An IBM internship student designed two different user interfaces (e.g. UI1 and U
ID: 3314402 • Letter: A
Question
An IBM internship student designed two different user interfaces (e.g. UI1 and UI 2) for help age of 6-7 students to solve simple math problems. She would like to know which interface is better for improving their skills in solving math problems. The collected data include the attributes as UI1 test score, UI2 test score, Number of Electric devices they have at home, Age, Student Gender
UI 1
UI 2
Devices
Age
Gender
5
18
4
6
boy
15
20
3
7
girl
15
21
4
7
boy
14
22
5
7
boy
12
15
4
6
girl
10
20
3
6
girl
14
23
3
7
boy
12
25
4
6
girl
16
21
4
7
girl
5
19
5
6
boy
6
18
3
6
boy
14
26
5
7
girl
15
25
4
7
girl
7
28
5
6
boy
9
22
3
6
girl
6
29
4
7
boy
15
24
4
6
girl
11
20
3
6
girl
6
21
5
6
boy
15
23
4
7
girl
15
25
6
7
boy
5
22
4
6
boy
7
24
3
6
girl
16
15
5
7
girl
15
24
5
7
boy
15
23
6
7
boy
For this assignment, you need to apply t-test to find answers to the following research questions (show how you get the answer with providing the p-value):
1. Is there a significant difference between UI1’s scores with respect to the number of devices they have at home?
2. Is there a significant difference between UI2’s scores with respect to the number of devices they have at home?
UI 1
UI 2
Devices
Age
Gender
5
18
4
6
boy
15
20
3
7
girl
15
21
4
7
boy
14
22
5
7
boy
12
15
4
6
girl
10
20
3
6
girl
14
23
3
7
boy
12
25
4
6
girl
16
21
4
7
girl
5
19
5
6
boy
6
18
3
6
boy
14
26
5
7
girl
15
25
4
7
girl
7
28
5
6
boy
9
22
3
6
girl
6
29
4
7
boy
15
24
4
6
girl
11
20
3
6
girl
6
21
5
6
boy
15
23
4
7
girl
15
25
6
7
boy
5
22
4
6
boy
7
24
3
6
girl
16
15
5
7
girl
15
24
5
7
boy
15
23
6
7
boy
Explanation / Answer
Question 1.
Given that,
mean(x)=11.3462
standard deviation , s.d1=4.1177
number(n1)=26
y(mean)=4.1538
standard deviation, s.d2 =0.9249
number(n2)=26
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.06
since our test is two-tailed
reject Ho, if to < -2.06 OR if to > 2.06
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =11.3462-4.1538/sqrt((16.95545/26)+(0.85544/26))
to =8.69
| to | =8.69
critical value
the value of |t | with min (n1-1, n2-1) i.e 25 d.f is 2.06
we got |to| = 8.68996 & | t | = 2.06
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 8.69 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 8.69
critical value: -2.06 , 2.06
decision: reject Ho
p-value: 0
we have evidence to say that significant difference between UI1’s scores with respect to the number of devices they have at home
Question 2.
Given that,
mean(x)=22.0385
standard deviation , s.d1=3.4465
number(n1)=26
y(mean)=4.1538
standard deviation, s.d2 =0.9249
number(n2)=26
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.06
since our test is two-tailed
reject Ho, if to < -2.06 OR if to > 2.06
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =22.0385-4.1538/sqrt((11.87836/26)+(0.85544/26))
to =25.5558
| to | =25.5558
critical value
the value of |t | with min (n1-1, n2-1) i.e 25 d.f is 2.06
we got |to| = 25.55579 & | t | = 2.06
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 25.5558 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 25.5558
critical value: -2.06 , 2.06
decision: reject Ho
p-value: 0
we have evidence to say that there a significant difference between UI2’s scores with respect to the number of devices they have at home
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