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In evaluating a forage crop, it is important to measure the concentration of various constituents in the plant tissue. In a study of the reliability of such measurements, a batch of alfalfa was dried, ground, and passed through a fine screen. Five small \((0.3 \mathrm{gm})\) aliquots of the alfalfa were then analyzed for their content of insoluble ash. The results \((\mathrm{gm} / \mathrm{kg})\) were as follows: $$ \begin{array}{lllll} 10.0 & 8.9 & 9.1 & 11.7 & 7.9 \end{array} $$ For these data, calculate the mean, the standard deviation, and the standard error of the mean.

Short Answer

Expert verified
The mean is 9.52 gm/kg, the standard deviation is 1.44 gm/kg, and the standard error of the mean is 0.64 gm/kg.

Step by step solution

01

Calculate the Mean

To calculate the mean (average) of the insoluble ash contents, sum up all the values and divide by the total number of measurements. The formula for the mean, denoted as \(\bar{x}\), is given by \(\bar{x} = \frac{\sum x_i}{n}\) where \(x_i\) represents each value and \(n\) is the number of observations. For the given data set, the calculation is \(\bar{x} = \frac{10.0 + 8.9 + 9.1 + 11.7 + 7.9}{5}\).
02

Calculate the Variance

To find the standard deviation, first calculate the variance. The variance is the average of the squared differences from the Mean. Use the formula \(s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}\), where \(s^2\) is the variance, \(x_i\) is each value, \(\bar{x}\) is the mean, and \(n\) is the total number of observations. Apply the values accordingly and compute the sum of the squares of the differences.
03

Calculate the Standard Deviation

The standard deviation (\(s\)) is the square root of the variance. Once we have the variance from Step 2, take its square root to obtain the standard deviation. Thus, \(s = \sqrt{s^2}\).
04

Calculate the Standard Error of the Mean

The standard error of the mean (SEM) provides an estimate of how much the sample mean is expected to fluctuate due to sampling variability. The SEM is calculated by dividing the standard deviation by the square root of the number of observations. Thus, the formula is \(SEM = \frac{s}{\sqrt{n}}\), where \(n\) is the total number of observations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean Calculation in Statistics
The mean, often referred to as the average, is a fundamental statistical measure that represents the central point of a data set. It is calculated by summing all the individual values of a dataset and then dividing by the number of observations. In life sciences, such as evaluating the concentration of constituents in plant tissue, the mean provides an indication of the overall level of a particular variable. For example, when measuring insoluble ash content in alfalfa samples, the mean gives us a sense of the typical ash concentration.

To compute the mean (\bar{x}) for a set of values \(x_i\), the formula \(\bar{x} = \frac{\sum x_i}{n}\) is used, where \(\sum x_i\) represents the sum of all measured values and \(n\) denotes the total number of samples. If we apply this to the alfalfa example provided, with values of 10.0, 8.9, 9.1, 11.7, and 7.9 grams per kilogram, the mean concentration of insoluble ash would be the sum of these values divided by 5, the total number of aliquots tested.
Variance in Statistics
Variance measures the dispersion of data points in a dataset around their mean. In other words, it quantifies how much the data numbers are spread out. This is particularly useful in the life sciences for assessing the consistency of experimental results, such as measurements of plant tissue constituents.

The formula to calculate the sample variance (\(s^2\)) is \(s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}\), where \(x_i\) are the individual values, \(\bar{x}\) is the mean, and \(n\) is the number of observations. The denominator \(n-1\) is used instead of \(n\) to correct for bias in the estimation of the population variance from a sample; this is known as Bessel's correction. The step of squaring the differences from the mean is crucial, as it ensures that negative deviations do not cancel out positive ones. Applying this to our example with the alfalfa measurements would involve subtracting the mean from each measured value, squaring the result, summing these squared differences, and finally dividing by 4 (since there are 5 samples).
Standard Deviation
Standard deviation is arguably one of the most important statistical tools and indicates how measurements spread around their mean. In practical terms, a low standard deviation means that the data points tend to be close to the mean, while a high standard deviation implies that the data points are spread out over a wider range. This is particularly relevant in life sciences when determining the reliability of various measurements.

