MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1

Tuesday, October 8, 2019
Abstract curve, painted by Wassily Kandinsky

Tasks
Data:
1. Using the list of 17 numbers at the top of the page, the median of this data, rounded to two decimal places, is: 2.8.

Steps to find the answer:
Raw Data: 7.2, 1.2, 1.8, 2.8, 18, -1.9, -0.1, -1.5, 13.0, 3.2, -1.1, 7.0, 0.5, 3.9, 2.1, 4.1, 6.5
Sort data to be:
-1.9, -1.5, -1.1, -0.1, 0.5, 1.2, 1.8, 2.1, 2.8, 3.2, 3.9, 4.1, 6.5, 7.0, 7.2, 13.0, and 18.0
Thus, median is: 2.8

Verify by R:
> x <-c(7.2, 1.2, 1.8, 2.8, 18, -1.9, -0.1, -1.5, 13.0, 3.2, -1.1, 7.0, 0.5, 3.9, 2.1, 4.1, 6.5)
> median (x)
[1] 2.8


2. If you find the median using the original method (paper and pencil), you have to arrange the values into numeric order (True/False): TRUE
Step to find answers:
Please see my answer at question 1 above


Colorful houses, painted by Wassily Kandinsky


3. The interquartile range for this data is (round each value to 3 decimal places): 6.55
Step to find answers:
Lower quartile (Q1) = 1/4(n + 1) = 1/4(17+1) th value = 4.5th value = (-0.1+0.5)/2 = 0.2
Upper quartile(Q3)= 3/4 (n + 1) th value = 3/4(17+1)th value= 13.5th value= (6.5+7.0)/2=6.75

Thus, interquartile range (IQR) = (Upper Quartile) – (Lower Quartile) = 6.75 – 0.2 = 6.55

Verify by R:
> IQR(x)
[1] 6
Hence, the result slightly different with doing by hand


4. The formula for calculating the interquartile range is:
(show the formula and a citation to the source that you used).
Answer:
Interquartile range (IQR) = (Upper Quartile) – (Lower Quartile) or IQR = Q3 – Q1

Reference
Stat Trek. (2017). Statistics and Probability Dictionary. Interquartile Range. Retrieved from http://stattrek.com/statistics/dictionary.aspx?definition=Interquartile%20range


Abstract, painted by Wassily Kandinsky


5. Using techniques that we have studied in this course, the upper and lower cutoff points (rounded to three decimal places) for identifying outliers (in the given data sample) are: 16.575 and -9.625
(this is not a request to show any outliers--just the cutoff points that would determine what constitutes an outlier).  You may round to three decimal places.

Steps to find answer:
Upper cutoff = Q3 + (1.5 * IQR) = 6.75 + (1.5*6.55) = 16.575
Lower cutoff = Q1 – (1.5 * IQR) = 0.2 – (1.5*6.55) = 0.2 - 9.825 = -9.625


6. The summary() command shows a list of outliers, if there are any (True/False): FALSE
Verify by R:
> x <-c(7.2, 1.2, 1.8, 2.8, 18, -1.9, -0.1, -1.5, 13.0, 3.2, -1.1, 7.0, 0.5, 3.9, 2.1, 4.1, 6.5)
> summary(x)
Min. 1st Qu. Median Mean 3rd Qu. Max.
-1.900 0.500 2.800 3.924 6.500 18.000
Thus, no list of outliers


Painted by Wassily Kandinsky


7. The list of outlier values is: 18
(if there are none, write "NA").
Steps to find answer:
First we find upper and lower cutoff as at question 5.
Since we have 16.575 and -9.625 as upper and lower cutoff, thus 18 is the only value outside upper and lower cutoff


8. The standard deviation of the list of 17 numbers is (round to 3 decimal places):
5.2495 = 5.25
Step to find answer:
We find standard deviation by following formula:

SD= 5.2495 = 5.25
Verify by R:
> sd(x)
[1] 5.249468






# To be continued to part 2

20 comments:

  1. Replies
    1. Thank you to visit:
      MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1
      -
      Abstract curve, painted by Wassily Kandinsky

      Delete
  2. I studied this so many years ago!! Nice to remember!!
    xoox

    marisasclosetblog.com

    ReplyDelete
    Replies
    1. Thank you to visit:
      MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1
      -
      Tasks
      Data:
      1. Using the list of 17 numbers at the top of the page, the median of this data, rounded to two decimal places, is: 2.8.

