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1 year ago

What is the only way to increase power of a significance test when lowering the probability of committing a Type I error

1 answer:

5 0

the probability of making a Type I error is equal to the significance level of power. To increase the probability of a Type I error, increase the significance level. Changing the sample size has no effect on the probability of a Type I error.

The electricity of a take a look at can be expanded in a number of methods, for example increasing the pattern length, reducing the standard errors, increasing the difference between the pattern statistic and the hypothesized parameter, or growing the alpha degree.

The chance of creating a kind I mistakes is α, that's the extent of importance you put for your hypothesis check. An α of 0.05 indicates which you are inclined to accept a five% chance which you are incorrect whilst you reject the null hypothesis. To lower this risk, you have to use a decrease cost for α

Learn more about power here: brainly.com/question/1634438

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Which matrix is equal to

polet [3.4K]

The one that look exactly the same it the third one

3 0

3 years ago

The length of a swimming pool is 3 feet longer than it’s width. The swimming pool is surrounded by a deck that is 2 feet wide an

Firlakuza [10]

Answer:

The Length of swimming pool is 8.35 feet

The width of swimming pool is 5.35 feet

Step-by-step explanation:

Given as :

The length of swimming pool is 3 feet longer that its width

Let The width of swimming pool = w feet

So, The Length of swimming pool = L = (w + 3) feet

Now, The swimming pool is surrounded by deck of 2 feet wide

The width of deck = w' = w + 2 + 2 = (w + 4) feet

The area of deck = 116 feet²

The length of deck = L' = L + 2 + 2 = (L + 4) feet

So, L' = (w + 3 + 4) feet

I.e L' = (w + 7) feet

∵ The area of deck = 116 feet²

So , L' × w' = 116 feet²

(w + 7)× (w + 4) = 116

Or, w² + 4 w + 7 w + 28 = 116

Or, w² + 11 w - 88 = 0

Solving the quadratic equation as

ax² + bx + c = 0

So, w = $\dfrac{-b\pm \sqrt{b^{2}-4\times a\times c}}{2\times a}$

Or, w = $\dfrac{-11\pm \sqrt{11^{2}-4\times 1\times (-88)}}{2\times 1}$

Or, w = $\dfrac{-11\pm \sqrt{473}}{2}$

or, w = $\dfrac{-11\pm 21.7}{2}$

or, w = 5.35 , -16.35

The width of swimming pool = w = 5.35 feet

The Length of swimming pool = L = (5.35 + 3) feet = 8.35 feet

Hence, The Length of swimming pool is 8.35 feet

And The width of swimming pool is 5.35 feet

Answer

3 0

3 years ago

PLEASE HELP! I HAVE A TIME CRUNCH! WILL MARK BRAINLIEST!! LOVE Y'ALL Fill in each * so that the statement is true.

siniylev [52]

Answer:

a, (x^2)^3 = x^6 = x^2.x^4

b, x^5·x^7 = x^12 = (x^3)^4

c, x^4·x^22 = x^26 = (x^2)^13

d, (x^2)^8 = x^16 = x^2.x^14

e, x^10/x^3 = x^7 = x^20.x^(-13)

f, x^(-3) = x^4.x^(-7)

g, 1/x^(-3) = x^3 = x^4.x^(-1)

Hope this helps!

6 0

3 years ago

Given Circle M where d = 11, determine<br> The circumference.

Advocard [28]

Answer:

In this particular case, Circle M has circumference 11π, or approximately 34.6 units.

Step-by-step explanation:

The circumference of a circle of diameter d is C = πd.

In this particular case, Circle M has circumference 11π, or approximately 34.6 units.

8 0

2 years ago

Which ordered pair is a solution to the system of linear equations One-half x minus three-fourths y = StartFraction 11 Over 60 E

s344n2d4d5 [400]

Answer:

$\left(\dfrac{2}{3},\dfrac{1}{5}\right)$

Step-by-step explanation:

Given the system of two equations:

$\left\{\begin{array}{l}\dfrac{1}{2}x-\dfrac{3}{4}y=\dfrac{11}{60}\\ \\\dfrac{2}{5}x+\dfrac{1}{6}y=\dfrac{3}{10}\end{array}\right.$

Multiply the first equation and the second equation by 60 to get rid of fractions:

$\left\{\begin{array}{l}30x-45y=11\\ \\24x+10y=18\end{array}\right.$

Now multiply the first equation by 4 and the second equation by 5:

$\left\{\begin{array}{l}120x-180y=44\\ \\120x+50y=90\end{array}\right.$

Subtract them:

$(120x-180y)-(120x+50y)=44-90\\ \\120x-180y-120x-50y=-46\\ \\-230y=-46\\ \\y=\dfrac{46}{230}=\dfrac{1}{5}$

Substitute it into the first equation:

$30x-45\cdot \dfrac{1}{5}=11\\ \\30x-9=11\\ \\30x=11+9\\ \\30x=20\\ \\x=\dfrac{2}{3}$

The solution is $\left(\dfrac{2}{3},\dfrac{1}{5}\right)$

4 0

3 years ago

Read 2 more answers