The standard deviation (\(s\)) is simply the square root of the variance, obtained using the formula \(s = \sqrt{s^2}\). Taking the square root is necessary to bring the units of measure back to the original data's units. In the case of the insoluble ash content in the alfalfa study, we first calculate the variance and then take its square root to get the standard deviation. This computed standard deviation informs us about the variation in ash content across the different aliquots tested.
Standard Error of the Mean
The standard error of the mean (SEM) is a statistic that represents how much the sample mean (average) is expected to fluctuate from the true population mean. The SEM is especially useful when dealing with samples of a population, as is commonly the case in life sciences studies. It is pivotal for understanding and quantifying the precision of the mean as an estimate of the expected true value.

The SEM can be calculated with the formula \(SEM = \frac{s}{\sqrt{n}}\), where \(s\) is the sample standard deviation and \(n\) is the number of observations. The denominator, \(\sqrt{n}\), normalizes this deviation by considering the size of the sample, which implies that as the number of measurements increases, the standard error decreases, boosting the confidence in the sample mean's accuracy. In our alfalfa example, once we have the standard deviation for the ash content measurements, we divide it by the square root of 5 to find the SEM and determine the reliability of our mean measurement.

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Most popular questions from this chapter

Data from two samples gave the following results: $$\begin{array}{|lcc|}\hline & \text { Sample 1 } & \text { Sample 2 } \\\\\hline n & 22 & 21 \\\\\bar{y} & 1.7 & 2.4 \\\\\text { SE } & 0.5 & 0.7 \\\\\hline \end{array}$$ Compute the standard error of \(\left(\bar{Y}_{1}-\bar{Y}_{2}\right)\).

For the 28 lamb birthweights of Example \(6.2 .2,\) the mean is \(5.1679 \mathrm{~kg},\) the \(\mathrm{SD}\) is \(0.6544 \mathrm{~kg},\) and the \(\mathrm{SE}\) is \(0.1237 \mathrm{~kg}\) (a) Construct a \(95 \%\) confidence interval for the population mean. (b) Construct a \(99 \%\) confidence interval for the population mean. (c) Interpret the confidence interval you found in part (a). That is, explain what the numbers in the interval mean. (Hint: See Examples 6.3 .4 and \(6.3 .5 .)\) (d) Often researchers will summarize their data in reports and articles by writing \(\bar{y} \pm \mathrm{SD}(5.17 \pm 0.65)\) or \(\bar{y} \pm \mathrm{SE}(5.17 \pm 0.12) .\) If the researcher of this study is planning to compare the mean birthweight of these Rambouillet lambs to another breed, Booroolas, which style of presentation should she use?

Suppose you are discussing confidence intervals with someone who has a limited statistics background. Briefly explain, without doing any calculations, why a \(100 \%\) confidence interval for a mean is infinitely wide.

In an experiment on soybean varieties, individually potted soybean plants were grown in a greenhouse, with 10 plants of each variety used in the experiment. From the harvest of each plant, five seeds were chosen at random and individually analyzed for their percentage of oil. This gave a total of 50 measurements for each variety. To calculate the standard error of the mean for a variety, the experimenter calculated the standard deviation of the 50 observations and divided by \(\sqrt{50}\). Why would this calculation be of doubtful validity?

A pharmacologist measured the concentration of dopamine in the brains of several rats. The mean concentration was \(1,269 \mathrm{ng} / \mathrm{gm}\) and the standard deviation was \(145 \mathrm{ng} / \mathrm{gm} .^{4}\) What was the standard error of the mean if (a) 8 rats were measured? (b) 30 rats were measured?

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