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Chapter 6: Problem 9

What Percent of Houses Are Owned vs }\end{array}\( Rented? The 2010 US Census \)^{4}\( reports that, of all the nation's occupied housing units, \)65.1 \%\( are owned by the occupants and \)34.9 \%$ are rented. If we take random samples of 50 occupied housing units and compute the sample proportion that are owned for each sample, what will be the mean and standard deviation of the distribution of sample proportions?

Short Answer

Expert verified
The mean of the distribution of sample proportions is \(0.651\) and the standard deviation is \(0.070\).

Step by step solution

01

Calculate the mean

The mean of the sample proportions is the population proportion. Since \(65.1 \%\) of the houses are owned, the mean of the sample proportions is \(0.651\).
02

Calculate the standard deviation

The standard deviation of the sample proportions is calculated using the formula: \(\sqrt{(population proportion)*(1-population proportion)/sample size}\). Here, the sample size is 50. So, the standard deviation is \(\sqrt{(0.651)*(1-0.651)/50}\).
03

Compute the standard deviation

After calculating the values in the standard deviation formula, you get \(\sqrt{(0.651)*(0.349)/50} = 0.070\).

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Key Concepts

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

Sample Proportion
When analyzing data from a sample, the sample proportion is an important concept in statistical analysis. Imagine you have a set of housing data. Specifically, a sample proportion refers to the fraction of the houses in your sample that are owned. For example, if 32 out of the 50 houses you randomly selected are owned, then the sample proportion is the ratio of those owned to the total sample size, calculated as \( \frac{32}{50} = 0.64 \).
  • The sample proportion provides insight into the larger population, helping us estimate the likelihood that a house is owned based on random sampling.
  • It's derived directly from your sampled data. In your exercise, the population proportion is given as 65.1%.
Understanding the sample proportion is crucial, especially when interpreting sample data against the population measure to ensure reliability and accuracy in predictions.
Standard Deviation
Standard deviation in statistics measures the amount of variation or spread in a set of values. When dealing with proportions, as in this exercise, the standard deviation tells you how much the sample proportion differs from the population proportion.
  • The formula for calculating the standard deviation of sample proportions is \( \sqrt{p(1-p)/n} \), where \( p \) is the population proportion and \( n \) is the sample size.
  • In the context of the 2010 Census data, it helps determine how much the proportion of house ownership in different samples of size 50 might vary.
  • To calculate: with \( p = 0.651 \) and sample size \( n = 50 \), we find \( \sigma = \sqrt{0.651 \times 0.349 / 50} = 0.070 \).
This calculation gives you an estimate of the variability you can expect when using sample data to make inferences about the whole population's housing ownership.
Census Data Analysis
Census data analysis involves examining comprehensive data collected about a population. This data is crucial because it offers detailed insights into the various demographic and socio-economic aspects of a nation. For the U.S. Census, understanding the distribution of housing ownership (owned vs. rented) helps planners and policymakers gauge housing trends and needs.
  • Such data serves as a reliable baseline, against which sample surveys can be evaluated.
  • It becomes a parameter in calculations such as determining the mean proportion of owned houses, which is directly linked to the population proportion stated as 65.1%.
Moreover, comparing sample survey data with census findings can validate the surveys' accuracy and guide adjustments in policy or research focus. This process underscores the importance of using comprehensive census data as a benchmark while evaluating sample-based findings and projections.

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Most popular questions from this chapter

In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion: (a) Find the mean and standard error of the distribution of sample proportions. (b) If the sample size is large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 50 from a population with proportion 0.25

NBA Free Throws In Exercise 6.10, we learn that the percent of free throws made in the \(\mathrm{NBA}\) (National Basketball Association) has been about \(75 \%\) for the last 50 years. If we take random samples of free throws in the NBA and compute the proportion of free throws made, what percent of samples of size \(n=200\) will have a sample proportion greater than \(80 \%\) ? Use the fact that the sample proportions are normally distributed and compute the mean and standard deviation of the distribution.

Standard Error from a Formula and a Bootstrap Distribution In Exercises 6.19 to \(6.22,\) use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of home team wins in soccer, with \(n=120\) and \(\hat{p}=0.583\)

Standard Error from a Formula and a Bootstrap Distribution In Exercises 6.19 to \(6.22,\) use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of lie detector trials in which the technology misses a lie, with \(n=48\) and \(\hat{p}=0.354\)

Use StatKey or other technology to generate a bootstrap distribution of sample differences in proportions and find the standard error for that distribution. Compare the result to the value obtained using the formula for the standard error of a difference in proportions from this section (and using the sample proportions to estimate the population proportions). Sample A has a count of 90 successes with \(n=120\) and Sample \(\mathrm{B}\) has a count of 180 successes with \(n=300\).
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