Finding expected frequencies for Chi-Square Genotype Total (+++) 85 (---) 109 (-++) 33 (++-) 59...
Finding expected frequencies for Chi-Square
Genotype
Total
(+++) 85
(---) 109
(-++) 33
(++-) 59
(+--) 161
(-+-) 49
(--+) 68
(+-+) 200
Total
764
Given this data, how would I calculate the expected frequency for a X^2 analysis?
Homework Answers
Add Answer to:
Finding expected frequencies for Chi-Square
Genotype
Total
(+++) 85
(---) 109
(-++) 33
(++-) 59...
For a particular gene associated with a degenerative neurological disorder, you surveyed genotype frequencies in a...
For a particular gene associated with a degenerative neurological disorder, you surveyed genotype frequencies in a population and found that 16 people had genotype AA, 97 had genotype Aa, 15 had genotype aa. a) Calculate the allele frequencies. b) Calculate the expected genotype frequencies under Hardy-Weinberg Equilibrium. Use the chi-square test to determine whether the observed number of people with different genotypes and expected number of people with different genotypes are significantly different (at an alpha level of 0.05). What...
I need serious help with the chi-square part of this problem (X2). I have posted this...
I need serious help with the chi-square part of this problem
(X2). I have posted this problem before and gotten the
incorrect answer, looked up previously answered questions that were
similar, and tried it myself more than a few times. If someone can
please help and show a fairly detailed step process as to how to
figure this out, that would be greatly appreciated.
Mapt Phenylketonuria (PKU) is caused by homozygosity for a recessive allele and can cause a severe...
For a chi-square test of homogeneity for this contingency table, what are the expected frequencies for the first row, rounded to two decimal places?
A clinic administers two drugs to two groups of randomly assigned patients to cure the same disease: 70 patients received Drug / and 80 patients received Drug II. The following table gives the information about the numbers of patients cured and those not cured by each of these two drugs.DrugCuredNot CuredI6010II5921For a chi-square test of homogeneity for this contingency table, what are the expected frequencies for the first row, rounded to two decimal places?
For a chi-square test of homogeneity for this contingency table, what are the expected frequencies for the second column, rounded to two decimal places?
A clinic administers two drugs to two groups of randomly assigned patients to cure the same disease: 70 patients received Drug 1 and 80 patients received Drug 11 . The following table gives the information about the numbers of patients cured and those not cured by each of these two drugs.DragCuredNet CuredI6010II5921For a chi-square test of homogeneity for this contingency table, what are the expected frequencies for the second column, rounded to two decimal places?
Hi fellow tutor, Please note that from part A-J have been answered. I would like some...
Hi fellow tutor, Please note that from part A-J have been answered. I would like some help please with parts K-N. Thank you Question 5: Applying Mathematical Analysis to Evolutionary Change (17 points) Coloration patterns of a species of snake are controlled by a single gene with two alleles (A= dominant, a=recessive). One population of snakes living in North Carolina was sampled and analyzed using RFLP analysis (restriction fragment length polymorphism), a technique that cuts DNA in specific regions that...
The occurrence of the NN blood group genotype in the US population is 1 in 400,...
The occurrence of the NN blood group genotype in the US population is 1 in 400, consider NN as the homozygous recessive genotype in this population. You sample 1,000 individuals from a large population for the MN blood group, which can easily be measured since co-dominance is involved (i.e., you can detect the heterozygotes). They are typed accordingly: BLOOD TYPE GENOTYPE NUMBER OF INDIVIDUALS RESULTING FREQUENCY M MM 490 0.49 MN MN 420 0.42 N NN 90 0.09 Using the...
Question 1 In a chi -square test of independence , the null hypothesis states that-----------. a....
Question 1 In a chi -square test of independence , the null hypothesis states that-----------. a. the two variables of interest are related in the population. b. the column frequencies equal the row frequencies c. the sum of the and the column frequencies equal the total frequency d. the two variables of interest are unrelated in the population Question 2 A survey asked people whether they identified as a morning person, night person, or had no preference. a. 270 replied...
