The variance isa. a measure of the variation around the mean.b. computed as a squared deviation.c. higher when there is less phenotypic variation.d. Both a and b are correct.
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The variance is
a. a measure of the variation around the mean.
b. computed as a squared deviation.
c. higher when there is less
d. Both a and b are correct.
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- A characteristic has a narrow-sense heritability of 0.6.a. If the dominance variance (VD) increases and all other variancecomponents remain the same, what will happen to narrow-senseheritability? Will it increase, decrease, or remain the same? Explain.b. What will happen to broad-sense heritability? Explain.c. If the environmental variance (VE) increases and all other variancecomponents remain the same, what will happen to narrow-senseheritability? Explain.d. What will happen to broad-sense heritability? Explain.If you perform a Chi-Squared analysis and determine that your result is greater than the critical value, what would you do? a. Reject H0, accept HA b. Accept H0 c. neitherA characteristic has a narrow-sense heritability of 0.6. a. If the dominance variance (VD) increases and all other variance components remain the same, what will happen to narrow-sense heritability? Will it increase, decrease, or remain the same? Explain. b. What will happen to broad-sense heritability? Explain. c. If the environmental variance (VE) increases and all other variance components remain the same, what will happen to narrow-sense heritability? Explain.
- Describe in your own words what variance measures. Then explain why the variance of both samples must be used in the calculation of a t-test (In other words, explain why we have to factor in the variance when comparing the means).I'm needing help with part 1 through 3, please The data of Sugar Water and Plain Water: mean # of drops variance degrees of freedom t stat t crit (from the table) plain water 25.3 17.221 38 -4.01 2.0261 sugar water 21.3 21.256 1. Make a statement about t stat in relation to t crit. And what is the p-value? 2. Is there a statistically significant difference between the mean values (mean # of drops) of the sugar and plain water? Give your "yes" or "no" answer and the supporting evidence. 3. Hypothesis: The surface tension of plain water is higher than the surface tension of sugar water. Do these results support or refute the hypothesis about the effect of dissolving salt in water on surface tension?If the environmental variance (VE) increases and all other variance components remainthe same, what will the effect be?a. Broad-sense heritability will decrease.b. Broad-sense heritability will increase.c. Narrow-sense heritability will increase.d. Broad-sense heritability will increase, but narrow-sense heritability will decrease.
- How are the standard deviation and variance related to each other? How are they different than just measuring the range? Can plot samples of dandelion cover from two different locations have the same mean but different variance? Explain. Can you answer question 3? Thanks.The heights of mothers and daughters are given in the following table: Height of mother (in) Height of daughter (in) 64 66 65 66 66 68 64 65 63 65 63 62 59 62 62 64 61 63 60 62 a. Calculate the correlation coefficient for the heights of the mothers and daughters. b. Using regression, predict the expected height of a daughter whose mother is 67 inches tall.One basic purpose of experimental design is to reduce the number of participants needed O to control variance to keep the alpha level below.05 O to observe contingencies
- Can you explain how to get the percentage in this problem? Thank you! V. The comparison of the class blood group with the national average distribution is given as follows like: O+ = 19 AB+ = 5 A+ = 7 B+ = 2 so the table for national average distribution is: 1.Blood type Percentage National Average Distribution A+ 21.21% 24.64% B+ 6.06% 23.84% AB+ 15.15% 46% O+ 57.57% 46.94%Thank you for your comprehensive answer. May I follow up with another question on how to interpret this Maximum Likelihood tree attached to this text?Example 3.5.7 Fish Vertebrae Consider the distribution of vertebrae given in Table 3.5.1. In Exam- ple 3.5.5 we found that the mean of Y is µy = 21.49. The variance of Y is VAR(Y) = of = (20 – 21.49) x Pr{Y = 20} 21) %3D + (21 – 21.49)? x Pr{Y + (22 – 21.49)? x Pr{Y = 22} + (23 – 21.49) x Pr{Y = 23} = (-1.49) x 0.03 + (-.49) x 0.51 + (0.51) x 0.40 + (1.51)² × 0.06 = 2.2201 x 0.03 + .2401 x 0.51 + .2601 x 0.40 + 2.2801 x 0.06 = 0.066603 + 0.122451 + 0.10404 + 0.136806 = 0.4299. The standard deviation of Y is ay = V0.4299 - 0.6557. %3D
