/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Home sales The housing market re... [FREE SOLUTION] | 91Ó°ÊÓ

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Home sales The housing market recovered slowly from the economic crisis of 2008 . Recently, in one large community, realtors randomly sampled 36 bids from potential buyers to estimate the average loss in home value. The sample showed the average loss from the peak in 2008 was \(\$ 9560\) with a standard deviation of \(\$ 1500\). a. What assumptions and conditions must be checked before finding a confidence interval? How would you check them? b. Find a \(95 \%\) confidence interval for the mean loss in value per home.

Short Answer

Expert verified
The necessary assumptions and conditions for this problem are randomness, normality, and independence. Under these assumptions and conditions, a \(95 \%\) confidence interval for the mean loss in value per home can be estimated to be between \$9060 and \$10060.

Step by step solution

01

Check Assumptions and Conditions

The assumptions and conditions that must be checked before finding a confidence interval include the following: randomness, normality, and independence. These can be checked as follows:1. Randomness: The problem mentions that the bids were randomly sampled. Therefore, the randomness condition is satisfied.2. Normality: While the problem doesn't explicitly mention that the distribution is normal, the Central Limit Theorem could be applied since the sample size is more than 30.3. Independence: Presumably the bids are independent since one bid should not affect another. However, the problem doesn't provide sufficient information to confirm this.
02

Calculation of Standard Error

The standard error (SE) can be calculated using the formula: SE = s/√n, where \('s'\) is the standard deviation and \('n'\) is the sample size.In this case, s = \$1500 and n = 36. Substituting these values into the formula, we get: SE = \$1500/ √36 = \$250.
03

Find the z-Score

For a \(95 \%\) confidence interval, the z-score corresponding to the critical value is ±1.96 (this value can be found in a standard z-table or it's a commonly known value in statistics).
04

Calculation of Confidence Interval

Now, we need to calculate the confidence interval using the formula: Confidence Interval = x̄ ± (z * SE), where x̄ is the sample mean.In this case, x̄ = \$9560, z = 1.96, and SE = \$250. So, Confidence Interval = \$9560 ± (1.96 * \$250).
05

Final Confidence Interval

Substituting the values into the equation gives: Confidence Interval = \$9560 - 1.96(\$250) to \$9560 + 1.96(\$250), which simplifies to approximately \$9060 to \$10060. Therefore, we are \(95 \%\) confident that the true mean loss in home value is between \$9060 and \$10060.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental concept in statistics that helps us understand the behavior of sample means. When we take multiple samples from a population, the distribution of these sample means will approach a normal distribution, even if the original population distribution is not normal. This magical property is what allows us to make inferences about a population based on sample data.

Here are the key points about CLT:
  • It applies when the sample size is sufficiently large, often considered to be at least 30.
  • The CLT justifies the normality condition required for constructing confidence intervals.
Because the CLT holds, we can assume that the average of our sample bids follows a normal distribution. This assumption is crucial because it allows us to use z-scores and calculate confidence intervals accurately. Without the CLT, it would be challenging to apply standard statistical techniques when the population distribution is unknown.
Standard Error
The standard error (SE) is a measure of how much the sample mean is likely to vary from the true population mean. It's a type of standard deviation but specifically for sample means. SE is crucial because it helps quantify the uncertainty associated with our sample mean estimate.

The formula to calculate SE is:
\[SE = \frac{s}{\sqrt{n}}\]
Where:
  • \(s\) is the sample standard deviation,
  • \(n\) is the sample size.
If our sample size is large, the standard error will be smaller. This means our sample mean is a more accurate estimate of the population mean. In our home sales example, the standard error calculation was:\[SE = \frac{\\(1500}{\sqrt{36}} = \\)250\]
This SE indicates how much we can expect the sample mean home value loss to fluctuate around the true mean. Smaller SE implies greater precision in our confidence interval estimate.
z-Score
The z-score is a statistical metric that describes how many standard deviations a data point or a sample mean is from the population mean. It is used when we want to determine the probability of a sample statistic under the standard normal distribution.

For constructing a confidence interval, we use the z-score to calculate the margin of error. The z-score matching a 95% confidence interval is typically 1.96. This value comes from z-tables, indicating where about 95% of the data falls within the standard normal distribution.
  • A z-score of 1.96 implies that every sample mean computed will fall within this range, 95% of the time, around the true mean, if we drew multiple samples.
In the context of our problem, multiplying the z-score by the standard error gives the margin of error. Thus:\[1.96 \times \\(250 = \\)490\]
This margin is added and subtracted from our sample mean to find our confidence interval.
Random Sampling
Random sampling is a fundamental principle in statistics that ensures that every individual or item in a population has an equal opportunity of being selected in the sample. This technique is essential to ensure the representativeness of a sample.

Random sampling helps:
  • Reduce bias in the selection process.
  • Ensure that the sample provides a good approximation of the wider population.
In the home sales exercise, random sampling was used to select 36 bids, helping satisfy the condition of randomness necessary for constructing a valid confidence interval. When random sampling is properly implemented, it enhances the reliability of statistical conclusions by minimizing selection biases. This way, we can be confident that the insights we derive are applicable to the broader community and not just to a specific subset."

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Most popular questions from this chapter

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