BIOL 300 Chapter Notes - Chapter 4: Sampling Distribution, Confidence Interval, Statistical Parameter

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May 22nd, 2018 BIOL 300

Week 1 Reading

Chapter 4: Estimating with Uncertainty

- The saple ea ( is used to estiate the true ea of the populatio µ)

- The sample standard deviation (s) is used to estimate the population standard deviation

(õ)

- The sample proportion (ñ) is an estimate of the population proportion (p)

- For an estimate of a population parameter to be useful, we also need to quantify its

precision. We need a measure of how far the estimate is likely to be from the target

parameter being estimated. If precision is high, then our uncertainty is low. We can be

reasonably confident that our estimate is close to the truth. If instead precision is low,

then our uncertainty is high ad e’ll eed ore data to redue it.

4.1 The sampling distribution of an estimate

- Estimation is the process of inferring a population parameter from sample data

o The value of an estimate calculated from data is almost never exactly the same

as the value of the population parameter being estimated, because sampling is

influenced by chance

- We use the sampling distribution of the estimate, which is the probability distribution of

all the values for an estimate that we might have obtained when we sampled the

population

o The sampling distribution is the probability distribution of all values for an

estimate that we might obtain when we sample a population

o The saplig distriutio represets the populatio of alues for a estiate

▪ It is not a real population

o Rather, the sampling distribution is an imaginary population of values for an

estimate

- Increasing sample size reduces the spread of the sampling distribution of an estimate,

increasing precision

4.2 Measuring the uncertainty of an estimate

- The standard deviation if the sampling distribution of an estimate is called the standard

error

o Because it reflects the differences between an estimate and the target

parameter, the standard error reflects the precision of an estimate

o Estimates with smaller standard errors are more precise than those with larger

standard errors

o The standard error of a estiate is the stadard deiatio of the estiate’s

sampling distribution

- We represet the stadard error of the ea ith the sol ō( because of its

remarkal straightforard relatioship ith ō the populatio stadard deiatio of Y)

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 



  

o Standard error decreases with increasing sample size, according to the equation

above

- The troule ith the forula for the stadard error of the ea ō() is that we almost

eer ko the alue of the populatio stadard deiatio ō, ad so e aot

calculate it

o The next best thing is to approximate the standard error of the mean by using

the saple stadard deiatio s as a estiate of ō

o To show that this is an approximation, we use SE(

 



  

o According to this relationship, all we need is one random sample to approximate

the spread of the etire saplig distriutio for (

o The standard error of the mean is estimated from data as the sample standard

deviation (s) divided by the square root of the sample size (n)

4.3 Confidence intervals

- The confidence interval is another common way to quantify uncertainty about the value

of a parameter

o It is a range of number calculated from the data that is likely to contain within its

span the unknown value of the target parameter

o A confidence interval is a range of value surrounding the sample estimate that is

likely to contain the population parameter

- An example is the 95% confidence interval for the mean

o This confidence interval is a range likely to contain the value of the true

population mean µ

o It is calculated from the data and extends above and below the sample estimate

(

- For the 95% confidence interval, you have a lover limit and an upper limit. This

alulatio allos us to sa e are 95% confident that the true mean lies between ____

ad _____

o All numbers falling between the lower and upper bounds of a confidence interval

can be regarded as the most plausible values for the parameter, given the data

sampled

- In general, the width of the 95% confidence interval is a good measure of our

uncertainty about the true value of the parameter

o If the confidence interval is broad, then uncertainty is high and the data are not

very informative about the location of the population parameter

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Document Summary

The sa(cid:373)ple (cid:373)ea(cid:374) (cid:894)((cid:895) is used to esti(cid:373)ate the true (cid:373)ea(cid:374) of the populatio(cid:374) (cid:894) ) The sample standard deviation (s) is used to estimate the population standard deviation. The sample proportion ( ) is an estimate of the population proportion (p) For an estimate of a population parameter to be useful, we also need to quantify its precision. We need a measure of how far the estimate is likely to be from the target parameter being estimated. If precision is high, then our uncertainty is low. We can be reasonably confident that our estimate is close to the truth. If instead precision is low, then our uncertainty is high a(cid:374)d (cid:449)e"ll (cid:374)eed (cid:373)ore data to redu(cid:272)e it. It is not a real population: rather, the sampling distribution is an imaginary population of values for an estimate. Increasing sample size reduces the spread of the sampling distribution of an estimate, increasing precision.

Related Questions

Recently I have found myself in a discussion with a guy thatisn't a creationist, but denies evolution. He cites several studiesthat show problems like this one, but I do not have the knowledgeto refute it properly. Whats the problem with the following:

One problem is that population genetics shows us evolutiondestroys functional sequences much faster than it creates them inany large-genome animal with low reproductive rates, becausedeleterious mutations arrive faster than selection can remove them.This would include just about all mammals, reptiles, and birds--anyof our would-be ancestors over the last 300m years.

Specifically, humans get 60 to 160 mutations per generation,10-20+% of our genome is sensitive to substitution, and this givesus 6-32 deleterious mutations per offspring (most slight). Evenwith the low estimate of 6, every child is less fit than theirparents and natural (or even artificial) selection can only choosethe least degenerate each generation. For every generation where amutation can be selected, random mutations destroy 5-32+ previouslyselected. Beneficial mutations like lactose tolerance (a jammedswitch) take 1000s of years to appear and fixate.