7. The table below shows the frequencies of the PMCT118 alleles in several different human populations....
7. The table below shows the frequencies of the PMCT118 alleles in several different human populations. Allele Frequency (n = number of alleles typed) Allele Allele U.S. Caucasians (n = 200) African Americans (n=214) U.S. Caucasians (n = 200) African Americans (n=214) Hispanic Americans (n=210) 30 0.008 0.008 0.055 0.08 0.058 0.013 0.003 0.048 0.005 0.005 0.073 Hispanic Americans (n=210) 0.003 0.003 0.013 0.263 0.005 0.02 0.025 0.028 0.003 0.318 0.055 0.008 0.048 0.098 0.003 0.033 0.115 0.088 0.023 0.193...
Window Help 12 2 Secion Exencise 9-10 (oritical vaue For the given table of observed frequencies: A 15 7 10 B 3 109 c 16 6 11 (a) Compute the row totals, the column totals, and the grand total (b...
Window Help 12 2 Secion Exencise 9-10 (oritical vaue For the given table of observed frequencies: A 15 7 10 B 3 109 c 16 6 11 (a) Compute the row totals, the column totals, and the grand total (b) Construct the corresponding table of expected frequencies. (c) Compute the value of the chi-square statistic (d) How mamy degrees of freedom are there? (e) tf appropriate, perform a test of independence, using the α-0.05 level of significance, trot appropriate, explain...
How do you answer if your data support the expected ratio? Use a chi-square analysis to...
How do you answer if your data support the expected
ratio?
Use a chi-square analysis to test the expected ratio for the F2 of Cross C. Phenotype Observed Expected Expected (O-E) (O-E) (O-E) IE Number Fraction Number Yellow 10082 10248.0625 - 1610.0625 2516,75391 2.69 13/16 Purple 2531 3/16 2364.9375 166.0625 27576.7539111966 Total = Chi-square 14.35 Number of degrees of freedom= 2-1 = 1 Probability of a match= p<0.05 On the basis of your chi-square test, does your data support the...
I need serious help with the chi-square part of this problem
(X2). I have posted this problem before and gotten the
incorrect answer, looked up previously answered questions that were
similar, and tried it myself more than a few times. If someone can
please help and show a fairly detailed step process as to how to
figure this out, that would be greatly appreciated.
Mapt Phenylketonuria (PKU) is caused by homozygosity for a recessive allele and can cause a severe...
7. The table below shows the frequencies of the PMCT118 alleles in several different human populations. Allele Frequency (n = number of alleles typed) Allele Allele U.S. Caucasians (n = 200) African Americans (n=214) U.S. Caucasians (n = 200) African Americans (n=214) Hispanic Americans (n=210) 30 0.008 0.008 0.055 0.08 0.058 0.013 0.003 0.048 0.005 0.005 0.073 Hispanic Americans (n=210) 0.003 0.003 0.013 0.263 0.005 0.02 0.025 0.028 0.003 0.318 0.055 0.008 0.048 0.098 0.003 0.033 0.115 0.088 0.023 0.193...
Window Help 12 2 Secion Exencise 9-10 (oritical vaue For the given table of observed frequencies: A 15 7 10 B 3 109 c 16 6 11 (a) Compute the row totals, the column totals, and the grand total (b) Construct the corresponding table of expected frequencies. (c) Compute the value of the chi-square statistic (d) How mamy degrees of freedom are there? (e) tf appropriate, perform a test of independence, using the α-0.05 level of significance, trot appropriate, explain...
How do you answer if your data support the expected
ratio?
Use a chi-square analysis to test the expected ratio for the F2 of Cross C. Phenotype Observed Expected Expected (O-E) (O-E) (O-E) IE Number Fraction Number Yellow 10082 10248.0625 - 1610.0625 2516,75391 2.69 13/16 Purple 2531 3/16 2364.9375 166.0625 27576.7539111966 Total = Chi-square 14.35 Number of degrees of freedom= 2-1 = 1 Probability of a match= p<0.05 On the basis of your chi-square test, does your data support the...
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