What is the critical F value when the sample size for the numerator is seven and the sample size for the denominator is six
1 answer:
Answer:
Critical F value = 4.9503
Step-by-step explanation:
Given that:
The sample size of the numerator = 7
The sample size of the denominator = 6
The degree of freedom for the numerator df = n -1
The degree of freedom for the numerator df = 7 - 1
The degree of freedom for the numerator df = 6
The degree of freedom for the denominator df = n - 1
The degree of freedom for the denominator df = 6 - 1
The degree of freedom for the denominator df = 5
The assume that the test is two tailed and using a level of significance of ∝ = 0.10
The significance level for the two tailed test = 0.10/2 = 0.05
From the standard normal F table at the level of significance of 0.05
Critical F value = 4.9503
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Answer:
a) 0.984
b) 20 people
Step-by-step explanation:
a)
If The probability that a person carries the gene is 0.1, then in a sample of 20 people, 2 should carry the gene.
Now, we want to know how many samples there are with this property.
Since we have 20 elements where 18 are alike (do not carry the gene) and 2 are alike (carry the gene), we have to compute the number of permutations of 20 elements in which 18 are alike and 2 are alike. This number is
In this 190 20-tuples there are only 3 where the 2 carriers of the gene are in the first 3 places, namely
(1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
(1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
(0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
where 1=carries the gene, 0=does not carry the gene.
So there are 190-3 = 187 elements in which the first 3 elements have no 2 carriers, hence the probability that 4 or more people will have to be tested until 2 of them with the gene are detected is 187/190 = 0.984 (98.4%) rounded to three decimal places.
b)
Given that the probability that a person carries the gene is 0.1, then in a sample of 20 people, 20*0.1 = 2 should carry the gene.