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Chapter 8: Problem 5

The reaction times for a random sample of 9 subjects to a stimulant were recorded as \(2.5,3.6,3.1,4.3,\) 2.9. \(2.3,2.6,4.1,\) and 3.4 seconds. Calculate (a) the mean; (b) the median.

Short Answer

Expert verified
The mean of the reaction times is 3.2 seconds and the median is 3.1 seconds.

Step by step solution

01

Calculate the Sum of all Values

The sum of all values can be calculated by adding together the individual reaction times: \(2.5 + 3.6 + 3.1 + 4.3 + 2.9 + 2.3 + 2.6 + 4.1 + 3.4\). The total sum equals 28.8 seconds.
02

Compute the Mean

The mean (\(\mu\)) is the sum (\(\Sigma\)) of all the values divided by the count (n) of the values. Here, the sum of all the values has been found to be 28.8 and the count of the values is 9. Therefore, \(\mu = \frac{\Sigma}{n} = \frac{28.8}{9} = 3.2\) seconds.
03

Sort the values in Ascending Order

To calculate the median, first sort the data points in ascending order: \(2.3, 2.5, 2.6, 2.9, 3.1, 3.4, 3.6, 4.1, 4.3\).
04

Calculate the Median

The median is the middle value of a data set. Given that there are 9 data points in total, the median falls on the 5th position in a sorted dataset. Therefore, the median of the dataset is 3.1 seconds.

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Key Concepts

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

Mean Calculation
Understanding how to calculate the mean, commonly known as the average, is essential in data analysis and is a fundamental aspect of descriptive statistics. The mean represents the central tendency of a dataset, which is the point where the data points tend to cluster around. To calculate the mean, you sum up all the individual numerical values in a dataset and then divide this total by the number of observations.

For example, let's consider the dataset from the exercise consisting of a series of reaction times to a stimulus: the sum of all reaction times is 28.8 seconds. Since there are 9 subjects, we divide 28.8 by 9, resulting in a mean reaction time of 3.2 seconds. This number provides a quick snapshot of the 'average' reaction time within this group, offering a simplistic yet powerful way to summarize the data with a single value.
Median Calculation
The median is another measure of central tendency, like the mean, but with a slightly different insight into the dataset. It is particularly informative when the data is skewed or contains outliers. The median is the middle value when a data set is ordered from smallest to largest. For an odd number of observations, the median is the number that falls right in the center.

In our exercise with the 9 recorded reaction times, once we arrange them in ascending order, the fifth value represents the median because it is the middle observation. In this case, the median reaction time is 3.1 seconds. If there had been an even number of observations, the median would be the average of the two central numbers. The median can sometimes provide a better sense of a 'typical' value in a skewed distribution, where the mean might be distorted by extremely high or low values.
Data Analysis
Data analysis involves examining, cleaning, transforming, and modeling data to discover useful information, inform conclusions, and support decision-making. Both mean and median calculations are part of the initial steps in descriptive statistics, which aim to describe the main features of a collection of data quantitatively.

In the context of the given exercise, once we have the mean and median, we can use these measures to compare the reaction times of different samples or to understand the distribution of reaction times within this sample. Descriptive statistics also include the computation of other parameters like mode, range, variance, and standard deviation, providing a comprehensive picture of the data. When analyzing data, it's crucial to understand each statistic's nuances and when it's appropriate to use each measure. For instance, in datasets with outliers or non-normal distributions, the median might be more representative than the mean.

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Most popular questions from this chapter

If the standard deviation of the mean for the sampling distribution of random samples of size 36 from a large or infinite population is \(2,\) how large must the size of the sample become if the standard deviation is to be reduced to \(1.2 ?\)

The scores on a placement test given to college freshmen for the past five years are approximately normally distributed with a mean \(\mu=71\) and a variance \(a^{2}=8\). Would you still consider \(\sigma^{2}=8\) to be a valid value of the variance if a random sample of 20 students who take this placement test this year obtain a value of \(s^{2}=20 ?\)

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The chemical benzene is highly toxic to humans. However, it is used in the manufacture of many medicine dyes, leather, and many coverings. In any production process involving benzene, the water in the output of the process must not exceed 7950 parts per million (ppm) of benzene because of government regulations. For a particular process of concern the water sample was collected by a manufacturer 25 times randomly and the sample average \(x\) was \(7960 \mathrm{ppm}\). It is known from historical data that the standard deviation \(a\) is \(100 \mathrm{ppm}\) (a) What is the probability that the sample average in this experiment would exceed the government limit if the population mean is equal to the limit? Use the central limit theorem. (b) Is an observed \(\bar{x}=7960\) in this experiment firm evidence that the population mean for the process exceeds the government limit? Answer your question by computing $$P\left(X^{-}>_{-} 7960 \mid p=7950\right).$$ Assume that the distribution of benzene concentration is normal.

For an F-distribution find (a) \(f_{0.05}\) with \(v_{1}=7\) and \(v_{2}=15\); (b) \(f_{0.05}\) with \(v_{1}=15\) and \(v_{2}=7\) : (c) \(f_{0.01}\) with \(v_{1}=24\) and \(v_{2}=19\); (el) \(f_{0.95}\) with \(v_{1}=19\) and \(v_{2}=24\) (e) \(f_{0.99}\) with \(v_{1}=28\) and \(v_{2}=12\).
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