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The sample mean is a critical concept in statistics, representing the arithmetic average of a set of values from a sample. It is denoted by \( \bar{x} \), and it serves as an estimate for the population mean, which is the average of all possible observations in a population. In simple terms, the sample mean gives us an idea of the central tendency or the typical value in the data set.

When dealing with a sample, we do not have access to all the data in the population, so the sample mean is our best estimate. In the original problem, the sample mean is given as 32. This means that the average value of the 50 items in the sample is 32.

The accuracy of the sample mean as an estimate of the population mean depends on the sample size. Larger samples tend to provide more accurate estimates of the population mean because they are more likely to capture the diversity within the entire population.

The standard error (SE) measures the precision of the sample mean as an estimate of the population mean. Specifically, it quantifies how much we might expect the sample mean to vary from sample to sample due to random sampling.

The formula for the standard error of the sample mean is: \( \text{SE} = \frac{\sigma}{\sqrt{n}} \) where \( \sigma \) is the population standard deviation and \( n \) is the sample size. In the given exercise, the standard deviation \( \sigma \) is 6 and the sample size \( n \) is 50. Consequently, the standard error is:

• \( \text{SE} = \frac{6}{\sqrt{50}} \approx 0.849 \)

This value indicates how much the sample mean of 32 could vary, assuming different samples of 50 items are taken from the population. A smaller SE indicates a more precise estimate of the population mean.

The Z-score is a statistical measurement that describes a value's position relative to the mean of a set of values. In confidence intervals, the Z-score represents the number of standard errors a value is away from the mean. For a given confidence level, the corresponding Z-score helps us determine the width of the confidence interval.

For instance, familiar Z-scores include:

• For a 90% confidence level, the Z-score is approximately 1.645.
• For a 95% confidence level, the Z-score is approximately 1.960.
• For a 99% confidence level, the Z-score is approximately 2.576.

These values are derived from standard normal distribution tables or computational tools, helping us determine how much variability we are accounting for in our calculations. The choice of confidence level, and thus the Z-score, affects the margin of error and the width of the confidence interval.

The margin of error (ME) helps us understand the range within which we expect the true population mean to lie, based on the sample data. It reflects the uncertainty inherent in using a sample mean as an estimate of the population mean.

The formula to calculate the margin of error is: \( \text{ME} = Z \times \text{SE} \). This means we multiply the Z-score by the standard error. This calculation tells us how far off our sample mean could realistically be from the true population mean, within a specific confidence level. For the given problem, the margin of error for different confidence levels are calculated as follows:

• 90% Confidence: \( \text{ME} = 1.645 \times 0.849 \approx 1.396 \)
• 95% Confidence: \( \text{ME} = 1.960 \times 0.849 \approx 1.664 \)
• 99% Confidence: \( \text{ME} = 2.576 \times 0.849 \approx 2.187 \)

These margins determine how wide the confidence intervals are, ensuring that they realistically capture where the true population mean likely falls. A larger margin of error results in a wider confidence interval.