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Red blood cell counts \(\left(10^{-3} \mathrm{X}\right.\) cells per \(\mathrm{mm}^{3}\) ) of 15 lizards had an average of \(843.4 .\) The SD and the SE were, in random order, 64.9 and \(251.2 .\) Which is the \(S D,\) and which is the SE? How do you know?

Short Answer

Expert verified
The SD is 251.2, and the SE is 64.9 because SE is calculated as SD divided by the square root of the sample size, and SE is always smaller than SD.

Step by step solution

01

Understanding SD and SE Concepts

Understand the definitions of Standard Deviation (SD) and Standard Error (SE). SD measures the amount of variability or dispersion from the average (mean) for a set of data. SE measures how far the sample mean (average) of the data is likely to be from the true population mean. The SE is always smaller than the SD because it is the SD divided by the square root of the sample size.
02

Comparing SE and SD Values

Given that SE is calculated as the SD divided by the square root of the sample size, it is always smaller than the SD unless the sample size is 1. In this case, two values are provided: 64.9 and 251.2, since the SE must be smaller than the SD, the SE must be 64.9 and the SD must be 251.2.
03

Verifying with Sample Size

To confirm, calculate the SE using the sample size (n = 15 lizards) and the SD. SE equals SD divided by the square root of n. The only way to achieve a smaller number from the value of SD (251.2) is to divide it by the square root of 15 (which is approximately 3.873). This would give us a number smaller than 251.2 and would be in the same range as 64.9.

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Key Concepts

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

Standard Deviation (SD)
Standard Deviation (SD) is a statistical measure that quantifies the amount of variation or dispersion within a set of data points. Imagine you have a group of students' test scores. If most students scored around the class average, the variability of scores is low, and so is the standard deviation. Now if the scores are widely spread out, with some very high and some very low scores, the standard deviation would be high. This tells us that the data points are more spread out from the mean. The formula to calculate SD is:
\[ SD = \sqrt{\frac{1}{N} \sum_{i=1}^{N}(x_i - \bar{x})^2} \]
where \( x_i \) is each score, \( \bar{x} \) is the average score, and \( N \) is the number of scores. In the exercise regarding red blood cell counts of lizards, since SD represents scatter of the individual data points around the mean, a value of 251.2 would adequately reflect higher variability as compared to 64.9.
Standard Error (SE)
Standard Error (SE) is closely related to standard deviation, but while SD measures the variability of individual data points, SE measures the precision of the sample mean as an estimate of the true population mean. SE is calculated using the standard deviation and the sample size, and it decreases as sample size increases. This is because with larger samples, we expect our estimate of the mean to be more accurate. The formula for SE is:
\[ SE = \frac{SD}{\sqrt{n}} \]
where \(SD\) is the standard deviation of the sample and \(n\) is the sample size. Referring back to the problem at hand, 64.9 corresponds to SE because it is the smaller number resulting from dividing the larger SD by the square root of the sample size, thus reflecting the expected variation of the sample mean from the true population mean.
Sample Size
Sample Size (n) plays a crucial role when it comes to statistical analysis. It refers to the number of observations or replicates included in a statistical sample. The size of the sample can greatly influence the reliability of conclusions drawn from the data. For instance, larger samples typically lead to more precise estimates of the population parameter, such as the population mean. This concept underpins the calculations for both standard deviation and standard error, with larger samples generally yielding a lower SE. In the given example with lizards, having 15 lizards in the sample allows us to calculate the SE, which would not be possible or meaningful with a sample size of one.
Variability in Data
Variability in data refers to how spread out or clustered individual data points are in a dataset. It is a fundamental concept in statistics as it impacts virtually every statistical measure, including SD and SE. Measures of variability, like the range, interquartile range, and standard deviation, provide insights into the consistency or predictability of the data. A dataset with low variability will have data points that are close to each other and the mean, whereas a dataset with high variability will have data points that are more spread out. Understanding variability helps in assessing the data's behavior and in making predictions. In the biological study mentioned, the variability of red blood cell counts, which has high SD, suggests a diverse health status among the lizards.
Population Mean Estimate
The population mean estimate is a statistical term used to describe the average value of a trait within a population. When researchers collect data from a sample, they use the average of that sample to estimate the average for the entire population. The sample mean is a good estimator for the population mean, but it is not perfect because it is based on a subset, not the entire population. The standard error gives insight into how good that estimate might be by providing a margin of error; smaller standard errors mean a more precise estimate. In the context of our lizard population, the sample mean provides an estimate of the average red blood cell count per cubic millimeter for all lizards of that type, while the SE indicates the reliability of this estimate.

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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 \(\left(\bar{Y}_{1}-\bar{Y}_{2}\right)\).

Over a period of about 9 months, 1,353 women reported the timing of each of their menstrual cycles. For the first cycle reported by each woman, the mean cycle time was 28.86 days, and the standard deviation of the 1,353 times was 4.24 days. \(^{52}\) (a) Construct a \(99 \%\) confidence interval for the population mean cycle time. (b) Because environmental rhythms can influence biological rhythms, one might hypothesize that the population mean menstrual cycle time is 29.5 days, the length of the lunar month. Is the confidence interval of part (a) consistent with this hypothesis?

For each of the following, decide whether the description fits the SD or the SE. (a) This quantity is a measure of the accuracy of the sample mean as an estimate of the population mean. (b) This quantity tends to stay the same as the sample size goes up. (c) This quantity tends to go down as the sample size goes up.

A study was conducted to determine whether relaxation training, aided by biofeedback and meditation, could help in reducing high blood pressure. Subjects were randomly allocated to a biofeedback group or a control group. The biofeedback group received training for 8 weeks. The table reports the reduction in systolic blood pressure \((\mathrm{mm} \mathrm{Hg})\) after eight weeks. \(^{42}[\) Note: Formula \((6.7 .1)\) yields 190 degrees of freedom for these data. (a) Construct a \(95 \%\) confidence interval for the difference in mean response. (b) Interpret the confidence interval from part (a) in the context of this setting. $$ \begin{array}{|lcc|} \hline & \text { Biofeedback } & \text { Control } \\ \hline n & 99 & 93 \\ \bar{y} & 13.8 & 4.0 \\ \text { SE } & 1.34 & 1.30 \\ \hline \end{array} $$

Prothrombin time is a measure of the clotting ability of blood. For 10 rats treated with an antibiotic and 10 control rats, the prothrombin times (in seconds) were reported as follows \(^{43}\) : $$ \begin{array}{|lcc|} \hline & \text { Antibiotic } & \text { Control } \\ \hline n & 10 & 10 \\ \bar{y} & 25 & 23 \\ s & 10 & 8 \\ \hline \end{array} $$ (a) Construct a \(90 \%\) confidence interval for the difference in population means. (Assume that the two populations from which the data came are normally distributed.) [Note: Formula (6.7.1) yields 17.2 degrees of freedom for these data.] (b) Why is it important that we assume that the two populations are normally distributed in part (a)? (c) Interpret the confidence interval from part (a) in the context of this setting.

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