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The Margin of Error (ME) is an important statistical concept that provides us with a range within which we can expect the true population parameter to lie. In this context, it refers to how much we can expect our sample mean to differ from the actual population mean. The formula to calculate the margin of error for a confidence interval is: \[ ME = z \cdot \frac{s}{\sqrt{n}} \] Here:

• The z-value represents how many standard deviations a data point is from the mean, and it depends on the confidence level we choose. For a 95% confidence level, the z-value is approximately 1.96.
• s is the sample standard deviation, which measures the variation or dispersion of the sample data, given as \(175 in this case.
• n is the sample size, which tells us how many observations are included, here being 600.

Substitute the values into the formula, we have: ME = 1.96 \times (175 / \sqrt{600}) \approx 14.00. The margin of error is \)14.00, meaning we can be 95% confident that the true population mean holiday expenditure is within $14.00 of the sample mean calculated.

The Population Mean is a crucial measure in statistics, representing the average of a set of values from an entire population. When the data for the whole population is not available, we estimate the population mean using sample data. In this exercise, the sample mean is \(649. However, due to possible sample variability, we need a confidence interval to give a range where the true population mean likely falls. A confidence interval gives us a range that is likely to contain the population mean with a specified level of confidence. The formula for the confidence interval is: \[ \bar{x} \pm ME \] where \( \bar{x} \) is the sample mean and ME is the margin of error. Here, the sample mean \( \bar{x} = 649 \) and ME = 14.00. Thus, the confidence interval is: 649 \pm 14 = (635, 663). This interval suggests that we can be 95% confident that the actual average expenditure per household during the holiday season lies between \)635 and \(663. With this interval, it is apparent that there is an increase from last year's mean of \)632, suggesting households may be spending more.

The Sample Standard Deviation, denoted by s, is a measure of the amount of variation or dispersion present within a set of data. It indicates how spread out the data points in a sample are relative to the sample mean. This is pivotal in understanding the consistency or variability within sampled data. In the problem given, the sample standard deviation is \(175. This means that on average, each household's spending differs from the sample mean of \)649 by about $175. A higher standard deviation indicates more spread in the data, while a lower standard deviation suggests that data points are closer to the mean. The sample standard deviation is crucial for calculating the margin of error as it helps assess the reliability of the sample mean as an estimator for the population mean. Using the formula: \[ ME = z \cdot \frac{s}{\sqrt{n}} \] The standard deviation (s) directly affects the margin of error. A higher standard deviation generally results in a larger margin of error, indicating less precision in our estimate of the population mean. Thus, understanding sample standard deviation helps in evaluating how sure we can be about our estimations.