Biology 2244A/B Lecture 13: Lecture 13 – Hypothesis Testing 3

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7 Feb 2017

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That our hypothesized value of p is equal to the real value of p. If the p value is really small, we reject the null hypothesis. If our p value is less than or equal to 0. 05, too unlikely to happen with chance alone reject null. Any p value that is not 0, it is possible to get a sample like that from a sampling distribution. Any proportions that happen 5% of the time or less we are going to reject. If you set your significance level to 5% Let"s assu(cid:373)e (cid:374)ull h(cid:455)pothesis is true: assuming p = p0, null hypothesis dictates the sampling distribution of the statistic that we are using to estimate that parameter. Null hypothesis is true = hypothesized vale of p = real value of p. If probability is very small = reject the null hypothesis = when less than or equal to significance level.

Related Questions

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

silversalmon27

How does the biology community currently feel with regards topublishing descriptive and effect size statistics rather thansignificance stats? Almost every journal article I read in the cellbiology field almost always reports things like P values and statstests to report statistical significance, but should effect sizesbe more important to a biologist? Do we even care if something isstatistically significant if the effect size is negligible? Ratherthan crunch for significance, could one get away with showingthings like confidence intervals, eta^2, Cohen's d, and r valuesinstead over P values? P values tell you the odds that if youassume the null hypothesis is true, then the observation you'remaking are only 5%(assuming of course P<0.05). However, this canlead to the logical fallacy as noted by Aristotle--theory Apredicts that changing X will cause Y. An experimenter thusperforms experiments to manipulate X and sees changes in Y,therefore he/she concludes theory A is supported, which is howevercompletely wrong. Theories B, C, D, E....... could all also predictthat X changes Y and may even be better at it. Even if you concludethat your findings "support" theory A, it's still weak because youhaven't ruled out all of the other possibilities.

So in order to avoid statistical significance relative to nullhypothesis that has all sorts of pitfalls, can one just usedescriptive and effect size statistics just as effectively, if notmore so?

Need help with evolution questions:

1.) A haplotype is best defined as the ________________.

	haploid genotypes of all the gametes produced by a diploid individual

ABO blood type conferred by an individual gamete

genotype of either the paternal or maternal chromosomal complement

multilocus genotype of a chromosome or gamete

2.) Which of the following statements regarding linkage disequilibrium is true?

Exists when D is less than zero, but not when it is greater than zero.

Is reduced by sexual reproduction.

Is increased by crossing-over during meiosis.

Is increased by any random sampling error that happens to create or destroy certain chromosome genotypes but not others.

Is reduced by selection that favors certain combinations of genotypes.

3.) Selection on multilocus genotypes in random-mating populations leads to linkage disequilibrium when _______________.

it reduces mutation rates

it eliminates all haplotypes from the population

it prevents crossing-over during meiosis

some allele combinations confer greater fitness than do others

4.) When two loci are in linkage equilibrium, the allele carried on a chromosome at one locus is independent of the allele carried on the chromosome at another locus. When this is the case, _________________.

allele frequencies will not change over time

the two loci (or, more specifically, their protein products) do not interact

the frequency of each haplotype can be calculated by multiplying the frequency of its two alleles

all possible haplotypes occur at equal frequency

5.) Which of the following accurately characterize(s) Muller's ratchet model for the selective advantage of sexual reproduction?

Drift establishes linkage disequilibrium because multilocus genotypes with few mutations are eliminated by chance events.

In populations without sexual reproduction, deleterious mutations accumulate, eventually imposing such a burden that the population goes extinct.

Sex is favored selectively because it recreates the zero-mutation genotypes lost due to drift.

All of the above correctly characterize Muller's ratchet model for the evolution of sexual reproduction.

6.) The genetic signature of recent positive selection on a locus is _______________.

high frequency and low linkage disequilibrium

low frequency and low linkage disequilibrium

low frequency and high linkage disequilibrium

high frequency and high linkage disequilibrium

7.) Sexual reproduction is advantageous both in changing environments and in constant environments, compared to asexual reproduction.

True

False

8.) Sexual reproduction causes genetic recombination via crossing-over, which combines genotypes from different parents.

True

False

9.) One reason sex may be maintained in natural populations is proposed by the "Red Queen Hypothesis", which states that.....

genotypes created by sex are created by natural selection over and over

recombination creating novel genotypes that can resist infection by parasites

mutation load is reduced by novel adaptive mutations

the ability of sex to fix multiple

Biology 2244A/B Lecture 13: Lecture 13 – Hypothesis Testing 3

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