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3 years ago

You conduct a hypothesis test for the mean of a population (H0 : p = 5) at the .05 significance level.

Mathematics

1 answer:

Anastaziya [24]3 years ago

8 0

Answer:

The correct option is a)

05; .73; .27 , type 1 error, type 11 error and the power of the test respectively.

Step-by-step explanation:

Alpha is the probability of a type 1 error, given the null hypothesis is true. Therefore alpha= 0.05

Type 11 error is the probability of accepting a false null hypothesis.

Beta= 0.73

The Power of a test is the probability of rejecting the null hypothesis, given it is false

Power= 1- beta

Power= 1-0.73

Power= 0.27

Therefore the type 1 error is 0.05

The type 11 error is= 0.73

The power of the test= 0.27

The right option is a)

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2 years ago

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3 years ago

Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (1, 5), F (3, 5), a

hram777 [196]

Answer:

$(1,1)$

Step-by-step explanation:

Given: E, F, G, H denote the three coordinates of the area fenced

To find: coordinates of point H

Solution:

According to distance formula,

length of side joining points $(x_1,y_1)\,,\,(x_2,y_2)$ is equal to $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

So,

$EF=\sqrt{(3-1)^2+(5-5)^2}=2\,\,units\\FG=\sqrt{(6-3)^2+(1-5)^2}=\sqrt{9+16}=5\,\,units\\GH=\sqrt{(x-6)^2+(y-1)^2}\\EH=\sqrt{(x-1)^2+(y-5)^2}$

Perimeter of a figure is the length of its outline.

$EF+FG+GH+EH=16\\2+5+\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16-2-5\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$