Sample Test 1 Solutions - Math 1342 Each question is worth 7

Sample Test 1 Solutions - Math 1342
Each question is worth 7 points. Your results, plots and comments will be typed into Microsoft Word and
then either submitted to the D2L dropbox as a Word File or as a pdf file.
The following data shows the number of grams per serving of 30 different cakes
32
47
51
41
46
30
46
38
34
34
52
48
48
38
43
41
21
24
25
29
33
45
51
32
32
27
23
23
34
35
1. Enter the data into StatCrunch and create a summary statistics table. Remember your data must go
down column 1 (var1) Paste your table into your Word document. You can use the default selection that
StatCrunch uses but also include the IQR as well.
ANS.
Summary statistics:
Column n
var1
Mean
Variance
Std. dev. Median Range Min Max Q1 Q3 IQR
30 36.766667 87.909195 9.3759904
34.5
31
21
52 30 46
16
2. Use StatCrunch to create a Stem and Leaf plot and paste it into your Word document.
ANS.
Decimal point is 1 digit(s) to the right of the colon.
2 : 1334
2 : 579
3 : 02223444
3 : 588
4 : 113
4 : 566788
5 : 112
Or shortened
2 : 1334579
3 : 02223444588
4 : 113566788
5 : 112
1|P age
3. Looking at the Summary Statistics and the Stem and Leaf plot why does the median have a fractional
part and not a whole number when all of the numbers in the data are whole numbers? What are the
two numbers on each side of the median?
ANS.
The data has an even number of data items and therefore the middle 2 numbers were averaged. The two
numbers that were averaged are 34 and 35
4. Use StatCrunch to create a Box Plot which goes side to side (not up and down) and paste it into your
Word document. Is the median equally spaced between Q1 and Q3? If not which one is it closer to?
ANS.
The Box Plot is shown below and the median is shifted toward Q1.
2|P age
5. Use your IQR to determine if the data has medium or extreme outliers. See page 176 question 18 in
the textbook.
ANS.
From StatCrunch
Min Max Q1 Q3 IQR
21
52
30
46
16
1.5 * IQR = 1.5 * 16 = 24
Q1 – (1.5 * IQR) = 30 – 24 = 6
Q3 + (1.5 * IQR) = 46 + 24 = 70
So if there were any outliers they would be outside of the range of 6 to 70.
Our Min = 21 which is not lower than 6 and
Our Max = 52 which is not higher than 70 so our data has NO outliers.
6. Create a Grouped Frequency Table from the data using 5 classes. Make sure the width is an odd
number and complete the following table. Examples -> 4.3 rounds to w = 5, 7.3 rounds to w = 9
Column
n
Mean
Variance Std. Dev. Std. Err.
var1
30 36.766666 87.909195
9.37599 1.7118138
Median Range Min Max Q1 Q3 IQR
34.5
31
21
52
30
46
16
ANS.
Range = 31
Width = 31/5 = 6.2 and round up to the closest odd number which is 7
Class 1 ->>> 21 up to 21 + (7 - 1) = 21 + 6 = 27
Class 2 up by 1 ->>> 28 up to 28 + 6 = 34
Class Limits w = 6
Class Boundaries True w = 7
21 to 27
28 to 34
35 to 41
42 to 48
49 to 55
20.5 to 27.5
27.5 to 34.5
34.5 to 41.5
41.5 to 48.5
48.5 to 55.5
3|P age
2 : 1334579
3 : 02223444588
4 : 113566788
5 : 112
Class
1
2
3
4
5
Class
Limits
Class
Boundaries
Frequency
Midpoints
Freq * Midpoints
Freq *Mid^2
21 to 27
20.5 to 27.5
6
24
144
28 to 34
35 to 41
27.5 to 34.5
34.5 to 41.5
9
5
31
38
279
190
42 to 48
41.5 to 48.5
7
45
315
49 to 55
48.5 to 55.5
3
52
156
Sums
30
3456
8649
7220
14175
8112
41612
1084
7. Use the Grouped Frequency Table to calculate the grouped mean and compare this number with the
mean from your StatCrunch summary of the ungrouped data. Are the results the same or are they
different? Which answer do you think is more accurate and why?
