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larisa [96]
3 years ago
8

According to the National Association of Realtors, it took an average of three weeks to sell a home in 2017. Suppose data for th

e sale of 39 randomly selected homes sold in Greene County, Ohio, in 2017 showed a sample mean of 3.6 weeks with a sample standard deviation of 2 weeks. Conduct a hypothesis test to determine whether the number of weeks until a house sold in Greene County differed from the national average in 2017. Use α = 0.05 for the level of significance, and state your conclusion.
a. State the null and alternative hypothesis.
b. Find the value of the test statistic.
c. Find the p-value.
d. State your conclusion.
Mathematics
1 answer:
mezya [45]3 years ago
4 0

Solution :

Here, given :

Sample size, n = 39

Sample mean, $\bar X$ = 3.6

Standard deviation of the sample, s =2

The population mean, $\mu_0 = 3$

The significance level, $\alpha = 0.05$

a). Therefore the hypothesis is :

  $H_0 : \mu = 3 \text{ Vs} \ H_a: \mu \neq 3$

b). The test statics is given as :

  $t = \frac{\bar X - \mu_0}{\frac{s}{\sqrt n}} \rightarrow t_{n-1}$

  $t = \frac{3.6-3}{\frac{2}{\sqrt {39}}} $

     = 1.873

c). The p- value is given by :

 $P(t_{d.f}>|t_{stat}|)$

$=P(t_{39-1}> 1.873)$

$=0.0688$

d). The conclusion :

  In this case, the p-value is $0.688 > \alpha=0.05$

So, we do not reject $H_0$.

Therefore, we conclude that it is not a statistically significant difference between national average time for selling a home and the mean time for selling in Greene County.

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tamaranim1 [39]

9514 1404 393

Answer:

  a ≈ 4.68

Step-by-step explanation:

The law of cosines tells you ...

  a² = b² +c² -2bc·cos(A)

  a = √(b² +c² -2bc·cos(A))

  a = √(5² +8² -2·5·8·cos(33°)) = √(25 +64 -80·0.83867) ≈ √21.906

  a ≈ 4.68

6 0
3 years ago
Read 2 more answers
A basketball coach will select the members of a five-player team from among 9 players, including John and Peter. If the five pla
vivado [14]

Answer:

Step-by-step explanation:

Given that a basketball coach will select the members of a five-player team from among 9 players, including John and Peter.

Out of nine players five are chosen at random.

The team consists of John and Peter.

Hence we can sort 9 players as I group, John and Peter and II group 7 players.

Now the selection is 2 from I group and remaining 3 from II group.

Hence no of ways of selecting a team that includes both John and Peter=2C2*7C3=35

Total no of ways = 9C5=126

= \frac{2C2*7C3}{9C5}=\frac{35}{126} =\frac{5}{18}

6 0
3 years ago
An archaeologist earns $9147 monthly. Determine the amount he is paid annually.
Mnenie [13.5K]
$9147 x 12 =190,764
12 is how many months there is in a year
4 0
3 years ago
A rectangle has perimeter 28cm. Its area is 42sq.cm. Determine the dimensions of the rectangle. Include a diagram in your soluti
yawa3891 [41]

The dimensions of the rectangle are: b=9.65 cm and h=4.35 cm or b=4.35 cm and h= 9.65 cm.

<h3>Quadrilaterals</h3>

There are different types of quadrilaterals, for example, square, rectangle, rhombus, trapezoid, and parallelogram.  Each type is defined accordingly to its length of sides and angles. For example, in a rectangle,  the opposite sides are equal and parallel and their interior angles are equal to 90°.

The area of a rectangle is equal to base x height (A=bh). The perimeter of a geometric figure is the sum of its sides. Thus, for a rectangle, the perimeter is 2b+2h.

<h3>System of Linear Equations</h3>

System of linear equations is the given term math for two or more equations with the same variables. The solution of these equations represents the point at which the lines intersect.

The question gives:

Perimeter=28cm

Area= 42 cm²

If the perimeter is the sum of all sides of the rectangle, you have:

Perimeter= 2b+2h=28

If the area of the rectangle is 42cm², you have:

Area=bh=42 cm²

You can write a system of linear equations.

2b+2h=28 (1)

bh=42 (2)

From equation 2, you have  b=\frac{42}{h}. Then, you can replace it in equation 1.

2*\frac{42}{h}+2h=28 \\ \\ 84+2h^2=28h\\ \\ 2h^2-28h+84=0 \; dividing\; by\; 2\\ \\ h^2-14h+42=0

Now, you should solve the quadratic equation.

h_{1,\:2}=\frac{-\left(-14\right)\pm \sqrt{\left(-14\right)^2-4\cdot \:1\cdot \:42}}{2\cdot \:1}

h_{1,\:2}=\frac{-\left(-14\right)\pm \:2\sqrt{7}}{2\cdot \:1}

h_1=\frac{-\left(-14\right)+2\sqrt{7}}{2\cdot \:1}=7+\sqrt{7}=9.65 cm

h_2=\frac{-\left(-14\right)-2\sqrt{7}}{2\cdot \:1}=7-\sqrt{7}=4.35cm

If h1=9.65 cm, then b_1=\frac{42}{9.65} =4.35 cm, and if  h2=4.35 cm, then b_1=\frac{42}{4.35} =9.65 cm.

Read more about the quadratic function here:

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8 0
2 years ago
for the opening day of a carnival, 800 admission tickets were sold. The receipts totaled $3775. Tickets for children cost $3 eac
zloy xaker [14]
Total tickets sold  = 800
Total revenue = $3775

Ticket costs:
$3  per child,
$8  per adult,
$5 per senior citizen.

Of those who bought tickets, let
x =  number of children 
y = number of adults
z = senior citizens

Therefore
x + y + z = 800                   (1)
3x + 8y + 5z = 3775           (2)

Twice as many children's tickets were sold as adults. Therefore
x = 2y                                (3)

Substitute (3) into (1) and (2).
2y + y + z = 800, or
3y + z = 800, or
z = 800 - 3y                        (4)
3(2y) + 8y + 5z = 3775, or
14y + 5z = 3775                 (5)

Substtute (4) nto (5).
14y + 5(800 - 3y) = 3775
-y = -225
y = 225
From (4), obtain
z = 800 - 3y = 125
From (3), obtain
x = 2y = 450

Answer:
The number of tickets sold was:
450 children,
225  adults,
125 senior citizens.
6 0
3 years ago
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