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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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A simple random sample of size nequals40 is drawn from a population. The sample mean is found to be 103.9​, and the sample stand
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Answer:

There is enough evidence to support the claim that the population mean is greater than 100

Step-by-step explanation:

<u>Step 1</u>: We state the hypothesis and identify the claim

H_0:\mu=100 and H_1:\mu \:>\:100 (claim)

<u>Step 2</u>: Calculate the test value.

t=\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }

\implies t=\frac{103.9-100}{\frac{21.9}{\sqrt{40} } }=1.1263

<u>Step 3</u>: Find the P-value. The p-value obtained from a calculator is using d.f=39 and test-value 1.126 is 0.134

<u>Step 4</u>: We fail to reject the null hypothesis since P-value  is greater that the alpha level. (0.134>0.05).

<u>Step 5</u>: There is enough evidence to support the claim that the population mean is greater than 100.

<u>Alternatively</u>: We could also calculate the critical value to obtain +1.685 for \alpha=0.05 and d.f=39 and compare to the test-value:

The critical value (1.685>1.126) falls in the non-rejection region. See attachment.

NB: The t- distribution must be used because the population standard deviation is not known.

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