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Chapter 9: Problem 473

Find the median of the samole 34. 29. 26. 37. 31 .

Short Answer

Expert verified
The median of the given sample is \(31\).

Step by step solution

01

Rearrange the sample in ascending order

Sort the given sample in ascending order: 26, 29, 31, 34, 37.
02

Find the median

There are five elements in the data set, which is an odd number. As a result, the median is the middle value. In this case, the median is the third value, which is 31. So, the median of the sample is 31.

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Key Concepts

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

Ascending Order
To find the median of any data set, the first step is to rearrange the data in ascending order. This means organizing the numbers from smallest to largest. Doing this ensures that each number is placed in a sequence, making it easier to locate the middle value.

For example, given the data set 34, 29, 26, 37, and 31, we rearrange it in ascending order to become:
  • 26
  • 29
  • 31
  • 34
  • 37
With this order established, determining the median becomes a lot more straightforward.
Odd Number Data Set
An odd number data set is characterized by having an uneven number of elements. For example, the data set 26, 29, 31, 34, 37 has five elements, clearly an odd number.

When dealing with an odd number of elements, finding the median is simpler because it will always be the middle element of the sorted sequence. This property makes such sets easier to analyze for their central tendency, as you do not have to average two central values like in even-numbered sets.
Middle Value
Identifying the middle value is crucial when calculating the median. In a data set sorted in ascending order, the middle value is the one that separates the lower half from the upper half.

For an odd-numbered data set, as seen in our example, the middle value is the third element, which is 31. Since there are five elements total, the middle position or the third value accurately represents the median.
  • First value: 26
  • Second value: 29
  • Median (middle value): 31
  • Fourth value: 34
  • Fifth value: 37
This single value is the median.
Sample Data Rearrangement
Rearranging data might seem like a simple task but plays a vital role in operations such as median calculation. It involves taking the original list of numbers and systematically changing their order so that they are laid out from the lowest to the highest value.

In our example, the original disorganized sample was 34, 29, 26, 37, and 31. Only by rearranging it into 26, 29, 31, 34, 37 can we effectively identify the middle, or median, value. Rearrangement not only brings clarity but also correctness in further data analysis processes.

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Most popular questions from this chapter

A random sample of 20 boys and 15 girls were given a standardized test. The average grade of the boys was 78 with a standard deviation of 6, while the girls made an average grade of 84 with a standard deviation of 8 . Test the hypothesis that \(\sigma_{1}^{2}=\sigma_{2}^{2}\) against the alternate hypothesis \(\sigma_{1}^{2}<\sigma_{2}^{2}\) where \(\sigma_{1}^{2}\) and \(\sigma_{2}^{2}\) are the variances of the population of boys and girls. Use a .05 level of significance.

A physchologist wishes to determine the variation in I.Q.s of the population in his city. He takes many random samples of size 64 . The standard error of the mean is found to be equal to \(2 .\) What is the population standard deviation?

If \(Z\) is a standard normal variable, use the table of standard normal probabilities to find: (a) \(\operatorname{Pr}(z<0)\) (b) \(\operatorname{Pr}(-12.54)\)

Suppose we have a binomial distribution for which \(\mathrm{H}_{0}\) is \(\mathrm{p}=1 / 2\) where \(\mathrm{p}\) is the probability of success on a single trial. Suppose the type I error, \(\alpha=.05\) and \(\mathrm{n}=100 .\) Calculate the power of this test for each of the following alternate hypotheses, \(\mathrm{H}_{1}: \mathrm{p}=.55, \mathrm{p}=.60, \mathrm{p}=.65, \mathrm{p}=.70\), and \(\mathrm{p}=.75 .\) Do the same when \(\alpha=.01\).

Let \(X\) possess a Poisson distribution with mean \(\mu\), 1.e. $$ \mathrm{f}(\mathrm{X}, \mu)=\mathrm{e}^{-\mu}\left(\mu^{\mathrm{X}} / \mathrm{X} ;\right) $$ Suppose we want to test the null hypothesis \(\mathrm{H}_{0}: \mu=\mu_{0}\) against the alternative hypothesis, \(\mathrm{H}_{1}: \mu=\mu_{1}\), where \(\mu_{1}<\mu_{0}\). Find the best critical region for this test.
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