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Ksivusya [100]
3 years ago
11

The housing market has recovered slowly from the economic crisis of 2008.​ Recently, in one large​ community, realtors randomly

sampled 45 bids from potential buyers to estimate the average loss in home value. The sample showed the average loss was ​$9411 with a standard deviation of ​$1505.
a. Find a 99​% confidence interval for the mean loss in value per home.
Mathematics
1 answer:
k0ka [10]3 years ago
4 0

Answer:

The 99​% confidence interval for the mean loss in value per home is between $5359 and $13463

Step-by-step explanation:

We are in posession of the sample's standard deviation, so we use the student's t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 45 - 1 = 44

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 44 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.99}{2} = 0.995([tex]t_{995}). So we have T = 2.6923

The margin of error is:

M = T*s = 1505*2.6923 = 4052.

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 9411 - 4052 = $5359

The upper end of the interval is the sample mean added to M. So it is 9411 + 4052 = $13463

The 99​% confidence interval for the mean loss in value per home is between $5359 and $13463

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Check the picture below.

\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2  = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

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