# ANALYSIS OF BIOLOGICAL DATA PDF

The Analysis of Biological Data - Whitlock and Schluter Solutions to assignment problems - PLEASE DO NOT POST Chapter 1 (a) Discrete (b) Continuous. This books (The Analysis of Biological Data [PDF]) Made by Michael C. Whitlock About Books Knowledge of statistics is essential in modern. Get this from a library! The analysis of biological data. [Michael Whitlock; Dolph Schluter].

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PDF | On Jan 8, , Michael C Whitlock and others published DOWNLOAD PDF The Analysis of Biological Data. The Analysis of Biological Data | 𝗥𝗲𝗾𝘂𝗲𝘀𝘁 𝗣𝗗𝗙 on ResearchGate | On Jan 1, , Dolph Schluter and others published The Analysis of Biological Data. This edition adds 4 chapters (20 to 23) that introduce the use of generalized linear models with S-Plus, particularly logistic regression and models for analyzing count data. The first 19 chapters use examples from Whitlock and Schluters excellent book, "The Analysis of Biological.

Chapter 1 Observational study. The researcher has no control over which women have miscarriages and which lose their fetus from other causes. If younger individuals are more likely to use a cell phone, omitting cell phones would bias the sample towards older individuals. The multiple trees within the same plot might not be independent, if they are related, of similar age, or share the same environment. The 60 samples are not a random sample.

Males from the two populations have the same probability of being chosen i. Females choose one type of male over the other i. Note that the estimate of the parameter differs from the null hypothesis that the proportion is 0. The study probably reported a P-value of 0. The correct interpretation is that under the null hypothesis, the probability is 0.

It is not inconceivable that snakes would choose the cooler site. Snakes have no temperature preference in resting sites. Snakes prefer one temperature over the other. Chapter 7 Those who had sustained injury may have been more likely to take time to tell their tale of horror than those who had not. Because there is no sample of the population, the interpretation of the confidence interval makes no sense, so standard confidence interval calculations are not warranted.

Confidence interval: With many sampled populations, some ca. For a two-tailed test, we must multiply the sum by 2. This does not meet the 0. In other words, the precision of the estimate of the proportion improves. With larger sample sizes, smaller proportional differences can be detected.

The probabilities that bound the test statistic are given below, along with the precise values calculated by Excel. Alternate hypothesis: The mean is 0. We must combine categories to avoid expected numbers of deaths per regiment year less than 1. Our test statistic is less than 5. Alternative hypothesis: You could calculate the probability of observing weekend births out of total using the binomial distribution.

However, the calculations would be tedious for such large numbers, as you would need to sum the probabilities of 0 to weekend births. As an alternative, you could use a goodness of fit test to calculate the test statistic and compare it to the chi-square distribution.

This is approximate, and in some cases the data may violate assumptions for the test although not in this particular case , but it is simple and fast to calculate. Plots with more than 3 truffles had to be combined with plots with three truffles to avoid expected frequencies of less than 1.

There are four categories, one estimated parameter, so 2 df. The truffles are clumped. There are more plots with high and low numbers of truffles than expected by the Poisson distribution, and too few with the mean number. The standard error is the standard deviation here, 0. Next, we calculate the probability of each number of male fish out of six fish, from 0 to 6 we need to calculate the probabilities for events even if they do not appear.

For instance, to calculate the probability of zero males, we use: Because the expected values ere not high enough, we combined the categories at the higher end. We can reject the null hypothesis that the males are distributed with equal and independent probability for each outcrop. So there is an excess of outcrops with exactly one male. It appears that hurricanes are approximately Poisson distributed: So, what does this say about forecasts for "a bad hurricane year"?

The odds ratio is 3.

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Duct tape is a better cure for warts than liquid nitrogen in this study: Use a contingency table to test whether the proportion of males surviving differs between hybrid and pure offspring.

Most pied flycatchers should mate with the more-common collared flycatchers if this were true. To cope with this, we might combine categories. However there is no biologically meaningful way to combine categories so that the expected value of all cells is 5 or more.

