Summative Assessment 7
FOR NUMBERS 1-10
Class interval
Frequency (f)
Midpoint(x)
f*x
30-34
3
32
96
35-39
6
37
222
40-44
7
42
294
45-49
12
47
564
50-54
10
52
520
55-59
5
57
285
60-64
4
62
248
65-69
2
67
134
70-74
1
72
72
n = 50
Mean = Σfx
n
= 2435
50
= 48.7
Answer: Mean = 48.7
Σfx = 2435
FOR NUMBER 11-20
Standard deviation = √ (Σ(x - x̄)2 /n - 1)
First we get the mean of the score
Mean (x̄) = sum of total scores/number of students
= (85 + 95 + 75 + 96 + 86 + 94 + 86 + 82 + 96 + 77 + 85 + 88 + 89 + 90 + 82 + 83 + 97) / 17
= 1486/17
= 87.411764706
Scores (x)
Mean (x̄)
(x - x̄)
(x - x̄)2
85
87.411764706
-2.411764706
5.816608997
95
87.411764706
7.588235294
57.58131488
75
87.411764706
-12.41176471
154.0519032
96
87.411764706
8.588235294
73.75778547
86
87.411764706
-1.411764706
1.993079585
94
87.411764706
6.588235294
43.40484429
86
87.411764706
-1.411764706
1.993079585
82
87.411764706
-5.411764706
29.28719723
96
87.411764706
8.588235294
73.75778547
77
87.411764706
-10.41176471
108.4048444
85
87.411764706
-2.411764706
5.816608997
88
87.411764706
0.5882352941
0.346020761
89
87.411764706
1.588235294
2.522491349
90
87.411764706
2.588235294
6.698961937
82
87.411764706
-5.411764706
29.28719723
83
87.411764706
-4.411764706
19.46366782
97
87.411764706
9.588235294
91.93425605
Σx = 1486
Σ(x - x̄)2 = 706.1176473
Standard deviation = √ (Σ(x - x̄)2 /n - 1)
= √(706.1176473/17-1)
= √(706.1176473/16)
= √44.132352956
= 6.643218569
Rounded off to 2 decimal places= 6.64
Answer: Standard deviation = 6.64
FOR NUMBERS 21-30
n
1
2
3
4
5
6
7
8
9
10
x
3
1
4
6
10
15
6
7
5
8
Σ=65
y
3.20
3.15
3.00
4.34
4.91
4.25
3.96
3.08
4.03
3.89
Σ=37.81
xy
9.6
3.15
12
26.04
49.1
63.75
23.76
21.56
20.15
31.12
Σ=260.23
x2
9
1
16
36
100
225
36
49
25
64
Σ=561
y2
10.24
9.9225
9.00
18.8356
24.1081
18.0625
15.6816
9.4864
16.2409
15.1321
Σ=146.7097
Solving for its correlation coefficient ,
r=
n(∑xy)−(∑x)(∑y)
√[n(∑x2)−(∑x) 2][n(∑y2)−(∑y) 2]
=
10(260.23)−65(37.81)
√[10(561)−652][10(146.7097)−37.812]
= 2,602.3-2,457.65
√(𝟏𝟑𝟖𝟓)(𝟑𝟕. 𝟓𝟎𝟎𝟗)
=144.65
√51,938.7465
=144.65
227.9007383
r=0.6347
There is a high positive correlation between years of service of the 10 employees and their corresponding
ratings.
It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim
that there is a significant difference between years in service of 10 employees and their corresponding
efficiency ratings, at the 𝜶 = 𝟎. 𝟎𝟓 𝐬𝐢𝐠𝐧𝐢𝐟𝐢𝐜𝐚𝐧𝐜𝐞 𝐥𝐞𝐯𝐞𝐥.
FOR NUMBER 31-40
Null and Alternative Hypothesis
Null Hypothesis: There is no significant difference between years in service of 10 employees and their corresponding
efficiency ratings.
H0: p=0
Alternative Hypothesis: There is a significant difference between years in service of 10 employees and their
corresponding efficiency ratings
.
Ha:p≠0
Rejection Region
Based on the information provided, the significance level is 𝛼 = 0.05,
and the degrees of freedom are computed as follows,
df= 10- 2 = 8
Hence, it is found that the critical value for this two- tailed test is 𝑡𝐶 = 2.306, for 𝛼 = 0.05, and df =8.
