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Following are four variables for 30 cases from the General Social Survey. Age is reported in years. The variable happiness consists of answers to the question "Taken all together, would you say that you are (1) very happy (2) pretty happy, or (3) not too happy?" Respondents were asked how many sex partners they had over the past five years. Responses were measured on the following scale: \(0-4=\) actual numbers; \(5=5-10\) partners; \(6=11-20\) partners; \(7=21-100\) partners \(; 8=\) more than 100 . a. For each variable, find the appropriate measure of central tendency and write a sentence reporting this statistical information as you would in a research report. b. (This problem is optional.) For the variable age, construct a frequency distribution with interval size equal to 5 and the first interval set at \(20-24\). Compute the mean and median for the grouped data and compare with the values computed for the ungrouped data. How accurate are the estimates based on the grouped data?

Short Answer

Expert verified
Calculate mean, median, and mode for age, and median or mode for happiness and number of sex partners. Construct frequency distribution with interval of 5 for age, then compare grouped and ungrouped data estimates.

Step by step solution

01

- Determine the Measure of Central Tendency for Age

For the variable 'Age', the appropriate measures of central tendency are the mean, median, and mode. Calculate the mean by summing all ages and dividing by the number of cases (30). Find the median by ordering the ages and selecting the middle value. The mode is the age that appears most frequently.
02

- Report Central Tendency for Age

In a research report, you might say: 'The average age of the respondents is X years, with a median age of Y years and a mode of Z years.'
03

- Determine the Measure of Central Tendency for Happiness

Since 'Happiness' is an ordinal variable, the median or mode is preferred. Calculate the mode by finding the most frequent happiness response (1, 2, or 3). Compute the median by ordering the responses and selecting the middle value.
04

- Report Central Tendency for Happiness

In a research report, you might say: 'The most common happiness level among respondents is X, and the median happiness level is Y.'
05

- Determine the Measure of Central Tendency for Number of Sex Partners

For 'Number of Sex Partners', consider using the median since the data is ordinal with bounded categories. Order the responses and find the median value.
06

- Report Central Tendency for Number of Sex Partners

In a research report, you might say: 'The median number of sex partners among respondents is X, indicating that most respondents fall into this category.'
07

- Construct a Frequency Distribution for Age

Divide the ages into intervals of 5 years starting from 20-24, 25-29, ..., until all ages are included. Count the number of ages that fall into each interval.
08

- Calculate Mean and Median for Grouped Data

Use the midpoint of each interval to estimate the mean. Sum the products of each interval's midpoint and frequency, then divide by the total number of cases. To find the median for grouped data, locate the interval containing the median position (15.5th value), then interpolate within that interval using the formula for the median in grouped data.
09

- Compare Grouped and Ungrouped Data Estimates

Compare the mean and median from the grouped data with those from the ungrouped data to evaluate the accuracy. The estimates are usually close, but slight discrepancies may occur due to grouping.

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Key Concepts

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

Mean Calculation
The mean is a widely used measure of central tendency. It represents the average value of a dataset. To calculate the mean, sum all observations and then divide by the number of observations.
For example, if we have a dataset of ages: 22, 25, 30, 35, 40. Add them up: 22 + 25 + 30 + 35 + 40 = 152. There are five ages, so divide the total by 5. The mean age would be \[ \text{Mean} = \frac{152}{5} = 30.4 \]. The mean gives an overall idea of the data distribution but can be affected by outliers (extreme values).
Median Calculation
The median is the middle value in a dataset when the values are arranged in ascending order. It is another measure of central tendency that is less affected by outliers compared to the mean.
To find the median, sort the data and select the middle value. If there is an odd number of observations, the median is the middle one. For an even number of observations, the median is the average of the two middle values.
For example, in the dataset: 22, 25, 30, 35, 40, 42. There are six values, so the median is the average of the third and fourth values: \[\text{Median} = \frac{30+35}{2} = 32.5\]. The median is useful in skewed distributions as it better represents the central location of the data.
Mode Calculation
The mode is the value that appears most frequently in a dataset. It is the only measure of central tendency that can be used with categorical data. A dataset can have one mode, more than one mode (bimodal or multimodal), or no mode at all if no value repeats.
For example, in the dataset: 22, 25, 25, 30, 35, 35, 35, 40. The number 35 is the mode since it appears the most frequently.
The mode is particularly useful in understanding the most common category or value in a dataset, especially in categorical and ordinal data.
Frequency Distribution
A frequency distribution is a summary of how frequently each value occurs in a dataset. It helps in understanding the distribution and patterns within the data.
To create a frequency distribution for the variable 'Age' with the provided intervals (e.g., 20-24, 25-29, etc.), count the number of observations falling into each interval.
  • 20-24: 3
  • 25-29: 6
  • 30-34: 8
  • 35-39: 7
  • 40-44: 4
  • 45-49: 2
Frequency distributions allow us to visualize and interpret data more effectively and can form the basis for constructing histograms and calculating other statistical measures.
Grouped Data Analysis
In grouped data analysis, data is divided into intervals or groups. This is particularly useful when dealing with large datasets.
To estimate the mean for grouped data:
  • Compute the midpoint for each interval.
  • Multiply each midpoint by the frequency of the interval.
  • Sum these products.
  • Divide the sum by the total frequency.
\[ \text{Mean}_{grouped} = \frac{\text{Sum of (midpoint × frequency)}_\text{interval}}{\text{Total frequency}} \].
To find the median in grouped data, locate the median interval and use the interpolation formula. Grouped data analysis helps in identifying patterns and trends in large datasets while simplifying the calculations.

