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Chapter 3: Problem 53

Consider the sample data in the following frequency distribution. $$\begin{array}{ccc} \text { Class } & \text { Midpoint } & \text { Frequency } \\ 3-7 & 5 & 4 \\ 8-12 & 10 & 7 \\ 13-17 & 15 & 9 \\ 18-22 & 20 & 5 \end{array}$$ a. Compute the sample mean. b. Compute the sample variance and sample standard deviation.

Short Answer

Expert verified
a. Mean = 13 b. Variance = 25, Standard Deviation = 5

Step by step solution

01

Calculate the Sample Mean

To find the sample mean, we use the formula for the mean of a frequency distribution: \( \bar{x} = \frac{\sum (x_i \cdot f_i)}{\sum f_i} \). Here, \( x_i \) represents the midpoints, and \( f_i \) are the frequencies.- For class 3-7: Midpoint = 5, Frequency = 4, Contribution = \( 5 \times 4 = 20 \).- For class 8-12: Midpoint = 10, Frequency = 7, Contribution = \( 10 \times 7 = 70 \).- For class 13-17: Midpoint = 15, Frequency = 9, Contribution = \( 15 \times 9 = 135 \).- For class 18-22: Midpoint = 20, Frequency = 5, Contribution = \( 20 \times 5 = 100 \).Sum of contributions = 20 + 70 + 135 + 100 = 325.Sum of frequencies = 4 + 7 + 9 + 5 = 25.Thus, the sample mean is \( \bar{x} = \frac{325}{25} = 13.\).
02

Calculate Each Squared Deviation

To calculate the variance, we need the squared deviations, \((x_i - \bar{x})^2 \).- For midpoint 5: \((5 - 13)^2 = 8^2 = 64\).- For midpoint 10: \((10 - 13)^2 = 3^2 = 9\).- For midpoint 15: \((15 - 13)^2 = 2^2 = 4\).- For midpoint 20: \((20 - 13)^2 = 7^2 = 49\).
03

Calculate Weighted Squared Deviations

Multiply each squared deviation from Step 2 by the respective frequency \(f_i\):- For midpoint 5: \(64 \times 4 = 256\).- For midpoint 10: \(9 \times 7 = 63\).- For midpoint 15: \(4 \times 9 = 36\).- For midpoint 20: \(49 \times 5 = 245\).Sum of weighted squared deviations = 256 + 63 + 36 + 245 = 600.
04

Compute Sample Variance

The formula for sample variance is given by \( s^2 = \frac{\sum (x_i - \bar{x})^2 f_i}{n - 1} \), where \(n\) is the total number of data points:\[ s^2 = \frac{600}{25 - 1} = \frac{600}{24} = 25. \]
05

Compute Sample Standard Deviation

The sample standard deviation \(s\) is the square root of the sample variance:\[ s = \sqrt{25} = 5. \]