The rejection region for this two-tailed test is R= {t :|t| > 2.306}
Test Statistics
𝑡=
𝑡=
𝑟
2
√1 − 𝑟
𝑛−2
0.6347
2
√1 − 0.6347
10 − 2
t=2.325
Decision about the null hypothesis
Since it is observed that |t| = 2.325 ≥ 𝑡𝐶 = 2.306, it is then concluded that the null hypothesis is rejected.
Using the P- value approach: the p value is p=0.0485, and since p=0.0485 ≤ 0.05. it is concluded that the null hypothesis
is rejected.
Conclusion
It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that there is a
significant difference between years in service of 10 employees and their corresponding efficiency ratings, at the 𝛼 =
0.05 significance level.
FOR NUMBER 41-50
Spearman Rank correlation test
x
3
9
4
2
5
7
1
4
2
8
7
2
y
4
8
5
6
3
2
3
5
1
7
6
4
X ranks
5
12
6.5
3
8
9.5
1
6.5
3
11
9.5
3
𝑟𝑠 = 1 −
𝑟𝑠 = 1 −
6𝛴 ⅆ
Y ranks
5.5
12
7.5
9.5
3.5
2
3.5
7.5
1
11
9.5
5.5
D=y-x
0.5
0
1
6.5
-4.5
-7.5
2.5
1
-2
0
0
2.5
𝑚 (𝑚2 −1)+𝑚2 (𝑚22 −1)+𝑚3 (𝑚32 −1)+𝑚4(𝑚42 −1)+𝑚5 (𝑚52 −1)….
2+{ 1 1
}
12
𝑛(𝑛2 − 1)
6(137.5)
+{
3(9−1)+2(4−1)+2(4−1)+2(4−1)+2(4−1)+2(4−1)+2(4−1)
}
12
12(122 − 1)
24+6+6+6+6+6+6
12
825+{
𝑟𝑠 = 1 −
12(144 − 1)
𝑟𝑠 = 1 −
825 + 5
12(143)
𝑟𝑠 = 1 −
830
1,716
𝑟𝑠 = 1 − 0.483682983
𝒓𝒔 = 𝟎. 𝟓𝟏𝟔𝟑𝟏𝟕𝟎𝟏𝟔
}
d2
0.25
0
1
42.25
20.25
56.25
6.25
1
4
0
0
6.25
137.5
FOR NUMBER 51-60
Online DL
80
85
74
75
65
66
80
77
95
66
55
79
85
75
Modular DL
66
98
87
84
56
59
25
53
45
66
52
85
Measurement
Mean
SD
Sample Size
Online DL
75.5
10.11282506
14
Modular DL
64.66666667
20.81229241
12
Null and Alternative Hypothesis
Null Hypothesis: There is no significant difference on the performance of two groups of students.
H0:μ1=μ2
Alternative Hypothesis: There is significant difference on the performance of two groups of students.
Ha:μ1≠μ2
This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with
unknown population standard deviation will be used
Rejection Region
Based on the information provided, the significance level is 𝛼 = 0.05,
and the degrees of freedom are computed as follows, assuming that the population variances are equal:
df= df1 + df2 = 13 +11 =24
Hence, it is found that the critical value for this two- tailed test is 𝑡𝐶 = 2.064, for 𝛼 = 0.05, and df =24.
The rejection region for this two-tailed test is R= {t :|t| > 2.064}
Test Statistics
𝑥1 − 𝑥̅2
̅̅̅
𝑡=
√
(𝑛 − 1)𝑠12 + (𝑛2 − 1)𝑠22 1
1
(𝑛 + 𝑛 )
𝑛1 + 𝑛2 − 2
1
2
75.5 − 64.66666667
𝑡=
2
2
√(14 − 1)10.11282506 + (𝑛2 − 1)20.81229241 ( 1 + 1 )
14 + 12 − 2
14 12
t=1.728
Decision about the null hypothesis
Since it is observed that |t| = 1.728 ≤ 𝑡𝐶 = 2.064, it is then concluded that the null hypothesis is not rejected.
Using the P- value approach: the p value is p=0.0968, and since p=0.0968 ≥ 0.05. it is concluded that the null hypothesis
is not rejected.
Conclusion
It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that there is a
significant difference on the performance of two groups of students, at the 𝛼 = 0.05 significance level.