{"id":34,"date":"2021-01-12T22:19:25","date_gmt":"2021-01-12T22:19:25","guid":{"rendered":"https:\/\/textbooks.jaykesler.net\/introstats\/chapter\/location-of-the-data\/"},"modified":"2024-02-06T16:27:20","modified_gmt":"2024-02-06T16:27:20","slug":"location-of-the-data","status":"publish","type":"chapter","link":"https:\/\/textbooks.jaykesler.net\/introstats\/chapter\/location-of-the-data\/","title":{"rendered":"Location of the Data"},"content":{"raw":"

The common measures of location are\u00a0quartiles<\/span>\u00a0and\u00a0percentiles<\/span><\/p>\r\n

Quartiles are special percentiles. The first quartile,\u00a0Q<\/em>1<\/sub>, is the same as the 25th<\/sup>\u00a0percentile, and the third quartile,\u00a0Q<\/em>3<\/sub>, is the same as the 75th<\/sup>\u00a0percentile. The median,\u00a0M<\/em>, is called both the second quartile and the 50th<\/sup>\u00a0percentile.<\/p>\r\n

To calculate quartiles and percentiles, the data must be ordered from smallest to largest. Quartiles divide ordered data into quarters. Percentiles divide ordered data into hundredths. To score in the 90th<\/sup>\u00a0percentile of an exam does not mean, necessarily, that you received 90% on a test. It means that 90% of test scores are the same or less than your score and 10% of the test scores are the same or greater than your test score.<\/p>\r\n

Percentiles are useful for comparing values. For this reason, universities and colleges use percentiles extensively. One instance in which colleges and universities use percentiles is when SAT results are used to determine a minimum testing score that will be used as an acceptance factor. For example, suppose Duke accepts SAT scores at or above the 75th<\/sup>\u00a0percentile. That translates into a score of at least 1220.<\/p>\r\n

Percentiles are mostly used with very large populations. Therefore, if you were to say that 90% of the test scores are less (and not the same or less) than your score, it would be acceptable because removing one particular data value is not significant.<\/p>\r\n

The\u00a0median<\/span>\u00a0is a number that measures the \"center\" of the data. You can think of the median as the \"middle value,\" but it does not actually have to be one of the observed values. It is a number that separates ordered data into halves. Half the values are the same number or smaller than the median, and half the values are the same number or larger. For example, consider the following data.\r\n<\/span>1; 11.5; 6; 7.2; 4; 8; 9; 10; 6.8; 8.3; 2; 2; 10; 1\r\n<\/span>Ordered from smallest to largest:\r\n<\/span>1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5<\/p>\r\n

Since there are 14 observations, the median is between the seventh value, 6.8, and the eighth value, 7.2. To find the median, add the two values together and divide by two.<\/p>\r\n$$\\frac{6.8+7.2}{2}=7$$\r\n

The median is seven. Half of the values are smaller than seven and half of the values are larger than seven.<\/p>\r\n

Quartiles<\/span>\u00a0are numbers that separate the data into quarters. Quartiles may or may not be part of the data. To find the quartiles, first find the median or second quartile. The first quartile,\u00a0Q<\/em>1<\/sub>, is the middle value of the lower half of the data, and the third quartile,\u00a0Q<\/em>3<\/sub>, is the middle value, or median, of the upper half of the data. To get the idea, consider the same data set:\r\n<\/span>1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5<\/p>\r\n

The median or\u00a0second quartile<\/strong>\u00a0is seven. The lower half of the data are 1, 1, 2, 2, 4, 6, 6.8. The middle value of the lower half is two.\r\n<\/span>1; 1; 2; 2; 4; 6; 6.8<\/p>\r\n

The number two, which is part of the data, is the\u00a0first quartile<\/span>. One-fourth of the entire sets of values are the same as or less than two and three-fourths of the values are more than two.<\/p>\r\n

The upper half of the data is 7.2, 8, 8.3, 9, 10, 10, 11.5. The middle value of the upper half is nine.<\/p>\r\n

The\u00a0third quartile<\/span>,\u00a0Q<\/em>3, is nine. Three-fourths (75%) of the ordered data set are less than nine. One-fourth (25%) of the ordered data set are greater than nine. The third quartile is part of the data set in this example.<\/p>\r\n

The\u00a0interquartile range<\/span>\u00a0is a number that indicates the spread of the middle half or the middle 50% of the data. It is the difference between the third quartile (Q<\/em>3<\/sub>) and the first quartile (Q<\/em>1<\/sub>).<\/p>\r\n

