Question

A random sample of \(n=12\) four-year-old red pine trees was selected, and the diameter (in inches) of each tree's main stem was measured. The resulting observations are as follows: \(\begin{array}{llllllll}11.3 & 10.7 & 12.4 & 15.2 & 10.1 & 12.1 & 16.2 & 10.5\end{array}\) \(\begin{array}{llll}11.4 & 11.0 & 10.7 & 12.0\end{array}\) a. Compute a point estimate of \(\sigma\), the population standard deviation of main stem diameter. What statistic did you use to obtain your estimate? b. Making no assumptions about the shape of the population distribution of diameters, give a point estimate for the population median diameter. What statistic did you use to obtain the estimate? c. Suppose that the population distribution of diameter is symmetric but with heavier tails than the normal distribution. Give a point estimate of the population mean diameter based on a statistic that gives some protection against the presence of outliers in the sample. What statistic did you use? d. Suppose that the diameter distribution is normal. Then the 90 th percentile of the diameter distribution is \(\mu\) t \(1.28 \sigma\) (so \(90 \%\) of all trees have diameters less than this value). Compute a point estimate for this percentile. (Hint: First compute an estimate of \(\mu\) in this case; then use it along with your estimate of \(\sigma\) from Part (a).)

Short answer

a. The sample standard deviation (s) is the point statistic used for estimating \(\sigma\). b. The median is the point statistic used to estimate the population median diameter. c. The trimmed mean is used to provide a robust estimate of the mean diameter. d. The point estimate for the 90th percentile of the diameter distribution can be found by adding 1.28 times the standard deviation (s) to the mean diameter.

Step-by-step solution

Arrange Data

First, arrange the given data in ascending order. This will assist to compute median and other statistics.

Compute Standard Deviation

The population standard deviation (σ) is a measure of the amount of variation or dispersion in a set of values. Use the formula for standard deviation. This involves subtracting the mean from each data point, squaring the result, adding them all together, dividing by the number of data points (n), and taking the square root.

Compute Median

To find the median, which is the point estimate for the diameter here, find the middle value in the ordered dataset. If there is an even number of observations, calculate the mean of the two middle values.

Compute Robust Mean

To compute a robust mean (a statistic that mitigates the impact of outliers), trim off a certain percentage of the largest and smallest values. Here, remove the smallest and largest value and then compute the arithmetic mean of the remaining data. This is known as a trimmed mean.

Compute 90% Percentile

Assuming a normal distribution, the value at the 90th percentile is equal to \(\mu + 1.28 \sigma\). First, compute the population mean (\(\mu\)) from the sample data, then use the standard deviation (\(\sigma\)) computed in step 2. Add 1.28 times the standard deviation to the mean to compute the 90th percentile.

Key concepts

Population Standard Deviation

When statisticians refer to the population standard deviation, they are talking about a measure that indicates the spread of a set of values in an entire population. It is a way to quantify the amount of variation or dispersion of a set of data points. To calculate it, each data point is subtracted from the mean to find its deviation, squared, summed together, and then divided by the number of data points. Finally, taking the square root gives us the population standard deviation, symbolized by the Greek letter \( \sigma \).

In practice, when we do not have access to the whole population, we derive an estimate of the population standard deviation from a sample. This estimate is based on the sample standard deviation, but the calculations involve a slight adjustment to account for the fact that we are estimating using a smaller group of values. This process was illustrated in the exercise, showing how to estimate \( \sigma \) for the population of tree diameters using a sample.

Sample Median

The median is the value separating the higher half from the lower half of a data set. Finding the median is straightforward: Arrange the data points from smallest to largest, and the middle value is the median. In cases where there is no single middle value because you have an even number of data points, the median is then calculated by taking the average of the two middle values.

The median is particularly useful as it is not affected by outliers or a skewed distribution as much as the mean, meaning it can often give a better idea of 'typical' values in a dataset. This property makes the median a reliable point estimate for the population median, which was determined in the exercise for the diameters of the pine trees.

Trimmed Mean

A trimmed mean is a method of averaging that helps eliminate the influence of outliers or data points that are significantly different from the rest of the data set. Outliers can sometimes distort the mean in a way that does not accurately reflect the typical values in a set. To calculate a trimmed mean, we first remove a certain percentage of the smallest and largest values from the dataset. Then, we calculate the arithmetic mean of the remaining values.

The number of values to trim depends largely on the dataset and the discretion of the analyst. The trimmed mean offers a compromise between the mean, which includes all data values, and the median, which is solely the middle value. The solution to the exercise suggests this method as a robust alternative, suitable for samples from populations with heavy tails, like the distribution of tree diameters mentioned.

90th Percentile Calculation

Percentiles are measures that divide your data into 100 equal parts. The 90th percentile is the value below which 90% of the data points fall. To calculate this, assuming a normal distribution, you can use the formula \(percentile = \mu + z * \sigma \) where \( z \) is the z-score associated with the 90th percentile, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. The z-score represents the number of standard deviations a given data point is from the mean.

For the 90th percentile, the z-score is 1.28. The exercise showed how to estimate the 90th percentile of tree diameters using this method. In real-world terms, calculating such a percentile could be used to infer that 90% of the trees would have a diameter less than this computed value, which can be useful in forestry management and planning.