BIOL 243 Lecture Notes - Lecture 10: Statistical Inference, Statistical Hypothesis Testing, Sampling Distribution

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carmineduck394

18 Nov 2016

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Lecture 10 (going back to module 6 & 8) p-values & significance. Regardless of the details, conceptually is a null distribution. Observed value critical value by type 1 error. A hypothesis test is done in which the alternative hypothesis is that more than 10% of the population is left handed. The p-value for the test is calculated to be 0. 25. [correct: the only way to figure this out is to set the null and alternative . Recall type 1 error is your alpha value. We do(cid:374)"t k(cid:374)o(cid:449) a(cid:271)out the type 2 error (cid:271)e(cid:272)ause we do(cid:374)"t k(cid:374)o(cid:449) the alter(cid:374)ati(cid:448)e distri(cid:271)utio(cid:374). We o(cid:374)ly k(cid:374)o(cid:449) the probability under the null. Case study: happiness (marital satisfaction bw on-line vs off-line meeting venues) (cid:862)those (cid:449)ho (cid:373)eet their spouses o(cid:374)li(cid:374)e are less likely to di(cid:448)or(cid:272)e (cid:894)p<0. 002(cid:895) a(cid:374)d (cid:373)ore likely to ha(cid:448)e high marital satisfaction (p<0. 001) than those who meet offline. Meeting online only reduced the divorce rate from 7. 67% to 5. 96%, and barely increased happi(cid:374)ess fro(cid:373)(cid:863)

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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

silversalmon27

In a small paragraph, compare and summarize these two articles.

Article 1 Description

This article is an observational study, which discusses how frequency dependent selection affects two phenotypes of a predator through differential guarding response. The researcher examines the ratio of handedness of two samples of P. microlepis collected from two adjacent sites along Lake Tanganyikaâs northwestern shoreline for a period of eleven years. The process of observations includes observing the correspondence between the handedness of P. microlepis and their targeted prey (Hori, 1993). The researcher carried observations to identify the direction from which P. microlepis attacked the flank of their prey. Moreover, the researcher used a gill net to catch each P. microlepis that attacked the lure. Catching the P. microbes made it possible for the researcher to observe its features and determine its handedness. The researcher determined the handedness of the scales of P. microlepis by observing the fish under a microscope.

During the period of study, Hori (1993) established that the ratio of handedness of P. microbes remained at 0.5 for each one-to-two year intervals throughout the eleven years of the study. Further investigations on temporal changes in handedness ratio showed that the ratio oscillated with amplitude of 0.15 within a five-year period (Hori, 1993). These observations were made for the two species of P. microlepis under investigation. The researcher concludes that the direction of mouth opening of scale eating cichlid fish in Lake Tanganyika is determined by the genetics of the fish. The researcher demonstrates that frequency-dependent natural selection determines the number of left-handedness or right-handedness fish. Based on researcherâs observations, when right-handed fish attack from behind, they snatch the scales from their prey from the left flank (Hori, 1993). On the other hand, when left-handed fish attack their prey from behind, they attack the right flank (Hori, 1993). Therefore, fish with rare phenotypes are more successful as predators than ones that have the more predominant phenotype. This study can be used to guide researchers who are interested in seeking ways of increasing the population of different species of fish. It can also be used as a basis for investigations on how fish type determines its survival tactics or predator tactics.

Article 2 Description

The study was an experimental analysis which sought to quantify phosphoglucose isomerase (Pgi) frequency change over a period of time and compare Pgi genotypes based on physiological responses to high temperatures and subzero thermal extremes encountered by beetles in nature. The research seeks to establish how the local climate in Bishop relates to changes in allele frequency at Pgi and other allozyme loci. It also seeks to establish if ambient temperature of beetle habitants rise to levels that expose beetles to stressful temperature. The researcher also seeks to establish how Pgi genotypes differ physiologically when exposed to subzero temperatures. By answering these questions, the researcher will establish the relationship between changes in enzyme genotype frequency and regional climate changes in addition to investigating major physiological mechanisms through which animals respond to change in climate.

