Observed Phenotypic Values for the Monohybrid Cross Green Albino Total Expected Observed Data Table 3: Photograph of Germinated Seeds Photograph: Data Table 4: Chi-square Values for the Monohybrid Cross Values: Answers: Chi-square value Degrees of freedom P-value Questions: A) If all of the yellow (albino) seedlings were removed from the population‚ would the next generation still have a chance of displaying the albino phenotype? Explain your answer. B) How did the expected ratio of green
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would be that differences would not exist. c. What are the degrees of freedom for 1) gender‚ 2) marital status‚ 3) interaction between gender and marital status‚ and 4) error or within variance? 1. A-1= There are 2 groups of gender; male and female‚ so it would be 2-1=1 as the degree of freedom. 2. B-1= There are 3 groups for marital status; married‚ single‚ and divorced‚ so it would be 3-1=2 as the degree of freedom. 3. (A-1)*(B-1) = That would be the answer from gender
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study‚ are Factor‚ Error‚ andTotal. The factor is the characteristic that defines the populations being compared. In the tire study‚ the factor is the brand of tire. In the learning study‚ the factor is the learning method. (2) DF means "the degrees of freedom in the source." (3) SS means "the sum of squares due to the source." (4) MS means "the mean sum of squares due to the source." (5) F means "the F-statistic." (6) P means "the P-value." Now‚ let’s consider the row headings: (1) Factor means
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< 2e-16 *** my -0.25862 0.01292 -20.018 < 2e-16 *** mz -0.36557 0.01292 -28.296 < 2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 1.809 on 2043 degrees of freedom Multiple R-squared: 0.8874‚ Adjusted R-squared: 0.8872 F-statistic: 4026 on 4 and 2043 DF‚ p-value: < 2.2e-16 > data=read.table("d:/111113/2.txt"‚header=T) Call: lm(formula = S ~ u_direction + mx + my + mz‚ data = data)
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Chi-Square Test Chi-square is a statistical test commonly used to compare observed data with data we would expect to obtain according to a specific hypothesis. For example‚ if‚ according to Mendel’s laws‚ you expected 10 of 20 offspring from a cross to be male and the actual observed number was 8 males‚ then you might want to know about the "goodness to fit" between the observed and expected. Were the deviations (differences between observed and expected) the result of chance‚ or were they due to
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temperature‚ pressure‚ also includes the rotary kiln cylinder of different period of body surface temperature and electric current of the motor and seal. Operators for temperature control‚ is generally 500 degrees before maintaining temperature at 20 degrees per hour‚ 500 degrees is controlled at 50 degrees per hour‚ and will fill in the details of the drying kiln in the note column. Drying kiln‚ drying kiln temperature curve is very important‚ are not allowed to make the high or low temperature of drying
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X+Y 7. Determine the degrees of freedom. This is the number of categories (red eyes or sepia eyes) minus one. For the data in Case 1‚ what is the number of degrees of freedom? There are 2 categories minus 1. The degrees of freedom is 1. 8. Find the probability (p) value for 1 degree of freedom in the 0.05 row. Compare this with the chi–square value you calculated in your Lab Notebook. What can you say about the null hypothesis? The probability value for 1 degree of freedom in the 0.05 row is 3
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genotype of the parental (before the F1 generation; not shown here) generation? X+X+ x XwY 7. Determine the degrees of freedom. This is the number of categories (red eyes or sepia eyes) minus one. For the data in Case 1‚ what is the number of degrees of freedom? There are 2 categories minus 1. The degrees of freedom is therefore 1. 8. Find the probability (p) value for 1 degree of freedom in the 0.05 row. Compare this with the chi–square value you calculated in your Lab Notebook. What can you say
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both the streamlined bob and the plastic sphere into a single dataset is a good idea. A good fit has: The chi-squared statistic approximately equal to the number of degrees of freedom. Residuals that are randomly distributed about zero. A poor fit has: A chi-squared much greater than or much less than the number of degrees of freedom. Residuals that show a systematic deviation from zero. One or more fitted parameters that are zero within
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Problem Description: The main purpose of this report is to show how to solve a 3-D finite element model of a cantilever I-shape beam‚ which is subjected to two concentrated loads (P = 1600 lb.) at the flanges of the free end along z axis. In this assignment‚ a convergence study will be used to determine the convergence of the solution with respect to mesh refinement. In addition‚ it will be used to achieve an accurate solution for problems that have sufficiently dense mesh‚ which cannot be solve
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