SourceSumМeanDegreesof FreedomFp-valueof Variationof SquaresSquareTreatmentsErrorTotal(f) At the a = 0.05 level of significance, test whether the means for the three treatments are equal.State the null and alternative hypotheses.O Ho: HA = HB = HCH: Not all the population means are equal.O Ho: At least two of the population means are equal.H: At least two of the population means are different.O Ho: HA = HB = HCH: HA # HB # HCHo: Not all the population means are equal.Ha: HA = HB = HCHo: HA# HB # HCHa: HA = HB = HCFind the value of the test statistic. (Round your answer to two decimal places.)Find the p-value. (Round your answer to four decimal places.)p-value =State your conclusion.O Do not reject Ho. There is not sufficient evidence to conclude that the means for the three treatments are not equal.O Reject Ho: There is not sufficient evidence to conclude that the means for the three treatments are not equal.Reject Ho. There is sufficient evidence to conclude that the means for the three treatments are not equal.Do not reject Ho. There is sufficient evidence to conclude that the means for the three treatments are not equal. TreatmentA162141125143157123165125139145143141149137150178155132Sample157143135meanSamplevariance186.8140.8106.0(a) Compute the sum of squares between treatments.(b) Compute the mean square between treatments.(c) Compute the sum of squares due to error.(d) Compute the mean square due to error. (Round your answer to two decimal places.)(e) Set up the ANOVA table for this problem. (Round your values for MSE and F to two decimal places, and your p-value to fourdecimal places.)MeanSourceof VariationDegreesof FreedomSumFp-valueof SquaresSquareTreatmentsErrorTotal(f) At the a = 0.05 level of significance, test whether the means for the three treatments are equal.State the null and alternative hypotheses.O Ho: HA = HB = HCH: Not all the population means are equal.O Ho: At least two of the population means are equal.H: At least two of the population means are different.O Ho: HA = HB = HCH: HA# HB # HcO Ho: Not all the population means are equal.H: HA = HB = HC%3DO Ho: HA + HB * HCH3: HA = HB = HcFind the value of the test statistic. (Round your answer to two decimal places.)Find the pn-value (Round vour answer to four decimal places

Treatment A B 162 141 125 143 157 123 165 125 139 145 143 141 149 137 150 178 155 132 Sample 157 143 135 mean Sample 186.8 140.8 106.0 variance (a) Compute the sum of squares between treatments. (b) Compute the mean square between treatments. (c) Compute the sum of squares due to error. (d) Compute the mean square due to error. (Round your answer to two decimal places.) (e) Set up the ANOVA table for this problem. (Round your values for MSE and F to two decimal places, and your p-value to four decimal places.) Degrees of Freedom Source Sum Mean p-value of Variation of Squares Square Treatments Error Total (f) At the a = 0.05 level of significance, test whether the means for the three treatments are equal. State the null and alternative hypotheses. O Ho: HA = HB = HC H: Not all the population means are equal. O Ho: At least two of the population means are equal. H: At least two of the population means are different. O Ho: HA = HB = HC H: HA# HB # HC Ho: Not all the population means are equal. H: HA = HB = Hc O Ho: HA + HB * HC H: HA = HB = Hc %3! Find the value of the test statistic. (Round your answer to two decimal places.)

Treatment A B 162 141 125 143 157 123 165 125 139 145 143 141 149 137 150 178 155 132 Sample 157 143 135 mean Sample 186.8 140.8 106.0 variance (a) Compute the sum of squares between treatments. (b) Compute the mean square between treatments. (c) Compute the sum of squares due to error. (d) Compute the mean square due to error. (Round your answer to two decimal places.) (e) Set up the ANOVA table for this problem. (Round your values for MSE and F to two decimal places, and your p-value to four decimal places.) Degrees of Freedom Source Sum Mean p-value of Variation of Squares Square Treatments Error Total (f) At the a = 0.05 level of significance, test whether the means for the three treatments are equal. State the null and alternative hypotheses. O Ho: HA = HB = HC H: Not all the population means are equal. O Ho: At least two of the population means are equal. H: At least two of the population means are different. O Ho: HA = HB = HC H: HA# HB # HC Ho: Not all the population means are equal. H: HA = HB = Hc O Ho: HA + HB * HC H: HA = HB = Hc %3! Find the value of the test statistic. (Round your answer to two decimal places.)

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)

6th Edition

ISBN:9781337111348

Author:Bruce Crauder, Benny Evans, Alan Noell

Publisher:Bruce Crauder, Benny Evans, Alan Noell

Chapter5: A Survey Of Other Common Functions

Section5.6: Higher-degree Polynomials And Rational Functions

Problem 5E: Population Genetics In the study of population genetics, an important measure of inbreeding is the...

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Population Genetics In the study of population genetics, an important measure of inbreeding is the proportion of homozygous genotypesthat is, instances in which the two alleles carried at a particular site on an individuals chromosomes are both the same. For population in which blood-related individual mate, them is a higher than expected frequency of homozygous individuals. Examples of such populations include endangered or rare species, selectively bred breeds, and isolated populations. in general. the frequency of homozygous children from mating of blood-related parents is greater than that for children from unrelated parents Measured over a large number of generations, the proportion of heterozygous genotypesthat is, nonhomozygous genotypeschanges by a constant factor 1 from generation to generation. The factor 1 is a number between 0 and 1. If 1=0.75, for example then the proportion of heterozygous individuals in the population decreases by 25 in each generation In this case, after 10 generations, the proportion of heterozygous individuals in the population decreases by 94.37, since 0.7510=0.0563, or 5.63. In other words, 94.37 of the population is homozygous. For specific types of matings, the proportion of heterozygous genotypes can be related to that of previous generations and is found from an equation. For mating between siblings 1 can be determined as the largest value of for which 2=12+14. This equation comes from carefully accounting for the genotypes for the present generation the 2 term in terms of those previous two generations represented by for the parents generation and by the constant term of the grandparents generation. a Find both solutions to the quadratic equation above and identify which is 1 use a horizontal span of 1 to 1 in this exercise and the following exercise. b After 5 generations, what proportion of the population will be homozygous? c After 20 generations, what proportion of the population will be homozygous?
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