1. Mean, Median, mode, and standard deviation for the following data set
4, 7, 8, 9, 5, 6, 9, 13, 19
Mean – 8.89
4 + 7 + 8 + 9 + 5 + 6 + 9 + 13 + 19
... [Show More] = 80
80/9 = 8.88888888889
Median – 8
4, 5, 6, 7, 8, 9, 9, 13, 19
Mode – 9
4, 5, 6, 7, 8, 9, 9, 13, 19
Standard Deviation – 4.36
Mean – 8.89
170.8889
9 = 18.98765432
Square Root (18.98765431)
= 4.357482567
x x-8.89 (x-8.89 )^2
4 -4.89 23.9121
5 -3.89 15.1321
6 -2.89 8.3521
7 -1.89 3.5721
8 -0.89 .7921
9 0.11 .0121
9 0.11 .0121
13 4.11 16.8921
19 10.11 102.2121
Total = 170.8889
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shared via CourseHero.com2. Do a t-test to see if there is a difference between these two groups.
I thought the t-test was very standardized, but I included both the results I got
when following the lecture and how I previously learned to be safe, please let me
know which method is correct or why my results aren’t matching. Sorry I didn’t
reach out ahead of the due date to confirm, I was supposed to have a tutor help
who was also confused!
X1 X2
6 7
7 8
7 8
5 8
10 12
If I follow video in lecture:
X1 Mean (6+7+7+5+10/5) =7
Sum of Squares (SS1) = 14
X2 Mean (7+8+8+8+12/5) =8.6
Sum of Squares (SS1) = 15.2
Mean1 (7)- Mean2 (8.6) = -1.6
SS1(14) +SS2(15.2) =29.2
N1(5) + N2(5) = 10
29.2/10 =2.92
N1(5) x N2(5) =25
10/25 =.4
2.92 x .4 = 1.168
t = (-1.6)/(sq rt (1.168))
sq rt (1.168) = 1.0807
t = -1.4805
The results show the difference is considered not statically significant.
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shared via CourseHero.comThe way I had learned and what result I kept getting when trying to confirm on other
sources:
X1 Mean (6+7+7+5+10/5) =7
Sum of Squares (SS1) = 14
Df1 = n-1 = 5-1 =4
(S^2)1 =SS1/(n-1) =14/(5-1)= 3.5
X2 Mean (7+8+8+8+12/5) =8.6
Sum of Squares (SS1) = 15.2
Df2 = n-1 = 5-1 =4
(S^2)2 =SS2/(n-1) =15.2/(5-1)= 3.8
Mean1 (7)- Mean2 (8.6) = -1.6
s2
p = ((df1/(df1 + df2)) * s21) + ((df2/(df2 + df2)) * s22)
(S^2)p =((4/8)x(3.5)) +((4/8)x(3.8)) =3.65
s2M1 = s2p/N1
s2M2 = s2p/N2
3.65/5 = .73
.73 +.73 = 1.46
t = (M1 - M2)/√(s2M1 + s2M2)
(-1.6)/(sq rt (1.46)) = -1.32
t = -1.32
p = .2220
The difference is not significant at p<.05
3. See how strong the relationship is between relational satisfaction and relational
maintenance.
Rel Sat Rel maintenance
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shared via CourseHero.com7 8
10 10
6 5
10 8
9 7
SS1 =13.2
SS2 =13.2
Sum of 1 =42
Sum of 2 = 38
N= 5
(42)(38) = 1596
1596/5 = 319.2
7 x 8 = 56
10 x 10 = 100
6 x 5 = 30
10 x 8 = 80
9 x 7 = 63
Total 329
Sum of products = 329 – 319.2
Sum of products = 9.8
R = (9.8)/(sq rt(13.2 x13.2))
R = (9.8)/(sq rt(174.24))
R = (9.8)/(13.2)
R = .742424
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