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2 years ago

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A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12

. A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard deviation of 15. At the .05 significance level, what is the p-value for testing variances for a one-tail test?

1 answer:

marissa [1.9K]2 years ago

7 0

Answer:

Step-by-step explanation:

Hello!

You have two random samples obtained from two different normal populations.

Sample 1

n₁= 15

X[bar]₁= 350

S₁= 12

Sample 2

n₂= 17

X[bar]₂= 342

S₂= 15

At α: 0.05 you need to obtain the p-value for testing variances for a one tailed test.

If the statistic hypotheses are:

H₀: σ₁² ≥ σ₂²

H₁: σ₁² < σ₂²

The statistic to test the variances ratio is the Stenecor's-F test. $F_{H_0}=(\frac{S^2_1}{Sigma^2_1}) * (\frac{S^2_2}{Sigma^2_2} )$ ~ $F_{n_1-1;n_2-1}$

$F_{H_0}= \frac{(12)^2}{(15)^2} * 1= 0.64$

The p-value is:

P( $F_{14;16}$ ≤0.64)= 0.02

I hope it helps!

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The question is missing parts. Here is the complete question.

Let M = $\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]$ . Find $c_{1}$ and $c_{2}$ such that $M^{2}+c_{1}M+c_{2}I_{2}=0$ , where $I_{2}$ is the identity 2x2 matrix and 0 is the zero matrix of appropriate dimension.

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$c_{2}=\frac{-214}{10}$

Step-by-step explanation: Identity matrix is a sqaure matrix that has 1's along the main diagonal and 0 everywhere else. So, a 2x2 identity matrix is:

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$\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+\left[\begin{array}{cc}6c_{1}&5c_{1}\\-1c_{1}&-4c_{1}\end{array}\right] +\left[\begin{array}{cc}c_{2}&0\\0&c_{2}\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]$

And the system of equations is:

$6c_{1}+c_{2} = -31\\-4c_{1}+c_{2} = -15$

There are several methods to solve this system. One of them is to multiply the second equation to -1 and add both equations:

$6c_{1}+c_{2} = -31\\(-1)*-4c_{1}+c_{2} = -15*(-1)$

$6c_{1}+c_{2} = -31\\4c_{1}-c_{2} = 15$

$10c_{1} = -16$

$c_{1} = \frac{-16}{10}$

With $c_{1}$ , substitute in one of the equations and find $c_{2}$ :

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$c_{2}=-31-6(\frac{-16}{10} )$

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$c_{2}=\frac{-310+96}{10}$

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<u>For the equation, </u> $c_{1} = \frac{-16}{10}$ <u> and </u> $c_{2}=\frac{-214}{10}$ <u />

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