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 = 4.5^{th}
value = (0.1+0.5)/2 = 0.2
Upper
quartile(Q3)= 3/4 (n + 1) th value = 3/4(17+1)th value= 13.5^{th}
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 outliersjust 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
Love those paintings!☺
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Abstract curve, painted by Wassily Kandinsky
I studied this so many years ago!! Nice to remember!!
ReplyDeletexoox
marisasclosetblog.com
Thank you to visit:
DeleteMATH 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
A lot of numbers, leave it them that knows.
ReplyDeleteThank you to visit:
DeleteMATH 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
nice paintings
ReplyDeleteThank you to visit:
DeleteMATH 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
I studied statistics at the university!
ReplyDeleteDon't Call Me Fashion Blogger
Facebook
Bloglovin'
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I remember taking statistics for one of my undergraduate courses for psychology!
ReplyDeleteThank you to visit:
DeleteMATH 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
Good evening,
ReplyDeleteI 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!
He is one of top world painters right now.
DeleteThank 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 outliersjust 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
I probably don't like statistics but I love to know it. It sounds great. Thanks.
ReplyDeleteThank you to visit:
DeleteMATH 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
Bonjour,
ReplyDeleteJe dirai que j'aime tout simplement ces oeuvres !
Gros bisous 🌸
Thank you to visit:
DeleteMATH 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
Great! Have a nice day
ReplyDeleteThank you to visit:
DeleteMATH 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