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Dominik [7]
3 years ago
15

Assume that a hypothesis test of the given claim will be conducted. Identify the type I or type II error for the test. The princ

ipal of a school claims that the percentage of students at his school that come from single- parent homes is 11 %. Identify the type II error for the test.
Mathematics
1 answer:
ololo11 [35]3 years ago
5 0

Answer:

Type I error: Concluded that p ≠ 11% when it is 11%.

Type II error: Concluded that p = 11% when it is not 11%.

Step-by-step explanation:

A <em>type I error</em> is the rejection of the null hypothesis (<em>H₀</em>) when indeed the null hypothesis is true. It is symbolized by <em>α</em>.

A <em>type II error</em> is failing to discard the null hypothesis when indeed the null hypothesis is false. It is symbolized by <em>β</em>.

In this case a  principal of a school claims that the percentage of students at his school that come from single-parent homes is 11%.

The hypothesis to test this claim is:

<em>H₀</em>: The proportion of students at the school that come from single-parent homes is 11%, i.e. <em>p</em> = 0.11.

<em>Hₐ</em>: The proportion of students at the school that come from single-parent homes is not 11%, i.e. <em>p</em> ≠ 0.11.

  • The type I error will be committed when it is concluded that the proportion of students at the school that come from single-parent homes is not 11% when in fact it is 11%.
  • The type II error will be committed when it is concluded that the proportion of students at the school that come from single-parent homes is 11% when it is not.
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Two arithmetic means between 3 and 24 are -
defon

The value of two arithmetic means which are inserted between 3 and 24 are 24/9 and 75/9.

<h3>What is arithmetic mean?</h3>

Arithmetic mean is the mean or average which is equal to the ratio of sum of all the group numbers to the total numbers.  

The two arithmetic means between 3 and 24 are has to be inserted.

3, A₂, A₃, 24

All the four numbers are in arithmetic progression. The nth terms of AM can be found using the following formula:

t(n)=a(n-1)d

Here, d is the common difference a is the first terms and n is the total term.  The first term, a=3 and t₄=24. Thus, the common difference is;

t(n)=a(n-1)d\\t(4)=3(4-1)d\\24=3(3)d\\24=9d\\d=\dfrac{24}{9}

The second and 3rd term are:

A₂=3+\dfrac{24}{9}=\dfrac{51}{9}\\ A₃=\dfrac{51}{9}+\dfrac{24}{9}=\dfrac{75}{9}

Thus, the value of two arithmetic means which are inserted between 3 and 24 are 24/9 and 75/9.

Learn more about the arithmetic mean here;

brainly.com/question/14831274

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A ball is thrown straight up into the air from the top of a building standing at 50 feet with an initial velocity of 65 feet per
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Answer:

0.5 seconds (at 100 feet in the air).

Step-by-step explanation:

So, the height of the ball can be modeled by the function:

h(t)=-16t^2+16t+96

Where h(t) represents the height in feet after t seconds.

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Therefore, the maximum height will occur at the vertex of our function.

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In our function, a=-16; b=16; and c=96.

Find the x-coordinate of the vertex:

x=-\frac{(16)}{2(-16)}=1/2

So, the ball reaches its maximum height after 0.5 seconds of its projection.

Notes:

To find it’s maximum height, we can substitute 1/2 for our function and evaluate. So:

h(1/2)=-16(1/2)^2+16(1/2)+96

Evaluate:

h(1/2)=-4+8+96=100

So, the ball reaches its maximum height of 100 feet 0.5 seconds after its projection.

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4/5 ÷ 5/3

4/5 * 3/5 

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