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A jumbo mortgage is a mortgage with a loan amount above the industry-standard definition of conyentional conforming loan limits. As of January 2009, approximately \(2.57 \%\) of people who took out a jumbo mortgage during the previous 12 months were at least 60 days late on their payments. Suppose that this percentage is based on a random sample of 1430 people who took out a jumbo mortgage during the previous 12 months. a. Construct a \(95 \%\) confidence interval for the proportion of all people who took out a jumbo mortgage during the previous 12 months and were at least 60 days late on their payments. b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?

Short Answer

Expert verified
a. After calculations, the \(95\%\) confidence interval for the proportion of people who were late on their payments based on the given sample can be determined. b. The width of this confidence interval can be reduced by either decreasing the confidence level, increasing the sample size, or reducing the population variability. The more favourable method would usually be to increase the sample size, as it increases precision without affecting the confidence level.

Step by step solution

01

Identify Given Information and Required Confidence Interval

Identify the given values: the sample size (\(n\)) is 1430, the proportion (\(p\)) of people who were late on their payments is 0.0257 (or 2.57%), and the confidence level is 95%. To construct the confidence interval, the standard formula is \(\hat{p} \pm Z\sqrt{\frac{{\hat{p}(1-\hat{p})}}{{n}}}\) where \(\hat{p}\) is the sample proportion, \(Z\) is the Z-score for the given confidence level (for 95%, \(Z = 1.96\)), and \(n\) is the sample size.
02

Plug in Values into the Confidence Interval Formula

Substitute the given values into the formula and calculate the margin of error. The interval is therefore \(0.0257 \pm 1.96\sqrt{\frac{{0.0257(1-0.0257)}}{{1430}}}\). Calculate the value inside the square root first, then multiply the result by 1.96. Subtract and add the result from/to \(\hat{p}\) to get the confidence interval.
03

Constructing the Confidence Interval

Solve the above expression to find the lower and upper limits of the confidence interval.
04

Alternatives to Reduce the Width of the Confidence Interval

The width of a confidence interval can be reduced by decreasing the confidence level, increasing the sample size, or reducing the variability in the population. However, the best method generally is to increase the sample size, as this provides more accuracy without sacrificing the confidence level.

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Key Concepts

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

Sample Size
In statistics, sample size refers to the number of observations or data points collected in a survey or experiment. In our context, it's the 1430 people sampled who took out a jumbo mortgage. The sample size can significantly impact the reliability of the results. A larger sample size tends to offer a more accurate picture of the whole population.
This is because it reduces the margin of error, making the confidence interval more precise. If you're looking to narrow down your confidence interval, increasing the sample size is often the best solution. By gathering more data, you gain better insights into the population's actual characteristics.
Z-score
The Z-score is a significant factor in constructing confidence intervals. It shows how many standard deviations a data point is from the mean. For a 95% confidence level, the Z-score is generally 1.96. This number represents the area under the standard normal curve, corresponding to the middle 95% of the data.
The choice of Z-score changes with different confidence levels, impacting the width of the confidence interval. If you decrease the confidence level, the Z-score will be smaller, and the interval narrower. However, this will also decrease the reliability of your estimates.
Margin of Error
The margin of error in a confidence interval is a critical concept that reflects the range within which the true population parameter is expected to fall. It's calculated using the Z-score, sample size, and sample proportion.
Mathematically, it shows up in the formula as: \[Z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\] where \(\hat{p}\) is the sample proportion. A smaller margin means a more precise confidence interval. You can decrease it by increasing the sample size or reducing variability.
Proportion
The proportion is the percentage of the sample that exhibits a certain characteristic. In this example, it's 2.57% of people who paid their mortgage late. Represented as \(\hat{p}\) in calculations, it reflects an estimate of the true population proportion.
Understanding the proportion can help in assessing how representative the sample is of the entire population. It's important to use accurate sample proportions to construct meaningful confidence intervals, which in turn help in making informed predictions or decisions about the broader population.
Variability
Variability refers to how much the data spread out from the mean. A higher variability means the data points differ significantly from the average, potentially leading to a wider confidence interval.
Reducing variability can make the confidence interval narrower, resulting in a more precise estimate. However, controlling variability often depends on factors like the nature of the data and external conditions.
  • Reducing variability can involve more precise measurement techniques or narrowing down the criteria for sampling.
Ultimately, understanding and managing variability is crucial to improving the quality of statistical inferences.

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Most popular questions from this chapter

A May 8,2008 , report on National Public Radio (www.npr.org) noted that the average age of firsttime mothers in the United States is slightly higher than 25 years. Suppose that a random sample of 60 first-time mothers taken recently produced an average age of \(25.9\) years and a standard deviation of \(3.2\) years. Calculate a \(90 \%\) confidence interval for the average age of all current first-time mothers.

A sample of 200 observations selected from a population produced a sample proportion equal to \(.91\) a. Make a \(90 \%\) confidence interval for \(p\). b. Construct a \(95 \%\) confidence interval for \(p\). c. Make a \(99 \%\) confidence interval for \(p\). d. Does the width of the confidence intervals constructed in parts a through \(\mathrm{c}\) increase as the confidence level increases? If yes, explain why.

The mean time taken to design a house plan by 40 architects was found to be 23 hours with a standard deviation of \(3.75\) hours. a. Construct a \(98 \%\) confidence interval for the population mean \(\mu\). b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Describe all possible alternatives. Which alternative is the best and why?

a. Find the value of \(t\) from the \(t\) distribution table for a sample size of 22 and a confidence level of \(95 \%\) b. Find the value of \(t\) from the \(t\) distribution table for 60 degrees of freedom and a \(90 \%\) confidence level. c. Find the value of \(t\) from the \(t\) distribution table for a sample size of 24 and a confidence level of \(99 \%\)

A department store manager wants to estimate at a \(90 \%\) confidence level the mean amount spent by all customers at this store. The manager knows that the standard deviation of amounts spent by all customers at this store is \(\$ 31\). What sample size should he choose so that the estimate is within \(\$ 3\) of the population mean?

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