi = sqrt[nd2 / 2ks2] = sqrt[(6)(10)2 / (2)(3)(25.081)] = sqrt[(600) / 150.486]] = 1.997 ~= 2.00
Power = .81 (via lookup table)
(8) If you had a negative ANOVA test (i.e, retention of H0), how would the above power calculation color your interpretation of your results. (Maintain the same assumed d used in question 7.) Explain your response. [1]
ANS: You can "trust" the test to uncover d = 10, since the probability of avoiding a false retention was greater than 80%.