One approach might be to combine all females that mated more than once together, and then do Fisher's exact test on the resulting 2x2 table. Males may be most attracted to the best-nourished females, which would also be those with the highest probability of giving birth. Moreover, females who mated more often may live longer. Randomized assignment of females to number of matings would be one way to explicitly test the connection between number of matings and probability of giving birth.

The contingency table presented in the question assumes that each climber has an independent probability of mortality. However, many factors affect mortality, and on high mountains, weather can have a very serious effect and would hit all members of a climbing party at once. Similarly, an entire team might have the same acclimation regime, which might affect mortality.

Therefore, this table is guilty of pseudo-replication: This would call for Fisher's exact test. All of the predicted outcomes will be less than five, which would violate the assumptions of the chi-square contingency test.

This would call for Fisher's exact test, as three of the four cells will have an expected value of less than five, violating one of the assumptions of the chi-square contingency test. You can see this without actually calculating the expected values: The odds ratio was 0. The confidence interval for this requires using the ln odds ratio, ln 0. The confidence interval overlaps one, so we are not confident that the two groups differ in their ability to detect lying. The researchers assume an order of cause and effect.

It may be that people who are depressed are more likely to keep a diary. It may also be that a third factor influences both of these variables. This is an observational study, so it is not possible to ascribe cause and effect. We convert to a standard normal deviate Roughly a quarter of American males were too tall to qualify. A negligible proportion of women are excluded for being too tall. We need to calculate a minimum height for women that excludes 0.

We look up the probability of 0. Z is negative as this is the fraction excluded at the bottom of the height class. Now, we need to calculate the minimum height, H, that will yield a Z of H - There are many possible solutions here, including bimodality, skew, finite distributions. As this is below the mean, Z will be negative.

Thus the 25th percentile occurs about 0. We expect the 25th percentile state to have approximately As the sample size grows, the SE will decrease since we are dividing by the square root of n , so we can assign the sample size based on the decreasing standard deviation.

We wish to test whether bees distinguish between flowers with crab spiders or not. What is the probability of observing 24 trials where bees choose flowers with the spider?

We convert this to a standard normal deviate: A proportion 0. So, bees do not choose at random: Also, the researchers might have had different sample sizes, so even if the standard deviation had been the same the standard error would differ. By chance, the larger sample may have had a much higher sample standard deviation, causing it to have a broader confidence interval. For 7 df, The mean mating index is We reject the null hypothesis that there is no assortative mating based on size in sticklebacks.

Mean weight is We test whether the mean sampled weight, The critical value for t0. Chapter 12 The difference in white blood cell count is 1. We reject the null hypothesis: The null difference in means is zero. The calculated degrees of freedom are The difference in means is The pooled sample variance is 8. We reject the null hypothesis that the copulation times are equal for the two circadian rhythm mutations.

Two-sample t-test: Reject the null hypothesis; males are more aggressive when mated with a neighboring female. There are twelve pairs of data, so we use 11 df in finding t for the confidence interval. The value for t0. The standard error of the difference is 0. The null hypothesis is that the relatedness does not differ between helpers and non-helpers.

The observed difference in relatedness is 0. The difference is 1. The standard error is 0. To calculate the standard deviation from the standard error, multiply by the square root of the sample size: We cannot reject the null hypothesis that the conductivity does not differ between adult and baby dolphin blubber.

The confidence interval is: There is clearly a difference in oxygen consumption during feeding dives. The mean difference is 2. The toughness varies depending on the direction. You would want to do this on a random sample of humans, taking several samples from each person and using them to calculate mean toughness in each direction for that person. Since each sample from a different person is not independent, you should use a paired t-test to compare the differences in toughness in each direction, treating the difference for each person as a single data point.

No, this is not a valid statement. Drug X had some effect on chilblains and drug Y did not have a statistically significant effect. However, this did not mean that drug Y had no effect, which is the implication of concluding that drug X is better than drug Y based strictly on the two independent tests.

Chapter 13 There are no ties for the PHA response, so assigning ranks is easy. The alternate hypothesis is that the median number of sexual partners differs between the two groups. U is the larger of U1 and U2, or We reject the null hypothesis. Were the sex ratios the same in the two groups? Sexual behavior and the reporting of it might well differ between the sexes. It would be useful to have the survey taken under conditions guaranteeing anonymity, to increase accuracy of the answers.

