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A real estate agent in the coastal area of Georgia wants to compare the variation in the selling price of homes on the oceanfront with those one to three blocks from the ocean. A sample of 21 oceanfront homes sold within the last year revealed the standard deviation of the selling prices was $45,600. A sample of 18 homes, also sold within the last year, that were one to three blocks from the ocean revealed that the standard deviation was $21,330. At the .01 significance level, find the F-value. Leave your answer in 2 decimal places

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Answer:

The F-value is approximately 4.57

Step-by-step explanation:

The given parameters are;

The number of oceanfront homes in the sample, n₁ = 21 homes

The standard deviation of the selling prices, s₁ = $45,600

The number homes in the sample of homes one to three blocks from the ocean, n₂ = 18 homes

The standard deviation, s₂ = $21,330

The level of significance, α = 0.01

The F-test is given as follows;

[tex]F = \dfrac{s_1^2}{s_2^2}[/tex]

Therefore, we get;

F = 45,600²/21,330² ≈ 4.57

The F-value ≈ 4.57

The critical value from the F-table = 1.86236.

lhmarianateixeira

In this exercise we have to use the knowledge about force to calculate the critical force, in this way we find that:

[tex]F-value: 4.57[/tex]

 

We know that the information given by the text is:

  • The number of oceanfront homes in the sample: [tex]n_1 = 21 homes[/tex]
  • The standard deviation of the selling prices: [tex]S_1 = $45,600[/tex]
  • The number homes in the sample of homes one to three blocks from the ocean: [tex]n_2 = 18 homes[/tex]
  • The standard deviation: [tex]S_2 = $21,330[/tex]
  • The level of significance: [tex]\alpha= 0.01[/tex]

The F-test is given as follows:

[tex]F=(S_1^2)/(S^2_2)[/tex]

So we have:

[tex]F=(45.600^2)/(21.330^2)=4.57\\[/tex]

See more about force at brainly.com/question/4456579

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