(a)
To explain:
Theconditions for a chi-square goodness of fit test of the given law.
(a)
Explanation of Solution
Given:
Phenotype | Yellow round | Yellow wrinkled | Green round | Green wrinkled |
Count |
Concept used:
Formula
Degree of freedom
Significance level
Calculation:
The sample is random.
The variable under steady is tall or dawn
The expected value of number of observations in each level of the variable is at least
All the conditions for chi-square goodness of fit test satisfied.
Since the sample size are large.
The condition for the two samples
(b)
To explain:
The null and alternative hypothesis fir this test of the given law.
(b)
Explanation of Solution
Given:
Phenotype | Yellow round | Yellow wrinkled | Green round | Green wrinkled |
Count |
Concept used:
Formula
Degree of freedom
Significance level
Calculation:
Since the sample size are large.
The condition for the two samples
Null hypothesis
Alternative hypothesis
Since
So the hypothesis is fail to reject null hypothesis
(c)
To find:
The expected counts under the null hypothesis test of the given law.
(c)
Explanation of Solution
Given:
Phenotype | Yellow round | Yellow wrinkled | Green round | Green wrinkled |
Count |
Concept used:
Formula
Degree of freedom
Significance level
Calculation:
Since the sample size are large.
The condition for the two samples
Applying chi-square test of independence
Since the sample size are large.
The condition for the two samples
Applying chi-square test of independence
Draw the table
Category | Frequency | Observed counts | ||
(d)
To find:
The chi-square statistic and the p-valueof the given law.
(d)
Answer to Problem 21.23E
p-value is greater than
Explanation of Solution
Given:
Tall plants
Dwarf plants
Concept used:
Formula
Degree of freedom
Significance level
Calculation:
Since the sample size are large.
The condition for the two samples
Applying chi-square test of independence
Chi-square
Degree of freedom
P-value
The p-value of the test is
Since p-value is greater than
So, the null hypothesis is fail to reject.
This result is not significant.
So, the data is not support the Mendel’s second law.
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