Alternatively, a Mann-Whitney U could be used if the data are still highly skewed after transformation. Average for non-territorial log GnRH: Average for territorial log GnRH: Histogram of differences in species number for climbing and non-climbing clades.

We will use a sign test to see if more the climbing clade has more species more often than would be expected by chance. Therefore, we could use Welch's t-test. The variance increases as the mean increases, so the log transformation might help. We can reject the null hypothesis that these two have the same skin color. The log-transformed data are approximately normal with roughly equal standard deviations, so we can use a two-sample t-test.

Yes, babies differ in their exposure to smoke. Babies were not assigned randomly to smoking or non- smoking households. We can use the sign test instead. The distribution is left skewed, and has both positive and negative values. These differences are not normally distributed. With a sign test: The study should have a control group receiving a placebo treatment.

Without it we cannot estimate the effect of the treatment. Replication; balance same numbers of treated and untreated eyes ; blocking treated and untreated eyes were paired ; control untreated eyes; sham surgery or transplant from a blind cave fish would have provided a more complete control ; randomization eye to be treated was chosen randomly on each fish.

Ironically, blinding was not used. No likely effect on sampling error. No effect on bias. No clear effect on sampling error. No effect on sampling error. This is a paired design, so use the sample size formula for paired t-test in Quick Formula Summary.

This variation will make it more difficult to detect a treatment effect in the completely randomized design in which different treatments are applied to separate groups of subjects.

The paired design eliminates the subject-to-subject variation, increasing the power of the test of treatment effect. Experimental studies randomly assign treatments to experimental units, reducing bias by breaking associations between confounding variables and the explanatory variable.

This allows the causal relationship between the explanatory and response variables to be assessed. Random assignment is not possible in observational studies, and therefore they can never completely eliminate the effects of confounding factors. The diclofenac gel treatment was the control for the leech treatment. This might have influenced their expectations of the benefits of treatment, and so their responses and the outcome of the experiment. The paired design would eliminate this source of sampling error on the estimate of reaction times.

You would also need to specify the desired width of the confidence interval. Sex is a confounding variable. Imbalance of sample size reduces power. A factorial design. By including all combinations of treatments, it allows the measurement of the effects of each variable age and diet restriction separately, and the effects of their interaction.

Chapter 15 If this fails, use the Kruskal-Wallis test if the distributions have equal shape. The variances are unequal in the two groups, and the data do not appear to be normally distributed in all the groups. These assumptions do not appear to be met in the present data. In flies given a first blood meal from a cow, the measurements do not appear to be normally distributed and the variance is low compared with flies given a first blood meal from a lizard.

The two panels on the right of the figure above show the proportions after transformation. This has indeed largely fixed the problem: First blood meal affects the mean proportion of flies taking their second blood meal from cows.

In addition, a large sample size increases the power of the test. In addition, a balanced design increases the power of the test compared with an unbalanced design having the same total sample size. Show the data. Mouse strains in the population do not differ in the mean number of minutes spent in the open. Mouse strains in the population differ in the mean number of minutes spent in the open.

Conclude that mouse strains in the population vary in the mean number of minutes spent in the open. Some treatments have much higher variance than others. The assumption that the measurements are normally distributed within populations might also be violated for some groups e. This assumption is clearly violated, so we cannot conclude that the means are different. All we can conclude from the Kruskal-Wallis result is that the distributions are different, but not necessarily their means or medians.

The answer is b. Nevertheless, it is possible that there is no difference among age groups. This is slightly less than the repeatability of the femur measurement, indicating that head size has a higher proportion of its total variation attributable to measurement error. Conclude that mean cone size differs between habitat types. The untreated mice provide a baseline measurement, allowing the researchers to determine the effect of the sham surgery.

ANOVA is not robust to the violation of equal standard deviations under these conditions. After transforming, the differences in the standard deviations are less extreme: Group Sample mean Y Standard deviation s Enhanced 5.

Conclude that mean dilution differs between treatments. Mean weight is We reject the null hypothesis: You would want to do this on a random sample of humans. There is clearly a difference in oxygen consumption during feeding dives.

