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anonymous

  • one year ago

WILL FAN AND MEDAL! The table below shows data from a survey about the amount of time students spend doing homework each week. The students were either in college or in high school:

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    Which of the choices below best describes how to measure the spread of this data? (Hint: Use the minimum and maximum values to check for outliers.) a. Both spreads are best described with the IQR. b. Both spreads are best described with the standard deviation. c. The college spread is best described by the IQR. The high school spread is best described by the standard deviation. d. The college spread is best described by the standard deviation. The high school spread is best described by the IQR.

  3. anonymous
    • one year ago
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    I think it's A? I know it's not B.

  4. anonymous
    • one year ago
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    I think it may be D, actually. Cause I saw a couple of people say that they got A and C wrong. I put B before and got it wrong, so it must be D? I don't know.

  5. anonymous
    • one year ago
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    it says use the min and max data to check for outliers

  6. anonymous
    • one year ago
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    I've already graphed it. I'll attach the box plots.

  7. anonymous
    • one year ago
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    Don't they both have an outlier?

  8. anonymous
    • one year ago
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    if the mean > median generally an outlier exists

  9. anonymous
    • one year ago
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    The median is more accurate than the mean

  10. anonymous
    • one year ago
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    I'm just having a hard time understanding when to use standard deviation and when to use IQR. I know my textbook says that if the distribution of data is symmetrical, then the best measurement to use is standard deviation. And if the data is asymmetrical, the best measurement is IQR. I don't know if I'm making any sense here, so I'm sorry. I really suck at math.

  11. anonymous
    • one year ago
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    beuase it takes things in order, so the mean is shifgted upwards or downwards depending on the data

  12. anonymous
    • one year ago
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    standard deviation is a relative measure (i.e. relative to the mean)

  13. anonymous
    • one year ago
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    I know, and IQR is relative to the median.

  14. anonymous
    • one year ago
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    Sorry it took me so long to type, I was looking at my textbook.

  15. anonymous
    • one year ago
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    IQR , think about it, "quartile" means "quarter" divide into 4 quarters

  16. anonymous
    • one year ago
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    the lower quartile is the value one way into the data the upper quartile is the value 3 quarters into the data the upper quartile minus the lower quartile is the IQR

  17. anonymous
    • one year ago
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    I keep thinking it's A because the data isn't evenly distributed, and it has outliers. But I remember somebody saying they got that wrong, so I'm just completely stuck.

  18. anonymous
    • one year ago
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    think about the mean.,. now what is going to upset the mean, only outliers

  19. anonymous
    • one year ago
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    outliers dont upset the median as much

  20. anonymous
    • one year ago
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    Sorry, I didn't read what you typed above, it didn't show because my computer froze for a couple minutes. I'm gonna read it now

  21. anonymous
    • one year ago
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    this is becuase the median will shift as the data extends

  22. anonymous
    • one year ago
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    So the IQR for the high school would be 5 and the IQR for the college would be 7.5

  23. anonymous
    • one year ago
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    Okay, so the outliers upset the mean, not the median. So it would be IQR, not standard deviation, right?

  24. anonymous
    • one year ago
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    take simple examples: use 1, 2, 3, 4 5 what is the mean?

  25. anonymous
    • one year ago
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    3

  26. anonymous
    • one year ago
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    correct, now what is the median?

  27. anonymous
    • one year ago
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    3

  28. anonymous
    • one year ago
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    correct also, now chose 1, 2, 3, 4 & 13 what is the mean and what is the median

  29. anonymous
    • one year ago
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    mean: 4.6 median: 3

  30. anonymous
    • one year ago
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    see what happens to the mean when you have an outlier?

  31. anonymous
    • one year ago
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    Yeah, like you said before, it messes up the mean.

  32. anonymous
    • one year ago
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    now you are gettin it, cool!

  33. anonymous
    • one year ago
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    Well thanks for helping so much :p

  34. anonymous
    • one year ago
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    now you can look at your table and guess your answer

  35. anonymous
    • one year ago
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    5-50 average should be about 25

  36. anonymous
    • one year ago
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    however, the mean is 13.8 which means there must be an outlier somewhere

  37. anonymous
    • one year ago
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    So it would be A? Because there's also an outlier in the high school plot.

  38. anonymous
    • one year ago
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    Cause outliers would cause the shape of the data to be asymmetric, and IQR is the best measure of spread for asymmetrical data.

  39. anonymous
    • one year ago
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    @BPDlkeme234

  40. anonymous
    • one year ago
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    which shows how to best measure the spread of the data, well you cant use measure of central tendency (mean, median etc) to show spread, you have to use measures of spread (IQR or standard deviation etc)

  41. anonymous
    • one year ago
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    Once again you need to be able to "see" the data, i.e relative to the mean, to be able to judge which one is best for judging this, this is the whole point of the exercise

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