The white "Spirit" black bear (or Kermode) Ursus americanus kermodei, differs from the ordinary black bear by a single amino acid change in the melanocortin 1 receptor gene (MC1R). In this population, the gene has two forms (or alleles): the "white" allele b and the "black" allele B. The trait is recessive: white bears have two copies of the white allele of this gene (bb), whereas a bear is black if it has one or two copies of the black allele (Bb or BB). Both color morphs and all three genotypes are found together in the bear population of the northwest coast of British Columbia. If possessing the white allele has no effect on growth, survival, reproductive success, or mating patterns of individual bears, then the frequency of individuals with 0, 1, or 2 copies of the white allele (b) in the population will follow a binomial distribution. To investigate, Hedrick and Ritland (2011) sampled and genotype 87 bears from the northwest coast:42 were BB
24 were Bb
21 were bb
Assume that this is a random sample.
A formal hypothesis test was carried out to compare the observed and expected frequencies of genotypes. The procedure obtained P = 0.0001.
1. "The frequency distribution of genotypes has a binomial distribution in the population" is the________ hypothesis, whereas "The frequency distribution of genotypes does not have a binomial distribution" is the _________ hypothesis.
2. The degrees of freedom for the test statistic are __________.
Say whether the each of the following statements is true or false solely on the basis of these results:
3. The difference between the observed and expected frequencies is statistically significant.________
4. The test statistic exceeds the critical value corresponding to α = 0.05. ___________
5. The test statistic exceeds the critical value corresponding to α = 0.01. ____________
Answers
Answer:
1) i ) Null hypothesis, ii) Alternate hypothesis
2) Degree of Freedom = 14.9011
3) True
4) True
5) True
Step-by-step explanation:
1) "The frequency distribution of genotypes has a binomial distribution in the population" is the NULL hypothesis, whereas "The frequency distribution of genotypes does not have a binomial distribution" is the ALTERNATE hypothesis.
2) Degrees of freedom: see attached calculation for
3) True, the difference between the observed and expected frequencies is statistically significant since p-value < ∝alpha(0.05)
4) True
when, 14.9011 > 3.8415 where ∝ = 0.05
when, p-value < alpha,TS > critical value
here p-value < 0.0001 < 0.05
5) True
when, 14.9011 > 6.635 where ∝ = 0.01
when, p-value < alpha,TS > critical value
here p-value < 0.0001 < 0.01
Final answer:
The null hypothesis states that the genotype frequency distribution follows a binomial distribution, with the alternative hypothesis stating the contrary. The degrees of freedom for the test statistic are two. The P value of 0.0001 is statistically significant, thus the test statistic exceeds the critical values corresponding to both α = 0.05 and α = 0.01.
Explanation:
1. The frequency distribution of genotypes has a binomial distribution in the population is the null hypothesis, whereas The frequency distribution of genotypes does not have a binomial distribution is the alternative hypothesis.
2. For this chi-square test, since there are 3 possible outcomes (BB, Bb, bb) the degrees of freedom are 3-1 = 2.
3. The difference between the observed and expected frequencies is statistically significant, that is true. A P value of 0.0001 is highly significant, clearly less than 0.05, and denotes a significant difference.
4. The test statistic exceeds the critical value corresponding to α = 0.05. This is true, as the statistically significant low P value indicates the test statistic is in the critical region.
5. The test statistic exceeds the critical value corresponding to α = 0.01. This is also true.
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how do i evaluate 0.05
Answers
Answers
Answer:
18 = number of players
Step-by-step explanation:
Giving the following information:
Members of a softball team raised $2039.50 to go to a tournament. They rented a bus for $1157.50 and budgeted $49 per player for meals.
To calculate the total number of players they can bring, we need to use the following formula:
Total amount of money= fixed cost + unitary variable cost*number of players
2,039.5= 1,157.5 + 49*number of players
882/49= number of players
18 = number of players
The team can bring a maximum of 18 players to the tournament given the amount they raised and the budgeted expenses.
Let's use "x" to represent the number of players the team can bring to the tournament.
The total amount raised by the team is $2039.50,
and they rented a bus for $1157.50. Each player's meal will cost $49.
The total amount spent on the bus and meals for x players can be represented as follows:
Total Expenses = Bus Cost + (Number of Players) * (Cost per Player's Meal)
= $1157.50 + x * $49
Since the team's total expenses should not exceed the total amount raised,
$2039.50 ≥ $1157.50 + x * $49
$2039.50 - $1157.50 ≥ x * $49
$882 ≥ x * $49
Now, divide both sides by $49 to solve for x:
x ≤ $882 / $49
x ≤ 18
So, the team can bring a maximum of 18 players.
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Answers
Step-by-step explanation:
inequality form would be x<-12 or >2
I hope this helps
Answer:
2
Step-by-step explanation:
(x+5)>7
x+5>7
x>7-5
x>2
Answers
Answer:
Variance: 322.4479999999996
Standard Deviation: 17.956837137981722
Final answer:
To calculate the variance and standard deviation for the given sample set of data, find the sample mean, calculate the squared differences, and then find the sample variance and standard deviation.
Explanation:
To calculate the variance and standard deviation for the given sample set of data (83.6, 92.3, 56.5, 43.8, 77.1, 66.7), follow these steps:
- Calculate the sample mean by adding all the values together and dividing by the total number of values: (83.6 + 92.3 + 56.5 + 43.8 + 77.1 + 66.7) / 6 = 69.8.
- Calculate the squared differences between each value and the sample mean: (83.6 - 69.8)^2, (92.3 - 69.8)^2, (56.5 - 69.8)^2, (43.8 - 69.8)^2, (77.1 - 69.8)^2, (66.7 - 69.8)^2.
- Calculate the sample variance by summing up the squared differences and dividing by (n-1), where n is the total number of values: (83.6 - 69.8)^2 + (92.3 - 69.8)^2 + (56.5 - 69.8)^2 + (43.8 - 69.8)^2 + (77.1 - 69.8)^2 + (66.7 - 69.8)^2 = 300.46. Sample variance = 300.46 / 5 = 60.1.
- Calculate the sample standard deviation by taking the square root of the sample variance: √60.1 = 7.79. Rounded to the nearest tenth, the sample standard deviation = 7.8.
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Answers
Answer:
9
Step-by-step explanation:
Expected number of defective is given as :
N x P = np
Where = number of the sample drawn from the production lot
P= percentage of bulbs In The lot which are defective
N = 15
P = 60% = 0.60
So when we do the multiplication,
We have:
NP = 15x0.60
= 9
So in conclusion the expected value of defective bulbs in the sample is 9.
Answers
Answer:
d. a reflection across the y-axis