Answers: Chi-square problems

Homework 6: Chi-square problems

Answers to odd number problems

1.
a. Classic “goodness of fit” problem, we got Mendel’s theory to guide
b. No, although sample size is small for some of the plants, e.g., plant #3 with just 7 wrinkled
c. Null: plant 1: Observed counts fit Mendelian ratio, 3:1; Alternate: Observed counts do not fit Mendelian ratio, 3:1
repeat null/alternate for each plant and for the meta analysis
d.
code example

plant <- c(147,48)
sum(plant)*c(0.75, 0.25)
chisq.test (plant, p =(c(0.75, 0.25)))

e.  One hundred ninety-five (i.e., the individual seeds).

f. The plants are like repeat experiments, the seeds are biological replicates

Remember that experimental (sampling) unit refers to the level at which the treatments are applied. The “treatment” was presence or absence of dominant allele. Each trial (plant) is done independently, i.e., different sampling units, so we can combine all into a single, bigger experiment.

3.
a. “goodness of fit” for dihybdrid cross expect 9:3:3:1 phenotype ratio
b. Null: Observed counts fit Mendelian ratio, 9:3:3:1; Alternate: Observed counts do not fit Mendelian ratio, 9:3:3:1

The test of the null
chisq.test (c(926, 288, 293, 104), p =(c(9/16, 3/16, 3/16, 1/16)))

Chi-squared test for given probabilities

data: c(926, 288, 293, 104)
X-squared = 1.4687, df = 3, p-value = 0.6895

we fail to reject the null hypothesis, which means we accept that Mendel’s theory predicts the outcome

5.
a. Contingency table problem, no theory
b. Null: No association between albumin excretion rates and mortality rates; Alternate: There is association between albumin excretion rates and mortality rates.

The test of null (note — you should rearrange the provided table to fit our format: columns = outcome, rows = treatment or exposure)

rows Deaths Survived
Low 55 73
Normal 59 17

X-squared = 23.24, df = 1, p-value = 0.00000143

We reject null hypothesis, there was difference mortality rates between albumin excretion rates