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Watch out for these little buggers. Each of these exercises involves some feature that is somewhat tricky. Find the (a) mean, \((b)\) median, (c) mode, (d) midrange, and then answer the given question. In a study of speed dating conducted at Columbia University, female subjects were asked to rate the attractiveness of their male dates, and a sample of the results is listed below \((1=\) not attractive \(; 10=\) extremely attractive). Can the results be used to describe the attractiveness of the population of adult males? $$ \begin{array}{llllllllllllllllllllllllll} 5 & 8 & 3 & 8 & 6 & 10 & 3 & 7 & 9 & 8 & 5 & 5 & 6 & 8 & 8 & 7 & 3 & 5 & 5 & 6 & 8 & 7 & 8 & 8 & 8 & 7 \end{array} $$

Short Answer

Expert verified
The mean is 6.692, median is 7, mode is 8, and midrange is 6.5. The results should not be used to describe the entire population of adult males.

Step by step solution

01

Find the Mean

To find the mean, add all the values together and then divide by the number of values. Sum = 5 + 8 + 3 + 8 + 6 + 10 + 3 + 7 + 9 + 8 + 5 + 5 + 6 + 8 + 8 + 7 + 3 + 5 + 5 + 6 + 8 + 7 + 8 + 8 + 8 + 7 = 174 Number of values = 26 Mean = Total sum / Number of values = 174 / 26 = 6.692 (rounded to 3 decimal places)
02

Find the Median

To find the median, first order the values from smallest to largest, and then find the middle value. If there is an even number of observations, the median will be the average of the two middle numbers. Ordered values: 3, 3, 3, 5, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 9, 10 Since there are 26 observations (even number), the median is the average of the 13th and 14th values. The 13th and 14th values are 7 and 7.Median = (7 + 7) / 2 = 7
03

Find the Mode

The mode is the value that appears most frequently in the data set. Count of each value: 3 occurs 3 times 5 occurs 5 times 6 occurs 3 times 7 occurs 4 times 8 occurs 8 times 9 occurs 1 time 10 occurs 1 time Mode = 8 (since it occurs 8 times)
04

Find the Midrange

The midrange is the average of the highest and lowest values in the data set. Lowest Value = 3 Highest Value = 10 Midrange = (Lowest Value + Highest Value) / 2 = (3 + 10) / 2 = 6.5
05

Describe the Attractiveness of the Population of Adult Males

The sample data can provide some insight into the perceived attractiveness of the male dates in the speed dating study, but it should not be used to generalize to the entire population of adult males. This is because the sample consists of a specific group of individuals and may not be representative of the broader population.

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Key Concepts

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

Understanding the Mean
The mean, or average, is a central concept in descriptive statistics. It is calculated by adding all the values in a dataset and then dividing by the number of values. For example, in the speed dating study, the mean attractiveness rating was found by summing all the ratings (which totaled 174) and then dividing by the number of ratings (26). This gives us a mean of approximately 6.692. The mean provides a general sense of the central tendency of the data, showing us the average attractiveness rating.
Understanding the Median
The median is another measure of central tendency and represents the middle value in an ordered dataset. To find the median, we first arrange the ratings from smallest to largest and then identify the middle value. In the provided example, with 26 ratings (an even number), the median is the average of the 13th and 14th values in the ordered list. Both these values are 7, so the median attractiveness rating is 7. The median is particularly useful as it is not affected by outliers or extreme values.
Understanding the Mode
The mode is the value that appears most frequently in a dataset. It gives us the most common rating in the respondent's data. In this speed dating dataset, the number '8' appears eight times, more frequently than any other number, making it the mode. This highlights that '8' was the most common attractiveness rating given by the female subjects to the male dates. The mode offers valuable insights, especially in datasets with repeated values.
Understanding the Midrange
The midrange is calculated by finding the average of the highest and lowest values in a dataset. This measure gives a sense of the range of the data and helps identify the central spread. In the given study, the lowest attractiveness rating was 3, and the highest was 10. The midrange is thus (3 + 10) / 2, equaling 6.5. While useful, the midrange can be heavily influenced by extreme values, so it’s often used alongside other measures.
Sample Data Analysis
Sample data analysis involves examining a subset, or sample, of a larger population to draw conclusions about the population as a whole. However, it's important to consider sample representativeness. In this speed dating study, the sample ratings provide insight into the perceived attractiveness of the male dates within the specific context. Nevertheless, these results should not be generalized to the entire population of adult males without caution, as the sample may not be representative. Various factors, such as the sample size and selection methods, influence the validity of such generalizations. Therefore, while sample analysis is valuable, it’s crucial to acknowledge its limitations.

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Most popular questions from this chapter

The 20 brain volumes \(\left(\mathrm{cm}^{3}\right)\) from Data Set 8 "IQ and Brain Size" in Appendix B have a mean of \(1126.0 \mathrm{~cm}^{3}\) and a standard deviation of \(124.9 \mathrm{~cm}^{3}\). Use the range rule of thumb to identify the limits separating values that are significantly low or significantly high. For such data, would a brain volume of \(1440 \mathrm{~cm}^{3}\) be significantly high?

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Consider a value to be significantly low if its score is less than or equal to \(-2\) or consider the value to be significantly high if its \(z\) score is greater than or equal to \(2 .\) Data Set 29 "Coin Weights" lists weights (grams) of quarters manufactured after 1964 . Those weights have a mean of \(5.63930 \mathrm{~g}\) and a standard deviation of \(0.06194 \mathrm{~g}\). Identify the weights that are significantly low or significantly high.

Refer to the frequency distribution in the given exercise and find the standard deviation by using the formula below, where \(x\) represents the class midpoint, \(f\) represents the class frequency, and \(n\) represents the total number of sample values. Also, compare the computed standard deviations to these standard deviations obtained by using Formula 3-4 with the original list of data values: (Exercise 37) 11.5 years; (Exercise 38) 8.9 years; (Exercise 39) 59.5; (Exercise 40) 65.4. $$s=\sqrt{\frac{n\left[\Sigma\left(f \cdot x^{2}\right)\right]-[\Sigma(f \cdot x)]^{2}}{n(n-1)}}$$ $$ \begin{array}{c|c} \hline \begin{array}{c} \text { Blood Platelet } \\ \text { Count of } \\ \text { Females } \end{array} & \text { Frequency } \\ \hline 100-199 & 25 \\ \hline 200-299 & 92 \\ \hline 300-399 & 28 \\ \hline 400-499 & 0 \\ \hline 500-599 & 2 \\ \hline \end{array} $$

Use the given data to construct a boxplot and identify the 5-number summary. Listed below are amounts of strontium-90 (in millibecquerels, or \(\mathrm{mBq}\) ) in a simple random sample of baby teeth obtained from Pennsylvania residents born after 1979 (based on data from "An Unexpected Rise in Strontium-90 in U.S. Deciduous Teeth in the \(1990 \mathrm{~s}\) "' by Mangano et. al., Science of the Total Environment). \(\begin{array}{lllllllllllllll}128 & 130 & 133 & 137 & 138 & 142 & 142 & 144 & 147 & 149 & 151 & 151 & 151 & 155\end{array}\) \(\begin{array}{llllll}156 & 161 & 163 & 163 & 166 & 172\end{array}\)

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