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3241004551 [841]

3 years ago

9

(6t4, 3) (-2tn 3) Simplify

1 answer:

Fofino [41]3 years ago

8 0

Idk the answer but someone with brain is gonna help you g

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Identifique el valor atípico en la siguiente lista. Luego encuentre la media y la mediana de la lista con y sin el valor atípico

Tju [1.3M]

Translate to English and I can help u, if I were to give u a answer it would be -8.5 based on the stuff given

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2 years ago

An equation is shown.

mr_godi [17]

Answer:

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Step-by-step explanation:

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2 years ago

What is the answer to this problem?<br>2(-7+x)>-2

mina [271]

Isolate the variable by dividing each side by factors that don't contain the variable.

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8 0

2 years ago

Let X1 and X2 be independent random variables with mean μand variance σ².

My name is Ann [436]

Answer:

a) $E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu$

So then we conclude that $\hat \theta_1$ is an unbiased estimator of $\mu$

$E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu$

So then we conclude that $\hat \theta_2$ is an unbiased estimator of $\mu$

b) $Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}$

$Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}$

Step-by-step explanation:

For this case we know that we have two random variables:

$X_1 , X_2$ both with mean $\mu = \mu$ and variance $\sigma^2$

And we define the following estimators:

$\hat \theta_1 = \frac{X_1 + X_2}{2}$

$\hat \theta_2 = \frac{X_1 + 3X_2}{4}$

Part a

In order to see if both estimators are unbiased we need to proof if the expected value of the estimators are equal to the real value of the parameter:

$E(\hat \theta_i) = \mu , i = 1,2$

So let's find the expected values for each estimator:

$E(\hat \theta_1) = E(\frac{X_1 +X_2}{2})$

Using properties of expected value we have this:

$E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu$

So then we conclude that $\hat \theta_1$ is an unbiased estimator of $\mu$

For the second estimator we have:

$E(\hat \theta_2) = E(\frac{X_1 + 3X_2}{4})$

Using properties of expected value we have this:

$E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu$

So then we conclude that $\hat \theta_2$ is an unbiased estimator of $\mu$

Part b

For the variance we need to remember this property: If a is a constant and X a random variable then:

$Var(aX) = a^2 Var(X)$

For the first estimator we have:

$Var(\hat \theta_1) = Var(\frac{X_1 +X_2}{2})$

$Var(\hat \theta_1) =\frac{1}{4} Var(X_1 +X_2)=\frac{1}{4} [Var(X_1) + Var(X_2) + 2 Cov (X_1 , X_2)]$

Since both random variables are independent we know that $Cov(X_1, X_2 ) = 0$ so then we have:

$Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}$

For the second estimator we have:

$Var(\hat \theta_2) = Var(\frac{X_1 +3X_2}{4})$

$Var(\hat \theta_2) =\frac{1}{16} Var(X_1 +3X_2)=\frac{1}{4} [Var(X_1) + Var(3X_2) + 2 Cov (X_1 , 3X_2)]$

Since both random variables are independent we know that $Cov(X_1, X_2 ) = 0$ so then we have:

$Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}$

7 0

2 years ago

Suppose that documentation lists the average sales price of a single-family home in the metropolitan Dallas/Ft. Worth/Irving, Te

Julli [10]

Answer:

"0.0373" seems to be the appropriate solution.

Step-by-step explanation:

The given values are:

n₁ = 43

n₂ = 35

$\bar{x_1}=217800$

$\bar{x_2}=204700$

$\sigma_1=30300$

$\sigma_2=33800$

Now,

The test statistic will be:

⇒ $Z=\frac{\bar{x_1}-\bar{x_2}}\sqrt{\frac{\sigma_1^2}{n_1} +\frac{\sigma_2^2}{n_2} }$

On substituting the given values in the above formula, we get

⇒ $=\frac{217800-204400}{\sqrt{\frac{(30300)^2}{43} +\frac{(33800)^2}{35} } }$

⇒ $=\frac{217800-204400}{\sqrt{\frac{918090000}{43} +\frac{1142440000}{35} } }$

⇒ $=\frac{13400}{7347.93}$

⇒ $=1.7828$

then,

P-value will be:

= $P(Z>1.7828)$

= $0.0373$

8 0

2 years ago