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xeze [42]
1 year ago
5

Sociologists want to test whether the number of homeless people in a particular urban area is increasing. In 2010, the average n

umber of homeless people per day who sought shelter was 42.3 (σ = 6.2). Data from the current year reveal that the mean number of people seeking shelter per day is 45.3. Compute the test statistic using an alpha = .01 and n = 1000.
Mathematics
1 answer:
Kobotan [32]1 year ago
7 0

Answer:

The test statistic value is 15.3.

Step-by-step explanation:

The hypothesis for this test is:

<em>H</em>₀: The average number of homeless people is not increasing, i.e. <em>μ</em> = 42.3.

<em>H</em>ₐ: The average number of homeless people is increasing, i.e. <em>μ</em> > 42.3.

Given:

\bar x=45.3\\\sigma=6.2\\n=1000

As the population standard deviation is provided use a single mean <em>z</em>-test for the hypothesis testing.

The test statistic is:

z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}=\frac{45.3-42.3}{6.2/\sqrt{1000}}=15.3

Thus, the test statistic value is 15.3.

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Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:
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Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

a) We need to find a relation R reflexive and transitive that contain the relation R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

  • 1R4 and 4R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 4R1 and 1R2, then 4 must be related with 2.

Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

  • since 1R2, then 2 must be related with 1.
  • since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

  • 1 must be related with 1,
  • 2 must be related with 2,
  • 3 must be related with 3,
  • 4 must be related with 4

For be symmetric

  • since 1R2, 2 must be related with 1,
  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

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