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The population mean is a central concept in statistics that represents the average of all possible values in a population. It helps us understand the central tendency in data. While we may not know the true population mean, we can estimate it using sample data. The sample mean, denoted as \(\bar{x}\), is often used to approximate the true population mean.
In scenarios where a sample is selected from a large population, like a study surveying car dealers, the average earnings or values recorded can provide an insight into the population mean. In our exercise, the sample mean was \(\$290\), indicating the average profit per car sold among the sampled dealers. Using this, along with other statistical tools, we construct a confidence interval to estimate the range in which the true population mean will fall with a certain level of confidence.

The t-distribution is a useful tool when working with small samples and unknown population variances. It is similar to the normal distribution but has heavier tails, which accounts for the additional uncertainty due to the smaller sample size.
In the given exercise, since the sample size is only 20, we apply the t-distribution. A critical aspect of using the t-distribution is determining the degrees of freedom, which is calculated as \(n-1\). For our problem, that’s \(20-1 = 19\) degrees of freedom. The t-score can then be obtained from a t-table, which for a 95% confidence level and 19 degrees of freedom, is approximately 2.093.
This t-score helps adjust for variability and provides a multiplier for the standard error, leading us toward calculating the confidence interval accurately.

The standard error of the mean is crucial when estimating how much the sample mean \(\bar{x}\) is expected to vary from the true population mean due to sampling variability. It provides a measure of the statistical accuracy of an estimate.
The standard error is calculated using the formula: \[ SE = \frac{s}{\sqrt{n}} \]Here, \(s\) represents the sample standard deviation, and \(n\) is the sample size.
In our example, with \(s = 125\) and \(n = 20\), the standard error calculated is approximately \(27.95\).
This value gives us an idea of how "spread out" our sample mean might be around the population mean. A smaller standard error indicates a more precise estimate of the population mean. The standard error is integral to determining the margin of error, which will further aid in building the confidence interval.

The margin of error quantifies the range that the sample mean is likely to lie within concerning the true population mean with a specified confidence level. It directly influences the width of the confidence interval.
To compute the margin of error, we multiply the standard error of the mean by the appropriate t-score: \[ ME = t \times SE \]In our problem, the previously calculated standard error is \(27.95\), and the t-score for a 95% confidence level with 19 degrees of freedom is 2.093.
Thus, the margin of error is calculated as \(2.093 \times 27.95 \approx 58.47\).
This margin of error depicts the maximum expected difference between the sample mean and the population mean at the 95% confidence level. By adding and subtracting this margin from the sample mean, we form the confidence interval, indicating the range likely containing the true population mean.