Summary of Statistics and Hypothesis Testing

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

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 thtermwhich 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 thterm with a corresponding value of 110