IQR<\/em>\u00a0=\u00a0Q<\/em>3<\/sub>\u00a0\u2013\u00a0Q<\/em>1<\/sub><\/p>\r\n

The\u00a0IQR<\/em>\u00a0can help to determine potential\u00a0outliers<\/strong>.\u00a0A value is suspected to be a potential outlier if it is less than (1.5)(IQR<\/em>) below the first quartile or more than (1.5)(IQR<\/em>) above the third quartile<\/strong>. Potential outliers always require further investigation. In the previous example, $Q_1=2$ and $Q_3=9$ so the $IQR=9-2=7$. So a value would be a potential outlier if it was (1.5)(7) = 10.5 below 2 (which would be -8.5) or 10.5 above 9 (which would be 19.5). If we saw the value 20 in our data, that is probably an outlier.<\/p>\r\n\r\n

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NOTE<\/span><\/h3>\r\n<\/header>
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A potential outlier is a data point that is significantly different from the other data points. These special data points may be errors or some kind of abnormality or they may be a key to understanding the data.<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n

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EXAMPLE\u00a0<\/span>2.13<\/span><\/h3>\r\n<\/header>
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For the following 13 real estate prices, calculate the\u00a0IQR<\/em>\u00a0and determine if any prices are potential outliers. Prices are in dollars.\r\n<\/span>389,950; 230,500; 158,000; 479,000; 639,000; 114,950; 5,500,000; 387,000; 659,000; 529,000; 575,000; 488,800; 1,095,000<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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Solution\u00a0<\/span>2.13<\/span><\/h4>\r\n
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Order the data from smallest to largest.\r\n<\/span>114,950; 158,000; 230,500; 387,000; 389,950; 479,000; 488,800; 529,000; 575,000; 639,000; 659,000; 1,095,000; 5,500,000<\/p>\r\n

M<\/em>\u00a0= 488,800<\/p>\r\n$Q_1=\\frac{230,500+387,000}{2}=308,750$\r\n\r\n$Q_3 = \\frac{639,000+659,000}{2}=649,000$\r\n\r\n$IQR = 649,000-308,750 = 340,250$\r\n\r\n$(1.5)(IQR)=(1.5)(340,250) = 510,375$\r\n\r\n$Q_1-(1.5)(IQR) = 308,750 \u2013 510,375 = \u2013201,625$\r\n\r\n$Q3 + (1.5)(IQR) = 649,000 + 510,375 = 1,159,375$\r\n

No house price is less than \u2013201,625. However, 5,500,000 is more than 1,159,375. Therefore, 5,500,000 is a potential\u00a0outlier<\/span>.<\/p>\r\n\r\n<\/div>\r\n

<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n

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TRY IT\u00a0<\/span>2.13<\/span><\/h3>\r\n<\/header>
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For the following 11 salaries, calculate the\u00a0IQR<\/em>\u00a0and determine if any salaries are outliers. The salaries are in dollars.<\/p>\r\n\\$33,000; \\$64,500; \\$28,000; \\$54,000; \\$72,000; \\$68,500; \\$69,000; \\$42,000; \\$54,000; \\$120,000; \\$40,500\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n

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EXAMPLE\u00a0<\/span>2.14<\/span><\/h3>\r\n<\/header>
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For the two data sets in the\u00a0test scores example<\/a>, find the following:<\/p>\r\n\r\n

    \r\n \t
  1. The interquartile range. Compare the two interquartile ranges.<\/li>\r\n \t
  2. Any outliers in either set.<\/li>\r\n<\/ol>\r\n
    Solution\u00a0<\/span>2.14<\/span><\/div>\r\n
    \r\n

    The five number summary for the day and night classes is<\/p>\r\n\r\n

    \r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nMinimum<\/th>\r\nQ<\/em>1<\/sub><\/th>\r\nMedian<\/th>\r\nQ<\/em>3<\/sub><\/th>\r\nMaximum<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    Day<\/strong><\/td>\r\n32<\/td>\r\n56<\/td>\r\n74.5<\/td>\r\n82.5<\/td>\r\n99<\/td>\r\n<\/tr>\r\n
    Night<\/strong><\/td>\r\n25.5<\/td>\r\n78<\/td>\r\n81<\/td>\r\n89<\/td>\r\n98<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
    Table<\/span>2.21<\/span><\/div>\r\n<\/div>\r\n

    Solution 2.14<\/h4>\r\n
      \r\n \t
    1. The IQR for the day group is\u00a0Q<\/em>3<\/sub>\u00a0\u2013\u00a0Q<\/em>1<\/sub>\u00a0= 82.5 \u2013 56 = 26.5\r\n

      The IQR for the night group is\u00a0Q<\/em>3<\/sub>\u00a0\u2013\u00a0Q<\/em>1<\/sub>\u00a0= 89 \u2013 78 = 11<\/p>\r\n

      The interquartile range (the spread or variability) for the day class is larger than the night class\u00a0IQR<\/em>. This suggests more variation will be found in the day class\u2019s class test scores.<\/p>\r\n<\/li>\r\n \t

    2. Day class outliers are found using the IQR times 1.5 rule. So,\r\n