ANS. Grouped Mean
= Sum of Column of Freq * Midpoints / Sample Size
= 1084 / 30
= 36.13
The mean from the original ungrouped data is 36.76 and the grouped mean is 36.13. The numbers are not the
same and the reason that they are not is that the grouped table has lost some of the precision due to the
midpoints being used to represent the data.
8. Use your Grouped Frequency Table to calculate the grouped variance and the grouped standard
deviation.
ANS. pg 139
n * sum(freq * midpoints^2) = 30 * 41612 = 1248360
( sum( freq * midpoints) )^2 = 1084^2 = 1175056
n * (n - 1) = 30 * 29 = 870
s^2 = grouped variance = (1248360 – 1175056) / 870 = 84.257
s = grouped std = sqrt(84.257) = 9.179
4|P age
9. Use StatCrunch and your Grouped Frequency Table to make a Histogram. Paste the graph into your
Word document.
ANS.
5|P age
10. Use the data shown below and StatCrunch to create a Pie Chart and also a Pareto Chart. Paste the
graphs into your Word document.
Favorite Colors
Red
Blue
Green
Yellow
Orange
Frequency
5
10
14
3
12
ANS.
6|P age
11. Use the data: 5, 8, 12, 15, 9, 11, 13, 20, 8, 10, 9, 8, 6, 20, 8
a. Calculate the z Score for 11? Show your work. Hint: Use StatCrunch to help.
b. Calculate the z Score for 8? Show your work. Hint: Use StatCrunch to help.
ANS.
First use StatCrunch with the data to get the Summary
Summary statistics:
Column n Mean Variance Std. Dev. Median Range Min Max Q1 Q3
var1
15
10.8
20.6 4.5387225
9
15
5
20
8 13
a. z score for 11 = (value – mean)/std = (11 – 10.8)/4.538 = 0.0441
b. z score for 8 = ( 8 – 10.8)/4.538 = -0.6170
12. Use the data: 5, 8, 12, 15, 9, 11, 13, 20, 8, 10, 9, 8, 6, 20, 8
a. Calculate the percentile rank for 12? Show your work.
b. Determine the value that corresponds to the 30th percentile? Show your work.
ANS. We need the data sorted
Data
Position
5
1
6
2
8
3
8
4
8
5
8
6
9
7
9
8
10
9
11
10
12
11
13
12
15
13
20
14
20
15
a. Percentile Rank (page 155)
How many numbers are under 12? 10 of them
Percentile = (10 + 0.5) / 15 * 100 = 10.5/15 * 100 = 70 th
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b. 30th Percentile (page 157)
c = (15 * 30) / 100 = 4.5
Since this is not a whole number round up to next whole number = 5
5 is NOT the answer, 5 is the position where the 30 th percentile is at. Position 5 number is 8.
Data
Position
5
1
6
2
8
3
8
4
8
5
8
6
9
7
9
8
10
9
11
10
12
11
13
12
15
13
20
14
20
15
13. Chebyshev’s Theorem - If the mean is 10 and the standard deviation is 4 what is the predicted
percentage of values that will live in the interval (1, 19)?
ANS.
From the theory we know the interval is always
( mean – k * standard deviation, mean + k * standard deviation )
So matching
mean = 10, standard deviation = 4, k = unknown
( 10 – 4k, 10 + 4k ) = (1 , 19)
So
10 – 4k = 1 and solve for k
-4k = -9
k = 9/4 = 2.25
The proportion is
1 - 1/k^2 = 1 – 1/(2.25^2) = 1 - 0.1975 = 0.8025
So about 80.25% of the data will live in the interval ( 1, 19)
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14. For the following data use StatCrunch to make a Scatter Plot. Paste the graph into your Word
document.
Age, X
34
22
46
56
62
Hours, y
5.5
7
3.5
3
1
ANS.
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