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Most popular questions from this chapter

The following table presents the annual personhours of time lost due to traffic congestion for a group of cities for 2007. This statistic is a measure of traffic congestion $$\begin{array}{lc} {\text { City }} & \text { Annual Person-Hours of Time Lost to Traffic Congestion per Year per Person } \\ \hline \text { Baltimore } & 25 \\ \text { Boston } & 22 \\ \text { Buffalo } & 5 \\ \text { Chicago } & 22 \\ \text { Cleveland } & 7 \\ \text { Dallas } & 32 \\ \text { Detroit } & 29 \\ \text { Houston } & 32 \\ \text { Kansas City } & 8 \\ \text { Los Angeles } & 38 \\ \text { Miami } & 27 \\ \text { Minneapolis } & 22 \\ \text { New Orleans } & 10 \\ \text { New York } & 21 \\ \text { Philadelphia } & 21 \\ \text { Pittsburgh } & 8 \\ \text { Phoenix } & 23 \\ \text { San Antonio } & 21 \\ \text { San Diego } & 29 \\ \text { San Francisco } & 29 \\ \text { Seattle } & 24 \\ \text { Washington, DC } & 31 \end{array}$$ a. Calculate the mean and median of this distribution. b. Compare the mean and median. Which is the higher value? Why? c. If you removed Los Angeles from this distribution and recalculated, what would happen to the mean? To the median? Why? d. Report the mean and median as you would in a formal research report.

A variety of information has been gathered from a sample of college freshmen and seniors, including their region of birth; the extent to which they support legalization of marijuana (measured on a scale on which \(7=\) strong support, \(4=\) neutral, and \(1=\) strong opposition); the amount of money they spend each week out-of-pocket for food, drinks, and entertainment; how many movies they watched in their dorm rooms last week; their opinion of cafeteria food \((10=\) excellent, \(0=\) very bad \()\); and their religious affiliation. Some results are presented here. Find the most appropriate measure of central tendency for each variable for freshmen and then for seniors. Report both the measure you selected as well as its value for each variable (e.g., "Mode \(=3\) " or "Median \(=3.5\) "). (HINT: Determine the level of measurement for each variable first. In general, this will tell you which measure of central tendency is appropriate. See the section "Choosing a Measure of Central Tendency" to review the relationship between measure of central tendency and level of measurement. Also, remember that the mode is the most common score, and especially remember to array scores from high to low before finding the median.) $$\text { Fresh Men}$$ \(\begin{array}{clccccl} \text { Student } & \text { Region of Birth } & \text { Legalization } & \text {Out-ofPocket Expenses } & \text { Movies } & \text { Food } & \text { Religion } \\ \hline \text { A } & \text { North } & 7 & 33 & 0 & 10 & \text { Protestant } \\\ \text { B } & \text { North } & 4 & 39 & 14 & 7 & \text { Protestant } \\ \text { C } & \text { South } & 3 & 45 & 10 & 2 & \text { Catholic } \\ \text { D } & \text { Midwest } & 2 & 47 & 7 & 1 & \text { None } \\ \text { E } & \text { North } & 3 & 62 & 5 & 8 & \text { Protestant } \\ \text { F } & \text { North } & 5 & 48 & 1 & 6 & \text { Jew } \\ \text { G } & \text { South } & 1 & 52 & 0 & 10 & \text { Protestant } \\ \text { H } & \text { South } & 4 & 65 & 14 & 0 & \text { Other } \\ \text { ? } & \text { Midwest } & 1 & 51 & 3 & 5 & \text { Other } \\ \text { J } & \text { West } & 2 & 43 & 4 & 6 & \text { Catholic } \end{array}\) \(\text { SENIORS }\) \(\begin{array}{clccccl} \text { Student } & \text {Region of Birth } & \text { Legalization } & \text {Out-ofPocket Expenses } & \text { Movies } & \text { Cafeteria Food } & \text { Religion } \\ \hline \mathrm{K} & \text { North } & 7 & 65 & 0 & 1 & \text { None } \\ \mathrm{L} & \text { Midwest } & 6 & 62 & 5 & 2 & \text { Protestant } \\ \mathrm{M} & \text { North } & 7 & 60 & 11 & 8 & \text { Protestant } \\ \mathrm{N} & \text { North } & 5 & 90 & 3 & 4 & \text { Catholic } \\ \mathrm{O} & \text { South } & 1 & 62 & 4 & 3 & \text { Protestant } \\ \mathrm{P} & \text { South } & 5 & 57 & 14 & 6 & \text { Protestant } \\ \mathrm{Q} & \text { West } & 6 & 40 & 0 & 2 & \text { Catholic } \\ \mathrm{R} & \text { West } & 7 & 49 & 7 & 9 & \text { None } \\ \mathrm{S} & \text { North } & 3 & 45 & 5 & 4 & \text { None } \\ \mathrm{T} & \text { West } & 5 & 85 & 3 & 7 & \text { Other } \\ \mathrm{U} & \text { North } & 4 & 78 & 5 & 4 & \text { None } \end{array}\)

