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2 years ago

14

We test for a hypothesized difference between two population means: H0: μ1 = μ2. The population standard deviations are unknown

but assumed equal. The number of observations in the first sample is 15 and 12 in the second sample. How many degrees of freedom are associated with the critical value?

1 answer:

Lina20 [59]2 years ago

4 0

Answer:

The degrees of freedom associated with the critical value is 25.

Step-by-step explanation:

The number of values in the final calculation of a statistic that are free to vary is referred to as the degrees of freedom. That is, it is the number of independent ways by which a dynamic system can move, without disrupting any constraint imposed on it.

The degrees of freedom for the t-distribution is obtained by substituting the values of n1 and n2 in the degrees of freedom formula.

Degrees of freedom, df = n1+n2−2

= 15+12−2=27−2=25

Therefore, the degrees of freedom associated with the critical value is 25.

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a line whose perpendicular distance from the origin is 4 units and the slope of perpendicular is 2÷3. Find the equation of the l

GrogVix [38]

Answer:

$\huge\boxed{y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{2}{3}x+\dfrac{4\sqrt{13}}{3}}$

Step-by-step explanation:

The equation of a line:

$y=mx+b$

We have

$m=\dfrac{2}{3}$

substitute:

$y=\dfrac{2}{3}x+b$

The formula of a distance between a point and a line:

General form of a line:

$Ax+By+C=0$

Point:

$(x_0,\ y_0)$

Distance:

$d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+b^2}}$

Convert the equation:

$y=\dfrac{2}{3}x+b$ |<em>subtract $y$ from both sides</em>

$\dfrac{2}{3}x-y+b=0$ |<em>multiply both sides by 3</em>

$2x-3y+3b=0\to A=2,\ B=-3,\ C=3b$

Coordinates of the point:

$(0,\ 0)\to x_0=0,\ y_0=0$

substitute:

$d=4$

$4=\dfrac{|2\cdot0+(-3)\cdot0+3b|}{\sqrt{2^2+(-3)^2}}\\\\4=\dfrac{|3b|}{\sqrt{4+9}}$

$4=\dfrac{|3b|}{\sqrt{13}}\qquad|$ |<em>multiply both sides by $\sqrt{13}$ </em>

$4\sqrt{13}=|3b|\iff3b=-4\sqrt{13}\ \vee\ 3b=4\sqrt{13}$ |<em>divide both sides by 3</em>

$b=-\dfrac{4\sqrt{13}}{3}\ \vee\ b=\dfrac{4\sqrt{13}}{3}$

Finally:

$y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{4\sqrt{13}}{3}$

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Ilia_Sergeevich [38]

Answer:

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Step-by-step explanation:

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mr_godi [17]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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