/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Data on weekday exercise time fo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Data on weekday exercise time for 20 males, consistent with summary quantities given in the paper "An Ecological Momentary Assessment of the Physical Activity and Sedentary Behaviour Patterns of University Students" (Health Education Journal [2010]\(: 116-125),\) are shown below. Calculate and interpret the values of the median and interquartile range. $$ \begin{array}{rrrrrrrrr} 43.5 & 91.5 & 7.5 & 0.0 & 0.0 & 28.5 & 199.5 & 57.0 & 142.5 \\ 8.0 & 9.0 & 36.0 & 0.0 & 78.0 & 34.5 & 0.0 & 57.0 & 151.5 \\ 8.0 & 0.0 & & & & & & & \end{array} $$

Short Answer

Expert verified
The median weekday exercise time for the 20 males is \(35.25\) minutes, indicating half of the males have exercise times less than or equal to this value, and half have times greater than or equal to it. The interquartile range (IQR) is \(70.0\) minutes, showing that the middle 50% of males have exercise times between \(8.0\) and \(78.0\) minutes, providing an insight into the variability of the exercise durations among these university students.

Step by step solution

01

Arrange the Data in Ascending Order

First, we sort the data in ascending order so it will be easier to find the median, first quartile, and third quartile. The sorted data: \[ 0.0,\, 0.0,\, 0.0,\, 0.0,\, 7.5,\, 8.0,\, 8.0,\, 9.0,\, 28.5,\, 34.5,\, 36.0,\, 43.5,\, 57.0,\, 57.0,\, 78.0,\, 91.5,\, 142.5,\, 151.5,\, 199.5, \]
02

Calculate Median

Now, we need to find the middle value. Since there are 20 data points, the median will be the average of the 10th and 11th value. Median = \(\frac{34.5 + 36.0}{2}\) = \(35.25\)
03

Find Quartiles

We need to find Q1 and Q3, by taking the middle values of the lower half and upper half of the sorted data, respectively. Q1 is the middle value of the lower half (first 10 values): \(Q1 = 8.0\) Q3 is the middle value of the upper half (last 10 values): \(Q3 = 78.0\)
04

Calculate Interquartile Range

Now, we will find the IQR by subtracting Q1 from Q3: IQR = \(Q3 - Q1 = 78.0 - 8.0 = 70.0\)
05

Interpret the Results

The median of the data is \(35.25\) minutes, which means that half of the males have weekday exercise times less than or equal to \(35.25\) minutes, and the other half have weekday exercise times greater than or equal to \(35.25\) minutes. This value represents the central tendency of the exercise time. The interquartile range (IQR) is \(70.0\) minutes, which gives us an idea of the spread of the data. With an IQR of \(70.0\) minutes, we can say that the middle 50% of the males have weekday exercise times ranging from \(8.0\) to \(78.0\) minutes. This value provides an indication of how varied the exercise times are among the university students.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Median Calculation
Calculating the median involves finding the middle value of a data set. The median provides a measure of central tendency, offering a sense of the typical value in the data. First, arrange all the values from smallest to largest. In our data set of 20 values, we order them:
  • 0.0, 0.0, 0.0, 0.0, 7.5, 8.0, 8.0, 9.0, 28.5, 34.5,
  • 36.0, 43.5, 57.0, 57.0, 78.0, 91.5, 142.5, 151.5, 199.5
With 20 data points, calculate the median by averaging the 10th and 11th numbers:
\[\text{Median} = \frac{34.5 + 36.0}{2} = 35.25\]Such a median of 35.25 minutes implies that half the students exercised less than or equal to this time, while the other half exercised more.
Interquartile Range
The interquartile range (IQR) measures the spread of the middle 50% of data points, acting as an indicator of variability. To find the IQR, determine the first (Q1) and third (Q3) quartiles. These are essentially the medians of the first and second halves of your data set:
  • Q1: The median of the first 10 sorted numbers: 0.0 to 34.5; Q1 = 8.0.
  • Q3: The median of the last 10 sorted numbers: 36.0 to 199.5; Q3 = 78.0.
Calculate the IQR as follows:
\[\text{IQR} = Q3 - Q1 = 78.0 - 8.0 = 70.0\]This IQR of 70.0 minutes indicates the range within which the middle 50% of exercise times spread out, from 8.0 to 78.0 minutes.
Data Interpretation
Interpreting data involves understanding what the descriptive statistics tell us about the underlying patterns and distributions.
  • Median of 35.25 minutes: This suggests that on average, a typical student spends about 35 minutes exercising during weekdays. The median is less affected by extremely high or low values and therefore gives a central value that's more representative of the group's behavior as a whole.
  • IQR of 70 minutes: A large IQR shows significant variation among students' exercise times. It reveals that while some students might just exercise for a few minutes, others engage in prolonged physical activities.
With these statistics, educators and health professionals can cultivate better health initiatives, understanding the different levels of engagement and promoting strategies that motivate lesser-active students while supporting more active ones.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The article "The Wedding Industry's Pricey Little Secret \(^{n}\) (June \(12,2013,\) www.slate.com, retrieved April \(19,\) 2017 ) stated that the widely reported average wedding cost is grossly misleading. The article reports that in \(2012,\) the average wedding cost was $$\$ 27,427$$ and the median cost was $$\$ 18,086$$ a. What does the large difference between the mean cost and the median cost tell you about the distribution of wedding costs in \(2012 ?\) b. Do you agree that the average wedding cost is misleading? Explain why or why not. c. The article also states "the proportion of couples who spent the 'average' or more was actually a minority." Do you agree with this statement? Explain why or why not using the reported values of the mean and median wedding cost.

The accompanying data are a subset of data read from a graph in the paper "Ladies First? A Field Study of Discrimination in Coffee Shops" (Applied Economics [April, 2008] ). The data are the waiting time (in seconds) between ordering and receiving coffee for 19 male customers at a Boston coffee shop. $$ \begin{array}{rrrrrrrrrr} 40 & 60 & 70 & 80 & 85 & 90 & 100 & 100 & 110 & 120 \\ 125 & 125 & 140 & 140 & 160 & 160 & 170 & 180 & 200 & \end{array} $$ Use these data to construct a boxplot. Write a few sentences describing the important characteristics of the boxplot.

Although bats are not known for their eyesight, they are able to locate prey (mainly insects) by emitting high-pitched sounds and listening for echoes. A paper appearing in Animal Behaviour ("The Echolocation of Flying Insects by Bats" [1960]: 141-154) gave the following distances (in centimeters) at which a bat first detected a nearby insect: $$ \begin{array}{lllllllllll} 62 & 23 & 27 & 56 & 52 & 34 & 42 & 40 & 68 & 45 & 83 \end{array} $$ a. Calculate and interpret the mean distance at which the bat first detects an insect. b. Calculate the sample variance and standard deviation for this data set. Interpret these values.

Suppose that your younger sister is applying to college and has taken the SAT exam. She scored at the 83 rd percentile on the verbal section of the test and at the 94 th percentile on the math section. Because you have been studying statistics, she asks you for an interpretation of these values. What would you tell her?

In a study investigating the effect of car speed on accident severity, the vehicle speed at impact was recorded for 5000 fatal accidents. For these accidents, the mean speed was 42 mph and the standard deviation was 15 mph. A histogram revealed that the vehicle speed distribution was mound shaped and approximately symmetric. a. Approximately what percentage of the vehicle speeds were between 27 and \(57 \mathrm{mph} ?\) b. Approximately what percentage of the vehicle speeds exceeded \(57 \mathrm{mph} ?\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.