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Question one: Determine if there is a significant difference between boys’ and girls’ UI1 test scores?

UI1 Boys: 5,15,14,14,5,6,7,6,6,15,5,15,15 n=13

UI1 Girls: 15,12,10,12,16,14,15,9,15,11,15,7,16 n=13

Step one: sum of the groups.

SUM: U11 BOYS=128

SUM: U11 GIRLS = 167

Step two: square sum of step one:

128

=16384

167

=27889

Step three: determine mean of the two groups.

Boys U11 =

128

⁄

=9.846

Girls U11=

167

⁄

= 12.846

Step four: square the individual scores then add them up.

UI Boys: 5

+15

+14

6+6

+15

= 1524

UI1 Girls: 15

+12

+10

+12

+16

+14

+15

+11

+15

+16

= 2247

Step five: let UI1 boys be B

Let UI1 girls be G

(∑B)

= 128

=16384

(∑G)

= 167

= 27889

b =

9.846

=12.846

∑B

= 1524

∑G

= 2247

=13

9.846-12.846

t= √4688

+ √

2.911

√13 √ 13

−3

1.530

⁄

t=-1.960

Step 6: N

-1 + N

- 1 = 13-1 + 13-1 =24

Step 7: from the t- table

Using 5% (0.05), 24 degrees of freedom.

Level of 0.05 = 2.0634

Step 8: the calculated t value -1.960 is less than 2.0639 from the table.

Conclusion

Therefore, this means that p>0.05

As p-value is greater than the alpha level, we cannot significantly conclude that there is a difference between boys

And girls.

Question two: Determine if there is a significant difference between 6-year-olds’ and 7-year-olds’ UI1 scores?

Data set A= 6-year-old: 5,12,10,12,5,6,7,9,15,11,6,5,7

Data set B= 7-year-old: 15,14,14,16,14,15,6,15,15,16,15,15

Step one: sum of two groups.

SUM: A= 5+12+10+…….7 =110

SUM: B= 15+15+14+……+15=185

Step two: square sum of step one:

110

=12100

185

= 34225

Step three: find the mean of the two groups.

6-year-old =

110

⁄

=8.462

7-year-old=

185

⁄ = 14.231

Step four: square the individual scores then add them up.

6-year-old: 5

+12

+10

+12

+15

+11

= 1060

7-year-old: 15

+15

+14

+16

+14

+15

+16

+15

= 2711

Step five:

(∑A)

= 12100

(∑B)

= 34225

a =

8.462

=14.231

∑A

=2711

∑B

= 1060

=13

NB=13

8.462 - 14.231

t= √3.282

+ √

2.555

√13 √ 13

−5.769

1.154

⁄

t= -5.001

Step 6: degrees of freedom= N

-1 + N

- 1 = 13-1 + 13-1 = 24

Step 7: from the t - table

Using 5% (0.05), 24 degrees of freedom.

Level of 0.05 = 2.0639

Step 8: the calculated t value of -5.001 is more than 2.0639 from the table.

Conclusion

Therefore, this means that p<0.05

we can conclude that there is a significant difference between 6 years and 7-year U11 scores.

Question 3: Determine if there is a significant difference between UI1’s scores concerning the number of

devices they have at home?

Scores of the UI at home.

3 devices score: 15,10,14,6,9,11,7

4 devices score: 5,15,12,12,16,15,6,15,15,5

5 devices score: 15,5,14,7,6,16,15.

6 devices score: 15,5.

Step one: sum of individual devices scores at home.

3 devices score: 15+10+14+6+9+11+7=72

4 devices score: 5+15+12+12+16+15+6+15+15+5=116

5 devices score: 15+5+14+7+6+16+15=77

6 devices score: 15+5= 20

Step two: find the grand mean. = 72+116+77+20/26= 10.962

Individual group mean:

3 devices score:

⁄

= 10.286

4 devices score:

116

⁄

= 11.6

5 devices score:

⁄

= 11.0

6 devices score:

⁄ = 10.0

Step three: analysis of variance (ANOVA) procedure.

= µ

versus

≠ µ

at ∝ =0.05 significant level.

Source

S.S- sum of square

D.F- degree of freedom

M.S-mean of sum

of squares

Regression

34.125

11.375

0.6062

Error

412.828

18.765

Total

446.953

= [(15-10.286)

+(10-10.286)

+…………+ (7-10.286)

] +

[(5-11.6)

+(15-11.6)

+…………. +(5-11.6)

[(14-11)

+(5-11)

+…………. +(15-11)

] +

[(15-10)

+(5-10)

SSE =67.428+184.4+136+25

=412.828

SS7 = [(15-10.962)

+(10-10.962)

+…………. + (5-10.962)

]

=446.953

Step 4: Degrees of freedom

M=4 therefore 4-1=3 degrees of freedom.

N= 26 therefore 26-1=25 degrees of freedom.

F value = F 0.05, 3, 25 =2.99

Step 5: F calculated = 0.06062

F ∝=2.99

<F∝

0.6062<2.99

We accept H

= µ

Conclusion

There is no significant difference between scores corresponding the number of devices they have at home.

Question 4: Determine if there a significant difference between students’ UI1 and UI2 scores?

UI2=5,15,15,14,12,10,14,12,16,5,6,14,15,7,9,6,15,11,6,15,15,5

UI1= 18,20,21,22,15,20,23,25,21,19,18,26,22,29,24,20,21,23,25,22,24,15,24,23.

Step one: sum of scores.

UI1= 295

UI2= 571

Step two: square sum of step one:

295

=87025

571

=326041

Step three: find the mean of the two devices.

UI1 =

295

⁄ =11.346

UI2=

571

⁄ = 21.961

Step four: square the individual scores then add them up.

UI1: 5

+15

+14

……………………. +15

= 3771

UI2: 18

+20

+21

+22

+…………………...+3

= 12817

Step five:

11.346 - 21.961

t= √4.118

√3.228

√26 √ 26

−10.615

√

1.0782

⁄

t= -9.845

Step 6: degrees of freedom= N

-1 + N

- 1 = 26-1 + 26-1 =50

Step 7: from the T- table

Using 5% (0.05), 50 degrees of freedom.

Level of 0.05 = 2.00856

Step 8: the calculated t value is -9.845

t value from the table is 2.00856

Conclusion

Therefore, this means that p<0.05

we can conclude that there is a significant difference between UI1 and U12 scores.

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