Since each sample from a different person is not independent. We cannot reject the null hypothesis that the conductivity does not differ between adult and baby dolphin blubber. To conclude that drug X is more effective. To calculate the standard deviation from the standard error. The null hypothesis is that there is no difference. The observed difference in relatedness is 0. The standard error is 0.

The null hypothesis is that the relatedness does not differ between helpers and non-helpers. The mean difference is 2. The toughness varies depending on the direction. The confidence interval is: There are twelve pairs of data. The difference is 1. Drug X had some effect on chilblains and drug Y did not have a statistically significant effect. The standard error of the difference is 0. The value for t0. It would be useful to have the survey taken under conditions guaranteeing anonymity.

There are no ties for the PHA response. Average for non-territorial log GnRH: For this sample size. Were the sex ratios the same in the two groups? Sexual behavior and the reporting of it might well differ between the sexes. Chapter 13 Histogram of differences in species number for climbing and non-climbing clades.

The alternate hypothesis is that the median number of sexual partners differs between the two groups. U is the larger of U1 and U2. Average for territorial log GnRH: Babies were not assigned randomly to smoking or non- smoking households. The distribution is left skewed. There are 48 clades. We will use a sign test to see if more the climbing clade has more species more often than would be expected by chance.

We can reject the null hypothesis that these two have the same skin color. The log-transformed data are approximately normal with roughly equal standard deviations. The variance increases as the mean increases.

We can use the sign test instead. Without it we cannot estimate the effect of the treatment. No likely effect on sampling error.

With a sign test: No effect on bias. This is a paired design. No clear effect on sampling error. Random assignment is not possible in observational studies. The study should have a control group receiving a placebo treatment. The paired design eliminates the subject-to-subject variation. Experimental studies randomly assign treatments to experimental units.

No effect on sampling error. These differences are not normally distributed. This variation will make it more difficult to detect a treatment effect in the completely randomized design in which different treatments are applied to separate groups of subjects.. This allows the causal relationship between the explanatory and response variables to be assessed. This might have influenced their expectations of the benefits of treatment.

A factorial design. Imbalance of sample size reduces power. Sex is a confounding variable. The variances are unequal in the two groups. Chapter 15 The paired design would eliminate this source of sampling error on the estimate of reaction times. By including all combinations of treatments. If this fails. You would also need to specify the desired width of the confidence interval. The diclofenac gel treatment was the control for the leech treatment.

This has indeed largely fixed the problem: In addition. The two panels on the right of the figure above show the proportions after transformation. In flies given a first blood meal from a cow. First blood meal affects the mean proportion of flies taking their second blood meal from cows.

Show the data. These assumptions do not appear to be met in the present data. Variance within groups. Mouse strains in the population do not differ in the mean number of minutes spent in the open. Conclude that mouse strains in the population vary in the mean number of minutes spent in the open.

## The analysis of biological data

The assumption that the measurements are normally distributed within populations might also be violated for some groups e. Mouse strains in the population differ in the mean number of minutes spent in the open. This assumption is clearly violated. The answer is b.

Some treatments have much higher variance than others. All we can conclude from the Kruskal-Wallis result is that the distributions are different. The untreated mice provide a baseline measurement. After transforming. Error 0. Since the observed F-ratio is greater. ANOVA is not robust to the violation of equal standard deviations under these conditions.

Conclude that mean cone size differs between habitat types. Group Sample mean Y Standard deviation s Enhanced 5. This is slightly less than the repeatability of the femur measurement.

Conclude that mean dilution differs between treatments. Conclude that second language proficiency and grey matter density are correlated. The repeatability is 0. Since t is greater than t0. Reject H0. Chapter 16 Conclude that we cannot reject the null hypothesis of zero correlation.

Since rS is not greater than or equal to rS 0. An experiment would be necessary to test whether proficiency affects grey matter. If transformations fail to remedy the problem. Perhaps individuals with high grey matter densities are able to achieve a high proficiency in a second language.. For example. The researchers did not assign subjects to different values of slow wave sleep increase.

A log transformation of homicide rate and an arcsine transformation of percent left handed after dividing by also gave a satisfactory outcome though the log appeared slightly better. The log transformation is always a good one to try when variables are right-skewed and values are greater than zero. Conclude that is a positive correlation between increase in slow-wave sleep and increase in performance.