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Key Concepts

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

Frequency Distribution
Understanding a frequency distribution helps in organizing data effectively. It shows us how frequently each possible value within a dataset occurs. In our example, frequencies are grouped into classes with midpoints.
For instance:
  • Class '3-7' has a midpoint of 5, with a frequency of 4 occurrences.
  • Class '8-12' with a midpoint of 10 appears 7 times.
  • Class '13-17' has a midpoint of 15 and occurs 9 times.
  • Class '18-22', midpoint 20, appears 5 times.
Using frequency distribution simplifies the calculation of the sample mean and variance, as data is organized neatly into distinct classes. This makes it easier to manage larger datasets without losing sight of individual occurrences.
Sample Variance
Sample variance measures the spread or variability of a set of data points in a sample. It tells us how much the data differs from the sample mean. The calculation starts by determining deviations for each class midpoint from the mean, then squaring these deviations.
In our case:
  • The deviation for midpoint 5 is \( (5 - 13)^2 = 64 \).
  • For midpoint 10, \( (10 - 13)^2 = 9 \).
  • For midpoint 15, \( (15 - 13)^2 = 4 \).
  • For midpoint 20, \( (20 - 13)^2 = 49 \).
Weighted squared deviations are then calculated by multiplying each squared deviation by its frequency. The sum of these weighted squared deviations is 600.
The sample variance, \( s^2 \), is calculated as:\[ s^2 = \frac{600}{25 - 1} = 25. \]This means on average, data points are spread 25 units squared around the mean.
Standard Deviation
Standard deviation is a statistic that captures the amount of variation in a dataset relative to the mean. It gives us an idea of how much the individual data points deviate from the mean on average.
It is derived by taking the square root of the sample variance (\( s^2 = 25 \)). Thus, the standard deviation is:\[ s = \sqrt{25} = 5 \]This implies each data point tends to differ from the mean by about 5 units on average.
The standard deviation is particularly useful because it shares the same unit as the data, making interpretation more intuitive when assessing variability.
Statistical Analysis
Statistical analysis involves collecting and scrutinizing every data sample in a set of items from which samples can be drawn. It helps to derive meaningful conclusions and inferences.
In the context of our exercise, it allows us to break down data distribution and provides insights on:
  • The central tendency through the sample mean, indicating the average value.
  • The dispersion via sample variance and standard deviation, assessing how data varies.
Statistical analysis not only helps in understanding the current state of data but also forms the basis for predicting future trends. By calculating these statistics, we learn more about the underlying characteristics of the data, informing decisions backed by numerical evidence.

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Most popular questions from this chapter

The following data were used to construct the histograms of the number of days required to fill orders for Dawson Supply, Inc., and J.C. Clark Distributors (see Figure 3.2) $$\begin{aligned} &\text { Dawson Supply Days for Delivery: } 11 \quad 10 \quad 9 \quad 10 \quad 11 \quad 11 \quad 10 \quad 11 \quad 10 \quad 1\\\ &\text { Clark Distributors Days for Delivery: } 8 \quad 10 \quad 13 \quad 7 \quad 10 \quad 11 \quad 10 \quad 7 \quad 15 \quad 12 \end{aligned}$$ Use the range and standard deviation to support the previous observation that Dawson Supply provides the more consistent and reliable delivery times.

Five observations taken for two variables follow. $$\begin{array}{r|rrrrr} x_{i} & 4 & 6 & 11 & 3 & 16 \\ \hline y_{i} & 50 & 50 & 40 & 60 & 30 \end{array}$$ a. Develop a scatter diagram with \(x\) on the horizontal axis. b. What does the scatter diagram developed in part (a) indicate about the relationship between the two variables? c. Compute and interpret the sample covariance. d. Compute and interpret the sample correlation coefficient.

The U.S. Department of Education reports that about \(50 \%\) of all college students use a student loan to help cover college expenses (National Center for Educational Studies, January 2006 ). A sample of students who graduated with student loan debt is shown here. The data, in thousands of dollars, show typical amounts of debt upon graduation. $$\begin{array}{lllllllll} 10.1 & 14.8 & 5.0 & 10.2 & 12.4 & 12.2 & 2.0 & 11.5 & 17.8 & 4.0 \end{array}$$ a. For those students who use a student loan, what is the mean loan debt upon graduation? b. What is the variance? Standard deviation?

Car rental rates per day for a sample of seven Eastern U.S. cities are as follows (The Wall Street Journal, January 16,2004 ). $$\begin{array}{lc} \text { City } & \text { Daily Rate } \\ \text { Boston } & \$ 43 \\ \text { Atlanta } & 35 \\ \text { Miami } & 34 \\ \text { New York } & 58 \\ \text { Orlando } & 30 \\ \text { Pittsburgh } & 30 \\ \text { Washington, D.C. } & 36 \end{array}$$ a. Compute the mean, variance, and standard deviation for the car rental rates. b. \(A\) similar sample of seven Western U.S. cities showed a sample mean car rental rate of \(\$ 38\) per day. The variance and standard deviation were 12.3 and \(3.5,\) respectively, Discuss any difference between the car rental rates in Eastern and Western U.S. cities.

Consider a sample with data values of \(10,20,12,17,\) and \(16 .\) Compute the range and interquartile range.
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