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The sample mean is a fundamental concept in statistics that represents the average of a set of observations from a sample. In this exercise, we're looking at the sediment density of a region where the mean, or sample mean, is a key focus. For a sample of 25 specimens, this sample mean will help determine the probability of certain average sediment densities occurring.

When we talk about the sample mean, symbolized as \( \overline{X} \), it's crucial to understand how it is calculated. We take all the observed values in our sample, add them together, and divide by the number of observations. This simple average gives us an estimate of the population mean, which is the central tendency of the overall data from which the sample was drawn.

The quality of the sample mean as an estimator increases as the sample size grows. It is essential because it allows us to make inferences about the overall population based on a smaller, manageable number of observations. This inferential process is one of the core tasks in statistical analysis and is crucial for determining probabilities, as seen in this example.

A z-score is a way of standardizing individual data points concerning the mean and standard deviation of a dataset. It indicates how many standard deviations away a data point is from the mean, allowing different data to be compared on a standard scale.

In the context of this exercise, calculating the z-score helps determine the probability associated with a particular sample mean. For example, when we calculated the z-score for a sample mean sediment density of 3.00, the formula used was: \[ z = \frac{\overline{X} - \mu}{\sigma/\sqrt{n}} \]Here, \( \overline{X} = 3.00 \) is the sample mean, \( \mu = 2.65 \) is the population mean, and \( \sigma/\sqrt{n} = 0.17 \) is the standard error of the mean based on our sample size of 25.

Using these values, we calculated a z-score of 2.06. This standardized value enables us to use a standard normal distribution table to find probabilities corresponding to the z-score, in turn helping us answer questions about the likelihood of certain outcomes.

Determining the right sample size is critical for achieving desired confidence levels in results. In statistical analysis, the sample size impacts the standard error, precision of estimations, and, subsequently, the reliability of conclusions drawn from a study.

For the example problem, we wanted to ensure that the probability of the sample mean being less than or equal to 3.00 was at least 0.99. This requires calculating the necessary sample size that meets the condition. We used the z-score formula rearranged to solve for the sample size:\[ n = \left(\frac{\sigma \times z}{\overline{X} - \mu}\right)^2 \]
Here, \( z \approx 2.33 \) ensures a probability of 0.99, \( \sigma = 0.85 \) is the population standard deviation, and the difference \((3.00 - 2.65)\) is how far the desired sample mean is from the population mean.
The calculation yielded \( n \approx 38.898 \), which we round up to 39 to ensure meeting the probability criteria. This step is vital because each additional sample size increases the confidence and precision of the estimated statistics.

Probability calculation helps in quantifying the likelihood of a particular outcome or range of outcomes in statistical terms. In our exercise, we determine probabilities associated with specific sample mean values of the sediment density.

The standard normal distribution table is instrumental for this. After converting the sample mean to a z-score, we use this table to find the corresponding probabilities. In solving part (a), where we wanted the probability of the sample mean being at most 3.00, we found \( P(Z \leq 2.06) \approx 0.9803 \). This tells us that there is a 98.03% chance the sample mean will be 3.00 or less.

Similarly, for a sample mean between 2.65 and 3.00, we used the probabilities:- - From the table: \( 3.00 \) gives us 0.9803- The mean 2.65 equates to a probability of 0.5 since it's the midpoint.The desired probability is then the difference: \( 0.9803 - 0.5 = 0.4803 \), or 48.03%, that the sample mean lies in this range. Such calculations are essential for making informed decisions based on statistical data.