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Damm [24]
3 years ago
15

For ANOVA, the test statistic is called an ____ test statistic (also called a ____-ratio), which is the variance (2) samples (a.

k.a., variation due to treatment) divided by the variance (3) samples (a.k.a., variation due to error or chance).
Mathematics
1 answer:
Eduardwww [97]3 years ago
4 0

Answer:

<h3>For ANOVA, the test statistic is called an <u>F-statistic</u> test statistic (also called a <u>F-statistic </u>-ratio), which is the variance (2) samples (a.k.a., variation due to treatment) divided by the variance (3) samples (a.k.a., variation due to error or chance).</h3>

Step-by-step explanation:

  • For Analysis of Variance the F-statistic is the Variation between Means of Sample or Variation within the Samples
  • For F-tests the F-statistic is the test statistic .
  • Generally, an F-statistic is also called as a ratio of two quantities between (here the variance (2) samples (a.k.a., variation due to treatment) divided by the variance (3) samples (a.k.a., variation due to error or chance)). which results an F-statistic of approx 1.

For ANOVA, the test statistic is called an <u>F-statistic</u> test statistic (also called a <u>F-statistic </u>-ratio), which is the variance (2) samples (a.k.a., variation due to treatment) divided by the variance (3) samples (a.k.a., variation due to error or chance).

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Step-by-step explanation:

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Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the
Rama09 [41]

<em><u>Question:</u></em>

Britney throws an object straight up into the air with an initial velocity of 27 ft/s from a platform that is 10 ft above the ground. Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the object to hit the ground?

1s   2s   3s   4s

<em><u>Answer:</u></em>

It takes 2 seconds for object to hit the ground

<em><u>Solution:</u></em>

<em><u>The given equation is:</u></em>

h(t) = -16t^2 + v_0t+h_0

Initial velocity = 27 feet/sec

h_0 = 10\ feet

Therefore,

h(t) = -16t^2 +27t+10

At the point the object hits the ground, h(t) = 0

-16t^2 +27t+10 = 0\\\\16t^2-27t-10=0

Solve by quadratic formula,

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=16,\:b=-27,\:c=-10:\quad \\\\t=\frac{-\left(-27\right)\pm \sqrt{\left(-27\right)^2-4\cdot \:16\left(-10\right)}}{2\cdot \:16}\\\\t = \frac{27 \pm \sqrt{1369}}{32}\\\\t = \frac{27 \pm 37}{32}\\\\We\ have\ two\ solutions\\\\t = \frac{27+37}{32}\\\\t = \frac{64}{32}\\\\t = 2\\\\And\\\\t = \frac{27-37}{32}\\\\t = -0.3125

Ignore, negative value

Thus, it takes 2 seconds for object to hit the ground

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Answer:

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Step-by-step explanation:

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