MATHEMATICS COLLEGE

A resident of Bayport claims to the City Council that the proportion of Westside residents (1) with income below the poverty level is lower than the proportion of Eastside residents
(2) The City Council decides to test this claim by collecting a random sample of resident incomes from the Westside of town and a random sample of resident incomes from the Eastside of town. Seventy-six out of 578 Westside residents had an income below the poverty level. Hundred-and-twelve out of 688 Eastside residents had an income below the poverty Specify the hypotheses.
Calculate the value of the test statistic (round to 4 decimal places).
Calculate the p-value (round to 4 decimal places).

Answers

Answer 1

Answer:

The test statistics is $z = -1.56$

The p-value is $p-value = 0.05938$

Step-by-step explanation:

From the question we are told

The West side sample size is $n_1 = 578$

The number of residents on the West side with income below poverty level is $k = 76$

The East side sample size $n_2=688$

The number of residents on the East side with income below poverty level is $u = 112$

The null hypothesis is $H_o : p_1 = p_2$

The alternative hypothesis is $H_a : p_1 < p_2$

Generally the sample proportion of West side is

$\^(p) _1 = (k)/(n_1)$

=> $\^(p) _1 = (76)/(578)$

=> $\^(p) _1 = 0.1315$

Generally the sample proportion of West side is

$\^(p) _2 = (u)/(n_2)$

=> $\^(p) _2 = (112)/(688)$

=> $\^(p) _2 = 0.1628$

Generally the pooled sample proportion is mathematically represented as

$p = (k + u)/( n_1 + n_2 )$

=> $p = (76 + 112)/( 578 + 688 )$

=> $p =0.1485$

Generally the test statistics is mathematically represented as

$z = \frac{\^ {p}_1 - \^(p)_2}{\sqrt{p(1- p) [(1)/(n_1 ) + (1)/(n_2) ]} }$

=> $z = \frac{ 0.1315 - 0.1628 }{\sqrt{0.1485(1-0.1485) [(1)/(578) + (1)/(688) ]} }$

=> $z = -1.56$

Generally the p-value is mathematically represented as

$p-value = P(z < -1.56 )$

From z-table

$P(z < -1.56 ) = 0.05938$

$p-value = 0.05938$

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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
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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.________
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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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Answers

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Step-by-step explanation:

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Answers

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This means, the given triangle is an equilateral triangle.

So, each angle of triangle measures 60°.

So, we get an equation,

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Step-by-step explanation:

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