Problem 37 - Basic Statistics for Business & Economics

Question

A study of 25 graduates of four-year public colleges revealed the mean amount owed by a student in student loans was 55,051. The standard deviation of the sample was 7,568. Construct a 90 % confidence interval for the population mean. Is it reasonable to conclude that the mean of the population is actually 55,000 ? Explain why or why not.

Step-by-Step Solution

Answer

Yes, 55,000 is within the 90% confidence interval, so it could be the population mean.

Step 1: Identify the Information

First, let's gather the information provided in the problem statement. We have:- Sample mean, \( \bar{x} = 55,051 \)- Sample size, \( n = 25 \)- Sample standard deviation, \( s = 7,568 \)- Confidence level, \( 90\% \)

Step 2: Determine the Critical Value

Since the sample size is less than 30, we'll use the t-distribution to find the critical value. For a 90% confidence interval and degrees of freedom \( df = n - 1 = 24 \), look up the t-table or use a calculator to find the critical value \( t_{\alpha/2} \approx 1.711 \).

Step 3: Calculate the Standard Error

The standard error (SE) is calculated using the formula: \[ SE = \frac{s}{\sqrt{n}} \]Substitute the given values: \[ SE = \frac{7,568}{\sqrt{25}} = \frac{7,568}{5} = 1,513.6 \]

Step 4: Calculate the Margin of Error

The margin of error (E) can be calculated using:\[ E = t_{\alpha/2} \times SE \]Substitute the values to find E:\[ E = 1.711 \times 1,513.6 \approx 2,589.1 \]

Step 5: Construct the Confidence Interval

The confidence interval is calculated as:\[ (\bar{x} - E, \bar{x} + E) \]Substitute the values to find the interval:\[ (55,051 - 2,589.1, 55,051 + 2,589.1) = (52,461.9, 57,640.1) \]

Step 6: Evaluate the Hypothesis

We need to decide if it's reasonable to conclude that the population mean is 55,000. The constructed confidence interval is (52,461.9, 57,640.1). Since 55,000 is within this interval, it is reasonable to conclude that 55,000 could be the population mean.

Key Concepts

t-distribution

In statistics, when we deal with smaller sample sizes, typically less than 30, we use the t-distribution instead of the normal distribution. The t-distribution is a type of probability distribution that is symmetric, similar to the normal distribution, but has heavier tails. This means it is more prone to producing values that fall far from its mean.

Why use the t-distribution? Because it accounts for the extra variability that can occur with smaller sample sizes, providing a more accurate critical value for calculating confidence intervals. The critical value can be found using a t-table or statistical software based on the desired confidence level and degrees of freedom, defined as the sample size minus one (i.e., \( df = n - 1 \)).

In our exercise, the sample size was 25, leading to 24 degrees of freedom. For a 90% confidence level, the critical value was approximately 1.711. This critical value is key when calculating the margin of error for the confidence interval.

standard error

The standard error (SE) is a crucial component in estimating how much the sample mean is expected to vary from the true population mean. It tells us the precision of the sample mean estimate.

Standard error is calculated using the formula:

• \( SE = \frac{s}{\sqrt{n}} \)

where \( s \) is the sample standard deviation, and \( n \) is the sample size.

In our example, with a standard deviation of 7,568 and a sample size of 25, the standard error calculates to \( SE = 1,513.6 \). This figure serves as an indicator of how much disparity there can be from one sample to another, or from the sample mean to the true population mean.

margin of error

The margin of error (E) expresses the range within which we expect the population mean to fall. It's a critical aspect of constructing a confidence interval, which ultimately helps us make inferences about the population based on sample data.

The margin of error is calculated using the formula:

• \(E = t_{\alpha/2} \times SE \)

where \( t_{\alpha/2} \) is the critical value from the t-distribution, and SE is the standard error.

In the exercise, with \( t_{\alpha/2} \approx 1.711 \) and \( SE = 1,513.6 \), the margin of error is \( E \approx 2,589.1 \). We then use this margin to create the confidence interval, determining that the true population mean lies within 2,589.1 units of the sample mean.

sample mean

The sample mean, denoted as \( \bar{x} \), is the average of all measurements in a sample. It is a simple yet powerful statistic used to estimate the population mean. The sample mean serves as the center of the confidence interval.

In our given exercise, the sample mean was 55,051. This value provides a starting point for our calculations and helps in constructing the confidence interval.

Though the sample mean is a robust estimate of the population mean, it does not account for variability among samples. That's why we consider the standard error and margin of error to provide a range (or interval) where the population mean likely resides.