Question

1-

Labrador retrievers can be one of three colours: black, chocolate brown, and yellow. The proportion of puppies born with each coat colour can be predicted by the genotypes of the parents. One set of parents produced 50 puppies (in six litters), and we've recorded the observed numbers of puppies with each coat color. We only know the coat colours of the parents and not their exact genotype at the locus determining coat colour, and we want to test our hypothesis about the parental genotypes by comparing the observed puppy coat colours to the expected numbers from our hypothesis. Complete the following table by calculating each chi-square statistic.

Coat colour	Expected Frequency	Observed Number	Chi-square statistic
Black	0.25	12	Answer
Chocolate	0.25	17	Answer
Yellow	0.5	21	Answer
Total	1	50	Answer

2- The colour of the fur of mice is determined by three genes, which can produce five different colour phenotypes (brown, black, piebald, Himalayan, and albino).

You want to run a chi-squared goodness of fit test to compare observed counts to expected frequencies under a specific inheritance scenario. What are the degrees of freedom for this analysis?

3- You have measured the diameters of 20 trees and have saved the data in the following vector in R. What are the correct ways to calculate the standard error of the mean in R?

diams<-c(4.61, 4.73, 2.50, 8.11, 4.14, 8.36, 4.48, 9.46, 6.71, 9.70, 3.45, 6.79, 3.44, 8.57, 2.80, 8.54, 3.82, 3.80, 7.22, 7.80) 

Select one or more:

a.

sem<-sqrt(var(diams)/length(diams))

b.

sd(diams)/sqrt(nrow(diams))

c.

sqrt(var(diams)/nrow(diams))

d.

var(diams)/length(diams)

e.

sd(diams)/sqrt(length(diams))

4- You have a sample of the diameters of 100 trees, and you know that the sample variance is equal to 4mm. What is the standard error of the mean?

Answer 1

Question (a)

Coat colour	Expected Frequency (E)	Observed Number (O)	O- E	(O-E)2	Chi-square statistic= (O-E)2/ E
Black	0.25	12	11.75	138.0625	552.25
Chocolate	0.25	17	16.75	280.5625	1122.25
Yellow	0.5	21	20.5	420.25	840.5
Total	1	50	49	2401	2401

Question (b)

The degrees of freedom for a chi-square test = (r-1) * (c-1)

Here r = 3 since there are three genes

c = 5 since there are five colour phenotypes

The degrees of freedom for a chi-square test = (3-1) * (5-1)

= 2 * 4

= 8

Question (c)

The formual for standard error = /

Where is the Standard Deviation SD

n is the sample size or length of the sample

So Answer is Option E which is sd(diams)/sqrt(length(diams))

Question (d)

Given Sample size n = 100

Variance = 4mm

Standard deviation SD =

=

= 2

Standard error = SD /

= 2 /

= 2/10

= 0.2

So Standard error of the mean = 0.2