      Steps to find the answer:
      Raw Data: 7.2, 1.2, 1.8, 2.8, 18, -1.9, -0.1, -1.5, 13.0, 3.2, -1.1, 7.0, 0.5, 3.9, 2.1, 4.1, 6.5
      Sort data to be:
      -1.9, -1.5, -1.1, -0.1, 0.5, 1.2, 1.8, 2.1, 2.8, 3.2, 3.9, 4.1, 6.5, 7.0, 7.2, 13.0, and 18.0
      Thus, median is: 2.8

      Verify by R:
      > x <-c(7.2, 1.2, 1.8, 2.8, 18, -1.9, -0.1, -1.5, 13.0, 3.2, -1.1, 7.0, 0.5, 3.9, 2.1, 4.1, 6.5)
      > median (x)
      [1] 2.8

      Delete
  3. A lot of numbers, leave it them that knows.

    ReplyDelete
    Replies
    1. Thank you to visit:
      MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1
      -
      2. If you find the median using the original method (paper and pencil), you have to arrange the values into numeric order (True/False): TRUE
      Step to find answers:
      Please see my answer at question 1 above

      Delete
  4. Replies
    1. Thank you to visit:
      MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1
      -
      3. The interquartile range for this data is (round each value to 3 decimal places): 6.55

      Step to find answers:

      Lower quartile (Q1) = 1/4(n + 1) = 1/4(17+1) th value = 4.5th value = (-0.1+0.5)/2 = 0.2

      Upper quartile(Q3)= 3/4 (n + 1) th value = 3/4(17+1)th value= 13.5th value= (6.5+7.0)/2=6.75



      Thus, interquartile range (IQR) = (Upper Quartile) – (Lower Quartile) = 6.75 – 0.2 = 6.55



      Verify by R:

      > IQR(x)

      [1] 6

      Hence, the result slightly different with doing by hand



      Delete
  5. Replies
    1. Thank you to visit:
      MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1
      -

      Delete
  6. I remember taking statistics for one of my undergraduate courses for psychology!

    ReplyDelete
    Replies
    1. Thank you to visit:
      MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1
      -
      4. The formula for calculating the interquartile range is:

      (show the formula and a citation to the source that you used).

      Answer:

      Interquartile range (IQR) = (Upper Quartile) – (Lower Quartile) or IQR = Q3 – Q1



      Reference

      Stat Trek. (2017). Statistics and Probability Dictionary. Interquartile Range. Retrieved from http://stattrek.com/statistics/dictionary.aspx?definition=Interquartile%20range

      Delete
  7. Good evening,
    I was pleasantly surprised to find my favorite artist Wassily Kandinsky in this post. I always was fascinated by his paintings.
    Wish you a nice week!

    ReplyDelete
    Replies
    1. He is one of top world painters right now.

      Thank you to visit:
      MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1
      -
      5. Using techniques that we have studied in this course, the upper and lower cutoff points (rounded to three decimal places) for identifying outliers (in the given data sample) are: 16.575 and -9.625

      (this is not a request to show any outliers--just the cutoff points that would determine what constitutes an outlier). You may round to three decimal places.



      Steps to find answer:

      Upper cutoff = Q3 + (1.5 * IQR) = 6.75 + (1.5*6.55) = 16.575

      Lower cutoff = Q1 – (1.5 * IQR) = 0.2 – (1.5*6.55) = 0.2 - 9.825 = -9.625

      Delete
  8. I probably don't like statistics but I love to know it. It sounds great. Thanks.

    ReplyDelete
    Replies
    1. Thank you to visit:
      MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1
      -
      6. The summary() command shows a list of outliers, if there are any (True/False): FALSE

      Verify by R:

      > x <-c(7.2, 1.2, 1.8, 2.8, 18, -1.9, -0.1, -1.5, 13.0, 3.2, -1.1, 7.0, 0.5, 3.9, 2.1, 4.1, 6.5)

      > summary(x)

      Min. 1st Qu. Median Mean 3rd Qu. Max.

      -1.900 0.500 2.800 3.924 6.500 18.000

      Thus, no list of outliers

      Delete
  9. Bonjour,

    Je dirai que j'aime tout simplement ces oeuvres !

    Gros bisous 🌸

    ReplyDelete
    Replies
    1. Thank you to visit:
      MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1
      -
      7. The list of outlier values is: 18

      (if there are none, write "NA").

      Steps to find answer:

      First we find upper and lower cutoff as at question 5.

      Since we have 16.575 and -9.625 as upper and lower cutoff, thus 18 is the only value outside upper and lower cutoff

      Delete
  10. Replies
    1. Thank you to visit:
      MATH 1280 Introduction to Statistics – Assignment Unit 4 – Part 1
      -
      8. The standard deviation of the list of 17 numbers is (round to 3 decimal places):

      5.2495 = 5.25

      Step to find answer:

      We find standard deviation by following formula:





      SD= 5.2495 = 5.25

      Verify by R:

      > sd(x)

      [1] 5.249468



      Delete