To answer research questions and test research hypothesis, the researchers carried out several experiments and analyses of data. The experimental study involved collecting adult beetles that had emergent recently from pupae for a period of one year between August 1988 and August 1989 (Rank & Dahlhoff, 2002). Their genotype was determined at five different polymorphic enzyme loci. Some of the things analyzed during the study include fluctuations in climate and allele frequency change; daily and annual mean values of air temperature, Effects of Pgi genotype on HSP70 expression, and quantification of Pgi genotype on HSP70 expression among other things (Rank & Dahlhoff, 2002). The study results revealed that although beetle populations were genetically different the FST-values for 1996 were greater for each enzyme locus than those for 1998 (Rank & Dahlhoff, 2002). Moreover, researchers established that the frequency of Pgi allele had increased by 11 percent, a trend that was not observed for all the other loci investigated. Therefore, the researchers conclude that Pgi genotype differences in thermal physiology account for Pgi allele frequency change in nature. Contrary to selected hypothesis that the frequency of Pgi allele should increase with decrease in temperature, the study established that the frequency of Pgi allele decrease with decrease in temperature. This work is important because it provides a basis for more research on how variations in temperature affect beetle populations. The study findings can be used to guide researchers who are interested in seeking ways of protecting endangered beetle populations in different climatic conditions.

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:

1.What occupations are most vulnerable to acquiring cirrhosis of the liver from heavy drinking?

2.List several of the occupations identified by the research studies that are more vulnerable to alcohol-induced, cirrhosis related deaths and that have the highest mortality rates.

3. What are possible confounding variables for the research in occupations and heavy drinking that lead to fatal diseases?

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% 2. Zika and microcephaly Since May 2015,

Brazil has experienced a significant outbreak of Zika virus. In recent years, Brazilian officials reported an increase in the number of babies born with microcephaly.

1. Briefly describe what we know and what we do not know about Zika virus.

2. What has been done and what should be done to prevent Zika virus epidemic in US and world-wide?

Multiple Choice

5. Which of the following statements describes the major advantages of a randomized clinical trial? a. It avoids observer bias b. It lends itself to ethical justification c. It yields results replicable in other patients d. It rules out self-selection of participants to the different treatment groups d. It enrolls representative patients

6. A survey conducted in England revealed that of 224 families in which there had been a known case of poliomyelitis, 56 maintained parakeets as a family pet. In another British survey, 30 out of 99 poliomyelitis patients questioned kept parakeets. The inference that there is some relationship between the presence of a parakeet in a household and the occurrence of poliomyelitis among household members is a. Correct b. Incorrect because of failure to distinguish between incidence and prevalence c. Incorrect because a proportionate ratio is used when a rate is required to support the inference d. Incorrect because a failure to recognize a possible cohort phenomenon e. Incorrect because there is no control or comparison group.

7. An investigator determines the correlation coefficient between triglyceride levels and degree of atherosclerosis in sampled blood vessels to be +1.67. On the basis of this you would conclude that: a. Triglyceride level is a good predictor of atherosclerosis b. Triglyceride level is not a good predictor of atherosclerosis c. High triglyceride levels cause atherosclerosis d. Atherosclerosis cause high triglyceride levels e. The investigator has incorrectly determined the correlation coefficient

8. A screening test of known sensitivity and specificity is applied to two populations. The prevalence of the disease being screened for is 10% in population A and is 1% in population B. Which of the following is true? a. The percent of all negative tests that have false-negative results is lower in population A than in population B. b. Specificity is lower in population A than in population B. c. Reliability is higher in population A than in population B. d. The percent of all positive tests that have false-positive results is lower in population A than in population B. e. Specificity is higher in population A than in population B.

9. Serum cholesterol levels are obtained for four healthy men. The probability that all will fall below the 10th percentile of the distribution of cholesterol for healthy males is: a. 0.4 b. 1 â (0.1)4 c. (0.1)4 d. (0.9)4 e. Cannot be determined from these data

10. In a diabetes screening program, the screening level for a positive blood sugar level in test 1 is set at 160 mg/dl, and in test 2 at 130 mg/dl. The sensitivity is: a. Greater in test 1 b. Greater in test 2 c. Equal in test 1 and 2 d. Dependent on the size of the population being evaluated

BIOL 243 Lecture Notes - Lecture 10: Statistical Inference, Statistical Hypothesis Testing, Sampling Distribution

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