If you know the genome size, mutation rate, and reproductiverate, you can calculate at what rate deleterious mutations arrivefaster than they can be removed:

"It has been estimated that there are as many as 100 newmutations in the genome of each individual human. If even a smallfraction of these mutations are deleterious and removed byselection, it is difficult to explain how human populations couldhave survived. If the effects of mutations act in a multiplicativemanner, the proportion of individuals that become selectivelyeliminated from the population (proportion of `genetic deaths') is1-e^-U, where U is the deleterious mutation rate per diploid, so ahigh rate of deleterious mutation (U>>1) is paradoxical in aspecies with a low reproductive rate." High genomic deleteriousmutation rates in hominids, (Nature, 1999)

They worry about a deleterious rate of U much greater than 1 asbeing prohibitively high and we're talking about it being 6-32+,since we now know the functional genome is much larger than wethought it was in 1999. Taking their Poisson probabilitydistribution and using U=6, that means 1-e^-6 = 99.752% of thepopulation will have to be selected away each generation for onelucky enough to have no deleterious mutations. For two to surviveand maintain constant population size that would require on average806 offspring per female--impossible for most mammals, birds, andreptiles. Actually much more since natural selection is notomnisciently efficient.

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Part II 1.The California Occupational Mortality study data set was employed to assess mortality data for the years 1979-1981. A 2% sample of employed persons from the 1980 census of California was used. It contains the occupation of each person who had died. This study, like any occupation-based study, is restricted to the work life span as far as age of subjects is concerned. This study used ages 16-64 years. Certain persons, because of their source of work, were excluded as subjects: homemakers, retired persons, students, disabled, military personnel, etc. Only main-stream-type employment was used in this study to establish occupations at risk for heavy alcohol drinking. Using the Table below, answer the following:

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4. Based on available research data, develop and construct a web of causation along with the appropriate decision trees for occupational group-related alcohol deaths. Table. Alcohol Related Deaths in California, 1979-1981 Cirrhosis 5.5% Digestive organ cancers 5.7% Injuries 13.7% Suicide and homicide 5.% All other causes 64.9%

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Introduction: A Chi-square test is used to compare observed data with expected data according to a hypothesis. For instance, if you were crossbreeding 2 heterozygous pea plants, you would expect to see a 3:1 phenotypic ratio in the offspring. In this case, if you were to breed 400 pea plants, you would expect to see 300 plants showing the dominant trait and 100 showing the recessive trait. But what happens if you observe only 260 plants with the dominant trait and 140 plants with the recessive trait? Does this mean something is wrong with Mendelian genetics or is this difference in expected results just due to chance (random sampling error)? These are the questions that can be answered using Chi-square statistics. The results of this statistical test is used to either reject or accept (fail to reject) the null hypothesis. The null hypothesis states there is no significant difference between the observed results and the expected results. This means that if the null hypothesis is accepted, the difference in observed and expected results was just a matter of chance and so the observed results basically "fit" with what was expected. Degrees of freedom (df) = number of independent outcomes (Y) being compared less 1 df = Y-1 At the 95% confidence interval we are 95% confident that there is a significant difference between the observed and expected results, therefore rejecting the null hypothesis. Probability Value - Is the decimal value determined from the X2 table and is the probability of accepting the null hypothesis. A 0.05 probability value equates to a 95% confidence interval.

The Chi-squared test formula is: Example: If we cross two pea plants that are heterozygous yellow pods, we would expect a 3:1 phenotypic ratio. So let's say we actually did the cross and got 280 plants with green pods and 120 plants with yellow pods. Question: Is this a 3:1 phenotypic ratio? This is the value of Chi-squared Test. We have a total of 400 plants and we expect a 300 green:100 yellow phenotypic ratio If the calculated Chi-squared value is less than the critical value listed in the Chi-squared table, then we accept the null hypothesis. This means that there is no significant difference between the observed and the expected values. Our degrees of freedom (df) = 2 outcomes - 1, or df = 1. Now we go the X2 table below and using the df = 1 and probability value of 0.05, our critical value is 3.84. Since our calculated X2 value is 5.33, and is larger than the critical value, we reject the null hypothesis and can say (at 95% confidence) that there is a significant difference between our observed and expected values.

The parent generation is yellowed podded and green podded pea plants. You cross a yellow podded pea plant with a green podded pea plant and you get 100% yellow podded plants in the F1 Generation (Phenotypic ratio 4 : 0, yellow to green). What will be the expected phenotypic ratio when you allow the F1 generation to reproduce?

Fill out the Punnett square.

If we actually did the cross and got 1150 yellow and 350 green. Would this be a consistent with what was expected?

Learning Outcomes Questions

1. Why would you run a Chi-squared test?

To determine if our data is consistent with expected results.
		a To determine if our data is consistent with expected results. b To determine if our data exactly matches the expected results.
		c To determine the expected results.
	d	To compare the phenotypic ratios to the genotypic ratios.

2. Determine the degrees of Freedom of the phenotypic ratio for this genetic cross.

a. 1

b. 2

c. 3

d. 4

e. 5

3. Using the data given, what is the result of your Chi-squared analysis? x2= ___.

	a.	2.22
	b	2.71
	c	4.36
	d	187.78
	e	448.27

4. Using the results of your Chi-squared analysis, do we fail to reject or reject the null hypothesis?

a.		Fail to reject the null
b.		Reject the null
c.		It cannot be determined from the data given

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BIOL 300 Chapter Notes - Chapter 4: Sampling Distribution, Confidence Interval, Statistical Parameter

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