Conclude that homicide rate and percent left-handed individuals in societies are correlated. This could explain the results of the second team of researchers despite using the same population and sample size. A log-transformation of distance improves matters considerably. Results using log transformation of homicide and arcsine of percent left-handed: Since t is not greater than or equal to t0.

A log transformation of recruitment helps matters a bit more. Results using the log-log transformation: Log transformation of both variables: Do not reject H0. Association between treatment a categorical variable with two groups is measured by the difference between the means of the two groups rather than with a correlation coefficient.

Conclude that the null hypothesis of no difference between the means of the two groups is not rejected by these data. Conclude that adding more nutrients reduces the number of plant species supported. Chapter 17 The response variable of interest Y is the number of plant species supported. Conclude that the slopes are not significantly different among penguin groups. Do not reject. Assume that the variance in Y is the same at every X. Conclude that the slope of the linear relationship exponent of the power function is not significantly different from 0.

The slope of the relationship between ln metabolic rate and ln mass is not 0. The slope of the relationship between ln metabolic rate and ln mass is 0.

Conclude that there is a positive relationship between fleck duration and growth. Assume that the relationship is linear. Square root transformed number of spores rises with increasing host longevity but then appears to decline at the longest host lifespans.

If that fails. As a result. Chapter 18 8. They give the most plausible range of values for the mean Y actual year of birth corresponding to each X estimated year of birth. The slope is close to 1. On this basis. But the extrapolation is risky: The predicted values from the model are plotted along the X axis. Conclude that the slope is significantly different from zero. In the scatter plot. It is often effective when the variable is a count.

If we extrapolate from the regression based non-human primates. This gives the most plausible range of values for a single Y measurement actual year of birth corresponding to each X estimated year of birth. That the relationship between CV and age of maturation is linear and has the same slope in both exploitation groups.

There is no interaction between the effects of species and body mass on brain size. Standard error bars could be added to this plot. There is an interaction between the effects of species and body mass on brain size Or. Being unable to reject the null hypothesis or no interaction. The predicted values lie along two parallel regression lines. The non-parallel lines in the interaction plot suggests an interaction is present between morning glory species and family the effect of morning glory species on development time is not the same in all families.

They are not repeatable a future study of the same population would not use the same families nor of direct interest. The morning glory species are repeatable groups of direct interest. They are not a random sample of morning glory species. The assumption of a normal distribution is violated. Confidence interval based on the t-distribution might be appropriate after transformation.

Since randomization tends to break up associations. TIME is the development time in days. Mann-Whitney U-test. B is a randomized data set because there is little or no relationship between the two variables. Chapter 19 Randomization test. It describes the most plausible values of the number of generations since the origin of the huntingtin mutation. Chapter 20 9. Each individual shrub had the same probability of being sampled and sampling of individuals was independent.

Conclude that compensatory mutations are clumped. The fraction of patients improving p is not 0. The fraction of patients improving p is 0. Bootstrapping is an excellent way to calculate the uncertainty of the trimmed mean. The results of bootstrap replicate estimates of the trimmed mean are shown in the following histogram.

The distribution includes outliers and the trimmed mean is an objective way to increase the precision of the estimate. Of the bootstrap estimates. For the light data. There is almost complete overlap with the confidence interval for males. The wasps prefer the mated females.

The Expected frequency is 16 for both categories. This is no coincidence: The proportion of mated females chosen by the wasps is not 0.

The proportion of mated females chosen by the wasps is 0. More patients improve than deteriorate. Meta-analysis has potentially greater objectivity. In this case we should be confident in the results.

Infants prefer the direct gaze over indirect gaze. If so. Flag for inappropriate content. Related titles.

## The analysis of biological data / Michael Whitlock and Dolph Schluter - Details - Trove

Jump to Page. Search inside document. No sneaker One sneaker Two or more sneakers Total Eggs eaten 61 18 16 95 No eggs eaten 17 4 Total 35 20 with the mode at a young age. Kevin Gian. Vishnu Prakash Singh.

## Whitlock and Schluter-The Analysis of Biological Data Solutions Manual (2008).pdf

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