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3 years ago

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Assume that you plan to use a significance level of α = 0.05 to test the claim that p1 = p2. Use the given sample sizes and numb

ers of successes to find the pooled estimate. Round your answer to the nearest thousandth.
n1 = 677 n2 = 3377
x1 = 172 x2 = 654

1 answer:

dezoksy [38]3 years ago

3 0

Answer:

The calculated value Z = 3.775 > 1.96 at 0.05 level of significance

Null hypothesis is rejected

The Two Population proportion are not equal

<u>Step-by-step explanation</u>:

<em>Given first sample size n₁ = 677</em>

<em>First sample proportion </em>

<em> </em> $p^{-} _{1} = \frac{x_{1} }{n_{1} } = \frac{172}{677} = 0.254$ <em></em>

Given second sample size n₂ = 3377

<em>second sample proportion </em>

<em> </em> $p^{-} _{2} = \frac{x_{2} }{n_{2} } = \frac{654}{3377} = 0.1936$ <em></em>

<u><em>Null Hypothesis : H₀ :</em></u><em> p₁ = p₂.</em>

<u><em>Alternative Hypothesis : H₁</em></u><em> : p₁ ≠ p₂.</em>

Test statistic

$Z = \frac{p_{1} ^{-}-p^{-} _{2} }{\sqrt{P Q(\frac{1}{n_{1} } +\frac{1}{n_{2} }) } }$

where

$P = \frac{n_{1} p_{1} + n_{2} p_{2} }{n_{1}+n_{2} } = \frac{677 X 0.254+3377 X 0.1936}{677+3377}$

P = 0.2036

Q = 1 - P = 1 - 0.2036 = 0.7964

$Z = \frac{0.254- 0.1936 }{\sqrt{0.2036 X 0.7964(\frac{1}{677 } +\frac{1}{3377 }) } }$

Z = 3.775

<em>Critical value ∝=0.05</em>

<em>Z- value = 1.96</em>

<em>The calculated value Z = 3.775 > 1.96 at 0.05 level of significance</em>

<em>Null hypothesis is rejected </em>

<em>The Two Population proportion are not equal</em>

<em></em>

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And assuming independence we have this:

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Step-by-step explanation:

Part a

For this case we are assuming $X_1, X_2 , ..., X_n \sim U(0,a)$

And we are are ssuming the following estimator:

$\hat a = max(X_i)$

For this case the value for $\hat a$ is always smaller than the value of a, assuming $X_i \sim Unif[0,a]$ So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

$E(a) - a= 0$ and that's not our case.

Part b

For this case we assume that the estimator is given by:

$E(\hat a) = \frac{na}{n+1}$

And using the definition of bias we have this:

$E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}$

Since is a negative value we can conclude that underestimate the real value a.

And when we take the limit when n tend to infinity we got that the bias tend to 0.

$\lim_{ n \to\infty} -\frac{1}{n+1}= 0$

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For this case we the followng random variable $Y = max (X_i)$ and we can find the cumulative distribution function like this:

$P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)$

And assuming independence we have this:

$P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n$

Since all the random variables have the same distribution.

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Now we can find the expected value for the random variable Y and we got this:

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$n(n+2) = n^2 + 2n > n +2n = 3n$ and that's satisfied for n>1.

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