In Part I, you will analyze the given data set. Using Excel, you will find measu
ID: 3301259 • Letter: I
Question
In Part I, you will analyze the given data set. Using Excel, you will find measures of center, measures, variation, and other descriptive statistics to describe the data.
You will represent the data visually with a modified boxplot and a histogram, and you will need to explain what they tell you.
Using Microsoft Excel, complete the following:
Determine the measures of center for the data (mean, median, and mode) and explain which one describes the center of this data best.
Determine measures of variation (range, standard deviation, and variance) and explain which one describes the spread of this data best.
Determine the quartiles and the IQR (Inter Quartile Range).
Explain the method of determining outliers in the data and list any outliers.
Create the modified boxplot of the data showing any outliers. (You may need to draw this using MS Word tools.) Make sure that your boxplot has a proper scale.
Using Excel, group the data into 6 to 10 groups and create a histogram for the data. (Excel uses bins to group the data. See the attached worksheet for instructions on how to create histograms using Excel.)
Present your results in a Word document. In this document, please include the following:Any graphs, charts, or tables from Excel (These should be cut and pasted into this document.)
Any graphs, charts, or tables from Excel (These should be cut and pasted into this document.)
Your explanations about your findings.
Raw data for the project (Include the data for YOUR VARIABLE only. This should be cut and pasted at the end of this document.) DO NOT SUBMIT ANY EXCEL FILES.
FEMALE
295
2739
2992
3745
4486
4488
4878
4880
4881
4835
4842
6225
8680
8681
12348
14651
16767
17765
19377
19378
19382
20278
21626
32233
33104
33106
33334
33335
34779
35035
35272
35273
35505
35506
35507
35984
35988
36115
36502
38089
Female Arm
23.6
26.3
26.3
32.6
29.2
26.4
27.9
33.0
38.6
26.5
34.4
23.7
28.4
34.0
35.2
24.7
27.0
35.0
33.1
39.6
27.0
43.8
23.6
34.3
34.4
23.3
35.6
31.8
27.0
32.8
31.0
27.0
41.2
25.5
30.9
27.9
26.5
27.8
23.0
26.4
FEMALE
295
2739
2992
3745
4486
4488
4878
4880
4881
4835
4842
6225
8680
8681
12348
14651
16767
17765
19377
19378
19382
20278
21626
32233
33104
33106
33334
33335
34779
35035
35272
35273
35505
35506
35507
35984
35988
36115
36502
38089
Explanation / Answer
Data entered in Excel :
FEMALE
Female Arm
295
23.6
2739
26.3
2992
26.3
3745
32.6
4486
29.2
4488
26.4
4878
27.9
4880
33
4881
38.6
4835
26.5
4842
34.4
6225
23.7
8680
28.4
8681
34
12348
35.2
14651
24.7
16767
27
17765
35
19377
33.1
19378
39.6
19382
27
20278
43.8
21626
23.6
32233
34.3
33104
34.4
33106
23.3
33334
35.6
33335
31.8
34779
27
35035
32.8
35272
31
35273
27
35505
41.2
35506
25.5
35507
30.9
35984
27.9
35988
26.5
36115
27.8
36502
23
38089
26.4
To find the descriptive statistics:
Data tab -> Data Analysis Tool pack -> Descriptive Statistics -> ok -> in the Input Range, select all the cells in which the data is present -> Grouped by : Columns -> tick the labels -> tick the summary statistics -> ok.
FEMALE
Female Arm
Mean
20572.15
Mean
30.1575
Standard Error
2156.892002
Standard Error
0.828335429
Median
19380
Median
28.15
Mode
#N/A
Mode
27
Standard Deviation
13641.38278
Standard Deviation
5.238853246
Sample Variance
186087324.3
Sample Variance
27.44558333
Kurtosis
-1.733481272
Kurtosis
-0.095713163
Skewness
-0.070811587
Skewness
0.718319879
Range
37794
Range
20.8
Minimum
295
Minimum
23
Maximum
38089
Maximum
43.8
Sum
822886
Sum
1206.3
Count
40
Count
40
Confidence Level(95.0%)
4362.725867
Confidence Level(95.0%)
1.675466551
Measures of central tendency :
For females column, Mean = 20572.15, Median=19380.
For female arm column, Mean=30.1575, Median=28.15, Mode=27.
For both the columns, mean describes the central tendency in the best way.
Measures of variation :
For females column, Standard Deviation=13641.38278, Sample Variance=186087324.3, Range=37794.
For female arm column, Standard Deviation=5.238853246, Sample Variance=27.44558333, Range=20.8.
For both the columns, standard deviation describes the variation in the best way.
Quartiles :
Type “=Quartile(<select cells containing data>, 1 (1st quartile) or 3 (3rd quartile))”
First Quartile : For female columns, Q1=4880.75, for female arm column, Q1 = 26.4.
Third Quartile : For female columns, Q3= 35094.25, for female arm column, Q3 = 34.075.
IQR : Female = 30213.5, Female arm = 7.675.
For outlier detection, calculate Q1 – 1.5*IQR and Q3 + 1.5*IQR. Any value not lying within this range is an outlier.
For female column, the boundaries are -40439.5,80414.5 and for female arm column, the boundaries are 14.8875, 45.5875.
There is no outlier in both the columns.
FEMALE
Female Arm
295
23.6
2739
26.3
2992
26.3
3745
32.6
4486
29.2
4488
26.4
4878
27.9
4880
33
4881
38.6
4835
26.5
4842
34.4
6225
23.7
8680
28.4
8681
34
12348
35.2
14651
24.7
16767
27
17765
35
19377
33.1
19378
39.6
19382
27
20278
43.8
21626
23.6
32233
34.3
33104
34.4
33106
23.3
33334
35.6
33335
31.8
34779
27
35035
32.8
35272
31
35273
27
35505
41.2
35506
25.5
35507
30.9
35984
27.9
35988
26.5
36115
27.8
36502
23
38089
26.4
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.