wt ANOVA Sum of Squares df Mean Square F Sig. Between Groups 707.725 3 235.908 2.335 .072 Within Groups 1009791.503 9996 101.020 Total 1010499.228 9999 Figure 3
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1). The p-value for this test is . 072. Do we reject the null hypothesis?
2). Is there sufficient evidence to conclude that there is a statistically significant difference in mean weight between the different NYHA groups?
3). Pretend for a moment that the resulting p-value was 0.02. Do we reject the null hypothesis in this case? Why or why not? If we were to reject the null hypothesis, we would conclude that there is sufficient evidence to suggest a statistically significant difference in mean weight between the NYHA groups. The NYHA variable has four levels: would we know which of the four NYHA group means gave the statistically significant result?
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- Sum 4 Please help mePretend that you are comparing the actual genotype distribution for a population with the distribution of genotypes predicted by the Hardy-Weinberg theorem. So you hypothesize that the population is in Hardy-Weinberg equilibrium (i.e. that actual population data fit the Hardy-Weinberg expectations). If you carry out chi-square goodness of fit test and calculate a total chi-square value of 0.03 with 1 degree of freedom (see table), what does this mean?Question 1: A 'wild-type' allele is the allele that is ..(1).... a population, and that encodes a ..(2)...... functional gene-product. (1) Select: least common in missing from most common in (2) Select: non partially fully Question 2. The 'wild-type' phenotype is the most common phenotype observed in a population Select True or False
- The equation p + 2pq + q = 1 representing the Hardy-Weinberg proportions examines genes with only two alleles in a population. a. Derive a similar equation describing the equilib- rium proportions of genotypes for a gene with three alleles. [Hint: Remember that the Hardy- Weinberg equation can be written as the binomial expansion (p + q)*.] b. A single gene with three alleles (r, 1", and i) is responsible for the ABO blood groups. Individuals with blood type A can be either A or ^ i; those with blood type B can be either /* 1 or * i; people with AB blood are r*, and type O individuals are ii. Among Armenians, the frequency of ^ is 0.360, the frequency of 1 is 0.104, and the frequency of i is 0.536. Calculate the frequencies of individuals in this population with the four possible blood types, assuming Hardy-Weinberg equilibrium. %3D5. In a population genetic study of a species of California wild oat, the following genotype frequencies were found for a trait (hairiness of the leaf sheath) which is controlled by a single locus: HH 0.571 Hh 0.071 hh 0.358 a) What are the expected frequencies of these genotypes under Hardy-Weinberg? b) Calculate F. c) Briefly explain what the value in b) implies regarding heterozygosity and suggest what might be going on in this population to explain this result.Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 90 individuals and H5/H5=85 individuals. Considering the Hardy Weinberg equilibrium and calculating the Chi-square statistic, do you reject or fail to reject the null-hypothesis? 1. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P<0.05. Hence, I reject the null hypothesis. 2. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P>0.05. Hence, I reject the null hypothesis. 3. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P<0.05. Hence, I fail to reject the null hypothesis. 4. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the…
- 17) In chi-squared analysis... A. If the observed data exactly agree with the predictions of the hypothesis, then we expect the p-value to be (1.0, 100, less than 0.05, 0, greater than 0.05) B. If the observed data closely agree with the predictions of the hypothesis, then we expect the p-value to be (1.0, 100, less than 0.05, 0, greater than 0.05) C. If the observed data do not at all agree with the predictions of the hypothesis, then we expect the p-value to be (1.0, 100, less than 0.05, 0, greater than 0.05) Chose one from the parenthesis for each question.please help me with this questionWhat are the assumptions required for Hardy-Weinberg equilibrium? Explain.
- Was the control group valid in the Tuskegee study? Why yes or why not? Can you please help me with this question63. In an isolated population, a single mutation allele of the cystic fibrosis transmembrane regulator (CFTR) gene is present at a frequency of 0.1. The normal CFTR allele and the mutation allele are in Hardy-Weinberg equilibrium in this population. Which of the following is the approximate frequency of homozygosity for the normal CFTR allele in this population? A) 0 OB) 0.01 C) 0.05 D) 0.2 E) 0.5 F) 0.8 G) 1.0 Сони back to(a) Explain sensitivity and selectivity in terms of true and false positives. (b) What is the advantage of a Needleman-Wunsch alignment compared to a seeded alignment? (c) What does the expectation parameter mean in local alignment? What will happen if the expectation value is increased from its default value of 10 to a 100?