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Chapter 10: Problem 18

Student loan debt has reached record levels in the United States. In a random sample of 100 individuals who have student loan debt, it was found the mean debt was 23,979 dollar with a standard deviation of 31,400 dollar . Data based on results from the Federal Reserve Bank of New York. (a) What do you believe is the shape of the distribution of student loan debt? Explain. (b) Use this information to estimate the mean student loan debt among all with such debt at the \(95 \%\) level of confidence. Interpret this result. (c) What could be done to increase the precision of the estimate?

Short Answer

Expert verified
The distribution of student loan debt is likely positively skewed. The mean student loan debt is estimated to be between 17,824.6 dollars and 30,133.4 dollars with 95% confidence. Increasing the sample size will make the estimate more precise.

Step by step solution

01

Determine Shape of Distribution

Observing the high standard deviation relative to the mean, it is likely that the distribution is right-skewed (positively skewed). This suggests there are fewer individuals with extremely high debt amounts pulling the mean upward.
02

Calculate the Standard Error of the Mean

The standard error of the mean (SE) is calculated using the formula: \[ SE = \frac{s}{\sqrt{n}} \] where \(s\) is the standard deviation and \(n\) is the sample size. Substitute the given values: \[ SE = \frac{31400}{\sqrt{100}} = 3140 \]
03

Determine Critical Value for Confidence Interval

For a 95% confidence level, the critical value (\(z\)-score) is approximately 1.96.
04

Calculate the Margin of Error

The margin of error (ME) is calculated using the formula: \[ ME = z \times SE \] Substitute the values: \[ ME = 1.96 \times 3140 = 6154.4 \]
05

Calculate Confidence Interval

The confidence interval (CI) is calculated using the mean \[ \text{CI} = \text{mean} \pm \text{ME} \] Substitute the values: \[ \text{CI} = 23979 \pm 6154.4 \] So, the confidence interval is: \[ 17824.6 \text{ dollars} \leq \mu \leq 30133.4 \text{ dollars} \]
06

Interpret Confidence Interval

This means we are 95% confident that the mean student loan debt among all individuals with such debt falls between 17,824.6 dollars and 30,133.4 dollars.
07

Increase Precision of Estimate

To increase the precision of the estimate, we can increase the sample size. A larger sample size decreases the standard error, thus narrowing the confidence interval.

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Key Concepts

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

Sample Distribution
When discussing the distribution of student loan debt, it鈥檚 crucial to consider the shape of the data. With a high standard deviation compared to the mean, we can infer that the distribution is likely right-skewed (positively skewed). This means there are fewer individuals with extremely high debt amounts pulling the average upwards. In simple terms, imagine most students having an average amount of debt, but a few owing significantly more. These few cases make the distribution 'stretch' to the right.
Confidence Interval
A confidence interval gives a range of values that likely contain the population mean. In this exercise, we want to estimate the mean student loan debt with a 95% confidence level. This means that if we were to take 100 different samples and calculate the confidence interval for each, about 95 out of 100 of those intervals would contain the true population mean. The confidence interval in this case is from \[17,824.6 \text{ dollars} \leq \mu \leq 30,133.4 \text{ dollars}\]. This range tells us where we believe the actual average loan debt lies, with 95% certainty.
Standard Error
The standard error (SE) measures the accuracy with which a sample represents a population. It鈥檚 calculated as the standard deviation of the sample divided by the square root of the sample size. The formula for standard error is \[SE = \frac{s}{\sqrt{n}}\].
The standard error helps in understanding how much the sample mean will differ from the population mean. In our example: \[SE = \frac{31400}{\sqrt{100}} = 3140\].
A smaller standard error means our sample mean is a more accurate reflection of the population mean.
Critical Value
The critical value is a factor used to compute the margin of error in confidence intervals. It depends on the desired confidence level. For a 95% confidence level, the critical value is approximately 1.96, which corresponds to the z-score in a standard normal distribution. To find a critical value:
  • Decide your confidence level (e.g., 95%)
  • Find the corresponding z-score (for 95%, it鈥檚 around 1.96)
The critical value ensures that our calculation accounts for the confidence level, influencing the breadth of our confidence interval.
Margin of Error
The margin of error (ME) represents the maximum amount by which the sample estimate is expected to differ from the true population value. It鈥檚 calculated by multiplying the standard error by the critical value: \[ME = z \times SE\].
In our exercise: \[ME = 1.96 \times 3140 = 6154.4\].
This tells us that our sample mean will differ from the population mean by up to 6154.4 dollars. A smaller margin of error means a more precise estimate. To achieve this, we can increase the sample size, which decreases the standard error.

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Most popular questions from this chapter

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