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olchik [2.2K]
3 years ago
9

A real estate developer is considering investing in a shopping mall on the outskirts of Atlanta, Georgia. Three parcels of land

are being evaluated. Of particular importance is the income in the area surrounding the proposed mall. A random sample of four families is selected near each proposed mall. Following are the sample results. At the 0.05 significance level, can the developer conclude there is a difference in the mean income

Mathematics
1 answer:
Arada [10]3 years ago
4 0

Answer:

Step 1: The null and alternative hypothesis for the one way analysis of variance is,

H₀ : There is no difference in the means of each area.

H₁ : At least one mean is different.

The objective of this question is to test whether the income of the three areas are significantly different.

Step 2: The problem states to use 0.05 significance level.

Step 3: The test statistic follows the F-distribution.

Step 4: Decision rule is based on the critical value.

Numerator degrees of freedom is,

K - 1 = 3 - 1

= 2

Denominator degrees of freedom is 2.

n - k = 12 - 3

= 9

Using the F-distribution table for ∝=0.05 with numerator and denominator degrees freedom, the critical value is 4.26. The decision rule is to reject H₀ if the computed value of F exceeds 4.26.

Step 5: Calculations.

Using Excel follow the steps mentioned below.

1. Import or type the data into spreadsheet.

2. Data--------> Data Analysis----------> Anova: Single Factor, click OK.

3. Select Input Range, make a tick mark on Labels in the first row, enter 0.05 for alfa.

4. Click OK.

The obtained output for the one-way ANOVA is,

[ find the figure in attachment]

From the obtained output, the computed test statistic value, F = 14.18 , which is greater than the critical value of 4.26. So, reject H₀

Step 6: Interpretation. The average income of each area is different. It can be concluded that the average income is significantly different .

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7 - 4 ( d - 3 ) = 23
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7 - 4 ( d - 3) = 23

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Subtract 12 from both sides,

7 - 4d = 11

Subtract 7 to both sides

- 4d = 4

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A store receives a shipment of 5,000 MP3 players. In a previous shipment of 5,000 MP3 players, 300 were defective. A store clerk
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Answer:

A store receives a shipment of 5,000 MP3 players. In a previous shipment of 5,000 MP3 players, 300 were defective. A store clerk generates random numbers to simulate a random sample of this shipment. The clerk lets the numbers 1 through 300 represent defective MP3 players, and the numbers 301 through 5,000 represent working MP3 players. The results are given.

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Based on this sample, how many of the MP3 players might the clerk predict would be defective?

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What is the following product? (2 square root 7+3 square root 6)(5 square root 2+4 square root 3)
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(2\sqrt7+3\sqrt6)(5\sqrt2+4\sqrt3)\qquad\text{use distributive property}\\\\=(2\sqrt7)(5\sqrt2)+(2\sqrt7)(4\sqrt3)+(3\sqrt6)(5\sqrt2)+(3\sqrt6)(4\sqrt3)\\\\=10\sqrt{14}+8\sqrt{21}+15\sqrt{12}+12\sqrt{18}\\\\=10\sqrt{14}+8\sqrt{21}+15\sqrt{4\cdot3}+12\sqrt{9\cdot2}\\\\=10\sqrt{14}+8\sqrt{21}+15\sqrt4\cdot\sqrt3+12\sqrt9\cdot\sqrt2\\\\=10\sqrt{14}+8\sqrt{21}+(15)(2)\sqrt3+(12)(3)\sqrt2\\\\=\boxed{10\sqrt{14}+8\sqrt{21}+30\sqrt3+36\sqrt2}

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A student solves the following equation and
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Answer:

Since the equation is undefined for -2

Therefore, NO SOLUTION for the given equation.

Step-by-step explanation:

Considering the expression

\frac{3}{a+2}-6\cdot \frac{a}{-4+a^2}=\frac{1}{a-2}

\frac{3}{a+2}-\frac{6a}{-4+a^2}=\frac{1}{a-2}

\mathrm{Find\:Least\:Common\:Multiplier\:of\:}a+2,\:-4+a^2,\:a-2:\quad \left(a+2\right)\left(a-2\right)

\mathrm{Multiply\:by\:LCM=}\left(a+2\right)\left(a-2\right)

\frac{3}{a+2}\left(a+2\right)\left(a-2\right)-\frac{6a}{-4+a^2}\left(a+2\right)\left(a-2\right)=\frac{1}{a-2}\left(a+2\right)\left(a-2\right)

as

  • \frac{3}{a+2}\left(a+2\right)\left(a-2\right):\quad 3\left(a-2\right)
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so equation becomes

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-3a-6=a+2

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\mathrm{Divide\:both\:sides\:by\:}-4

\frac{-4a}{-4}=\frac{8}{-4}

a=-2

\mathrm{Verify\:Solutions}

\mathrm{Take\:the\:denominator\left(s\right)\:of\:}\frac{3}{a+2}-6\frac{a}{-4+a^2}-\frac{1}{a-2}\mathrm{\:and\:compare\:to\:zero}

\mathrm{Solve\:}\:a+2=0:\quad a=-2

\mathrm{Solve\:}\:-4+a^2=0:\quad a=2,\:a=-2

\mathrm{Solve\:}\:a-2=0:\quad a=2

So the following points are undefined

a=-2,\:a=2

Since the equation is undefined for -2

Therefore, NO SOLUTION for the given equation.

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