You have compiled the following information on each of the graduates voted "most likely to succeed" by a local high school for a 10 -year period. For each variable, find the appropriate measure of central tendency. \(\begin{array}{ccccc} \text { Case } & \text {Present Income } & \text { Marital Status } & \text { Owns a Home? } & \text { Years of Education Post-High School } \\ \hline \text { A } & 104,000 & \text { Divorced } & \text { Yes } & 8 \\ \text { B } & 68,000 & \text { Divorced } & \text { No } & 4 \\ \text { C } & 54,000 & \text { Married } & \text { Yes } & 4 \\ \text { D } & 45,000 & \text { Married } & \text { No } & 4 \\ \text { E } & 40,000 & \text { Single } & \text { No } & 4 \\ \text { F } & 85,000 & \text { Separated } & \text { Yes } & 8 \\ \text { G } & 30,000 & \text { Married } & \text { No } & 3 \\ \text { H } & 27,000 & \text { Married } & \text { No } & 1 \\ \text { I } & 93,000 & \text { Married } & \text { Yes } & 6 \\ \text { J } & 48,000 & \text { Single } & \text { Yes } & 4 \end{array}\)

A sample of 25 freshmen at a major university completed a survey that measured their degree of racial prejudice (the higher the score, the greater the prejudice). a. Compute the median and mean scores for these data. \(\begin{array}{lllll} 10 & 43 & 30 & 30 & 45 \\ 40 & 12 & 40 & 42 & 35 \\ 45 & 25 & 10 & 33 & 50 \\ 42 & 32 & 38 & 11 & 47 \\ 22 & 26 & 37 & 38 & 10 \end{array}\) b. These same 25 students completed the same survey during their senior year. Compute the median and mean for this second set of scores, and compare them to the earlier set. What happened? \(\begin{array}{lllll} 10 & 45 & 35 & 27 & 50 \\ 35 & 10 & 50 & 40 & 30 \\ 40 & 10 & 10 & 37 & 10 \\ 40 & 15 & 30 & 20 & 43 \\ 23 & 25 & 30 & 40 & 10 \end{array}\)

The following table lists the median family incomes for 13 Canadian provinces and territories in 2000 and 2006. Compute the mean and median for each year and compare the two measures of central tendency. Which measure of central tendency is greater for each year? Are the distributions skewed? In which direction? $$ \begin{array}{lcc} {\text { Province or Territory }} & 2000 & 2006 \\ \hline \text { Newfoundland and Labrador } & 38,800 & 50,500 \\ \text { Prince Edward Island } & 44,200 & 56,100 \\ \text { Nova Scotia } & 44,500 & 56,400 \\ \text { New Brunswick } & 43,200 & 54,000 \\ \text { Quebec } & 47,700 & 59,000 \\ \text { Ontario } & 55,700 & 66,600 \\ \text { Manitoba } & 47,300 & 58,700 \\ \text { Saskatchewan } & 45,800 & 60,500 \\ \text { Alberta } & 55,200 & 78,400 \\ \text { British Columbia } & 49,100 & 62,600 \\ \text { Yukon } & 56,000 & 76,000 \\ \text { Northwest Territories } & 61,000 & 88,800 \\ \text { Nunavut } & 37,600 & 54,300 \end{array} $$

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