Statistical Analysis Assignment: Descriptive Stats & Correlation

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Added on 2023/04/21

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This assignment solution covers a range of statistical concepts and techniques. It includes calculating measures of central tendency (mean, median, mode, and range) for a given dataset, analyzing the number of letters received per day using descriptive statistics, and constructing a cumulative frequency curve to determine quartiles and the number of villages exceeding a certain population threshold. Further, the assignment involves calculating variance, standard deviation, and mean deviation for temperature data. A comparison of two datasets is conducted using mean, median, and range. The impact of a charge on car occupancy is evaluated using mean calculations. Correlation analysis is performed on tyre data, including plotting a scatter diagram, determining the line of best fit, and interpreting the slope and intercept. Spearman's rank correlation coefficient is calculated and interpreted for judge rankings. Finally, a t-test is conducted to determine if there is a significant difference in leaf thickness between north and south-facing walls, and a conclusion is drawn based on the p-value.

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Question 1
a. Given a set of data (4, 6, 2, 4, 8, 2,1, 8, -3, -5, 2, -8,3, 6, 8, 10)
∑ x
n = 4+6+2+4+8+2+1+8−3−5+2−8+3+6+8+10
16 =3
b. Mode
x f
-8 1
-5 1
-3 1
1 1
2 3
3 1
4 2
6 2
8 3
10 1
Mode = 2 and 8
c. Median=
n
2 + n+2
2
2 = 4+3
2 =3.5
Question 2
Number of letters Number of Days
cumulative number
of day
0 6 6
1 19 25
2 13 38
3 9 47
4 6 53
5 0 53
6 0 53
7 2 55
a. Modal number of letters received per day=1 letter

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b. Median number of letters received per day, since n is odd, median = n+1
2 = 55+1
2 =28 hence
median =2 letters
c. Mean number letters per day=∑ fx
∑ f = 110
55 =2
d. Range of letters received=7-0=7
e. Interquartile range
Q1= n+1
4 = 56
4 =14nth term which corresponds to 1 letter.
Q3=3 ( n+1
4 )=3∗14=42 th termwhich corresponds to 3 letters.
Interquartile range=Q3-Q1=3-1=2
Question 3
range of
x f cf
50-100 7 7
100-150 24 31
150-200 29 60
200-250 18 78
250-300 12 90
0 50 100 150 200 250 300 350
0
10
20
30
40
50
60
70
80
90
100
Cummulative frequency curve

a. N for Medium =N/2 and N/2 +1
90
2 +( 90
2 +1)
2 =45.5 th term
From the graph this corresponds to about Q0=180
b. For the lower quartile, N1= N +1
4 =22.75 th term with a corresponding value of 110

c. For the upper quartile N3= N +1
4 ∗3=68.25 with a corresponding with a corresponding value of
215
d. Interquartile range=Q3-Q1=215-110=105
e. Q greater than 260
From the graph, Q260 =79
The number of number villagers with more than 260 villagers =90-79=11 villages.
Question 4
a.
From the cumulative frequency column,
x x-u (x-u)^2
21 0 0
23 2 4
24 3 9
19 -2 4
19 -2 4
20 -1 1
21 0 0
sum 22
b. Population mean deviation STD=
√ ( xi−x )2
n = √ 22
7 =1.77281
Question 5
Comparison of Tom and Sam’s performance
Sam Tom
30 36
24 20
48 25
13 39
25 20
mean 28 28
Median 27.5 28
Range 35 19

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Question6
a. Mean before the charge was introduced =1.5176
b. Mean after the charge was introduced =1.8008
c. Introducing the charges increased the average number of occupants per car.
Question 7
x f1 f2 xf1 xf2
1 658 450 658 450
2 275 388 550 776
3 86 125 258 375
4 25 47 100 188
5 4 8 20 40
6 1 1 6 6
Sum 1049 1019 1592 1835
mean 1.517636 1.800785

a.
b. The arrow shows the coordinates of point M
c. The scatter diagram there is a strong negative correlation between the number of miles and
depth of depth of thread in mm
d. Line of best fit is shown in the graph
e. From the regression line,
y = -0.1128x + 7.8326
To become illegal, y=1.6mm,
Substituting in the equation,
1.6== -0.1128x + 7.8326
X=55.25 miles
f. For 20 thousand miles, x=20
Substituting in the equation,
y = -0.1128*20 + 7.8326=5.5766mm
Question 8
a.

1 2 3 4 5 6 7 8 9 10 11
0
2
4
6
8
10
12
14
f(x) = 0.758375333530981 x + 3.84974799881411
R² = 0.997903345253215
A scatter graph of( € )against (miles)
X in miles
Y €
b. There is a strong positive correlation between the two variables
c. Coordinates of the mean point
X=∑ x
n = 60
10 =6 and Y=∑ y
n = 84
10 =8.4 (6,8.4) shown with an arrow in the graph
d. The line of best fit is shown in the graph
e. The line of best fit is of the form
y = 0.7584x + 3.8497
When y=12,
Substituting in the equation,
12= 0.7584x + 3.8497
Hence x=10.7467 miles
f. When x=1, substituting in the estimated equation,
y= 0.7584*1 + 3.8497=4.6081 €
g. The equation of the line is in the form of y=mx+c
Compared to
Y= 0.7584x + 3.8497
M=0.7584
C=3.8497
h. In this context, m=0.7584 indicates an increase in cost € for a distance increase in 1 mile
C=3.8497 is the constant cost incurred when no distance is covered.
Question 9
a. The Spearman’s rank correlation coefficient between two oral test of +.76.

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This shows that the scores in the two oral tests are positively and strongly related to one
another.
b. The Spearman’s rank correlation coefficient between two oral test of -0.06
This shows that there is a weak negative relationship between the scores in these two tests.
c.
Judge 1 Rank1 Judge 2 Rank2 d dsquare
4 3 4 3 0 0
5 2 6 1 1 1
6 1 5 2 -1 1
2 5 3 4 1 1
1 6 1 6 0 0
3 4 2 5 -1 1
Sum 4
=1− 6 × 4
6 ( 36−1 ) =0.8857
d. The rankings given by the two judges are strongly and positively correlated hence.
Question 10
Summary of the statistics
South North
300 180
270 210
220 220
250 160
210 210
250 190
290 160
190 180
220 200
270 210
Mean 247 192
STD
36.2246
1
21.4993
5
n 10 10

Hypothesis
H0: μs= μn
H1: μs≠ μn
Assuming a null hypothesis,
t = μs−μn
√ Ss2
ns
+ Sn2
nn
= 247−192
√ 36.2252
10 + 21.4992
10
=4.1289
P(|t|≥4.1289) =2*(0.00446) from t -tables=0.000892
Since the significance level was 5%=0.05
The p-value <p-critical hence reject the null hypotheses hence there is a significant difference between
the leaves thickness.