Test 2 practice
Mussel physiology
Researchers recorded observations of standardized anterior adductor muscle (AAM) scar length for mytilus trossulus mussels from five populations. The objective of the study was to identify physiological differences between populations. A plot of the data is provided below.
- [L9] Identify the assumptions for ANOVA and comment on whether they seem plausible for this data.
The output below shows an ANOVA model fitted to the data.
Call:
aov(formula = aam.length ~ location, data = mussels)
Terms:
location Residuals
Sum of Squares 0.004519674 0.005394906
Deg. of Freedom 4 34
Residual standard error: 0.01259658
Estimated effects may be unbalanced
[L9] How many mussels were measured in the study?
[L9] Construct the ANOVA table.
[L9] Report the outcome of the omnibus test following conventional style. (Assume a 5% level test.)
[L9] Estimate the effect size
.
The output below shows pairwise comparisons for the five mussel populations.
contrast estimate SE df t.ratio p.value
magadan - newport 0.00321 0.00630 34 0.510 1.0000
magadan - tillamook -0.00219 0.00598 34 -0.366 1.0000
magadan - tvarminne -0.01769 0.00680 34 -2.600 0.1370
magadan - petersburg -0.02543 0.00652 34 -3.901 0.0043
newport - tillamook -0.00540 0.00598 34 -0.904 1.0000
newport - tvarminne -0.02090 0.00680 34 -3.072 0.0417
newport - petersburg -0.02864 0.00652 34 -4.394 0.0010
tillamook - tvarminne -0.01550 0.00650 34 -2.383 0.2291
tillamook - petersburg -0.02324 0.00621 34 -3.744 0.0067
tvarminne - petersburg -0.00774 0.00701 34 -1.105 1.0000
P value adjustment: bonferroni method for 10 tests
- [L9] Identify which populations have significantly different mean AAM lengths at the 1% level.
The output below shows interval estimates for the mean AAM length of each population.
location emmean SE df lower.CL upper.CL
magadan 0.0780 0.00445 34 0.0659 0.0902
newport 0.0748 0.00445 34 0.0626 0.0870
tillamook 0.0802 0.00398 34 0.0693 0.0911
tvarminne 0.0957 0.00514 34 0.0817 0.1097
petersburg 0.1034 0.00476 34 0.0905 0.1164
Confidence level used: 0.95
Conf-level adjustment: bonferroni method for 5 estimates
- [L4] Interpret the interval for the largest estimated mean AAM length in context.
Creativity and motivation
This problem uses data from an experiment on the effect of intrinsic vs. extrinsic motivation on creativity. A random sample of 47 creative writing students at an unnamed university were randomly assigned to one of two groups. Each subject was instructed to write two short poems, but the instructions differed by group: one set of instructions alluded to external motivations for writing (e.g., recognition or reward), and the other alluded to internal motivations for writing (e.g., self-expression or personal satisfaction). Poems were scored by judges for creativity on a 40-point scale.
The output below shows a visualization of creativity scores by group, followed by inference on the difference in means.
Welch Two Sample t-test
data: score by treatment
t = -2.9153, df = 43.108, p-value = 0.005618
alternative hypothesis: true difference in means between group Extrinsic and group Intrinsic is not equal to 0
99 percent confidence interval:
-7.9749541 -0.3134517
sample estimates:
mean in group Extrinsic mean in group Intrinsic
15.73913 19.88333
[L5] Based on the boxplots, do the assumptions for the inference shown seem plausible?
[L5] Write the hypothesis and alternative tested in statistical notation.
[L5] Report the result of the 1% level test following conventional style.
[L4] Interpret the confidence interval in context.
[L4] Construct a 95% confidence interval for the difference in means.
Hippocampal volume among schizophrenia patients
Studies have provided evidence that the hippocampus is smaller in schizophrenic patients on average. The dataset for this problem contains data on volumes of the left hippocampus in cubic centimeters for pairs of monozygotic twins; one twin in each pair was affected by schizophrenia and the other was not. The output below shows inference on the mean difference (affected - unaffected) in hippocampal volume between twins in each pair.
One Sample t-test
data: hvolume.diff
t = -3.2289, df = 14, p-value = 0.003031
alternative hypothesis: true mean is less than 0
99 percent confidence interval:
-Inf -0.03718909
sample estimates:
mean of x
-0.1986667
[L5] Write the hypothesis and alternative tested in statistical notation.
[L5] How many pairs of twins were included in the study?
[L5] Report the result of the test following conventional style. (Use the level corresponding to the interval provided.)
[L4] Interpret the confidence interval in context.
[L5] Compute the standard error of the point estimate for the mean difference in hippocampal volumes.