# May june 2016 memo

MEMO FOR MAY/JUNE 2016 SECTION 1 (suggested solutions) 1 .True 2. Median = 𝑁+1 8+1 2 = 2 Therefore median = =4.5 9...

MEMO FOR MAY/JUNE 2016 SECTION 1 (suggested solutions) 1 .True 2. Median =

𝑁+1 8+1 2

=

2

Therefore median =

=4.5

9+11 2

=10

3.range of the distribution =highest score – lowest score=13-4 Therefore range = 9 4.N=8 5.True 20

6. prob friend winning first price =150 =0.13 35

7. prob self winning first price=150 =0.23 8.True 9.dependable variable 10.∑x² = 195 11. .∑y² =306 12.mode for X distribution = 7 13. False 14.False 15. Sample 16. Null hypothesis 17. critical value of t0.01 (39)=2.7079 18. Yes 19. Interval 20.negative

SECTION 2 (suggested solutions) QUESTION 1 (a) 0.62 moderate positive relationship (b) deduction : If students perform well in assignment 1 ,they can also perfom well in assignment 2. (c) common variance: / r = 0.62 r²= 0.3844 x 100% r²= 38.44%

38.44%

(d) intercept: a = Ÿ -bX = 4.67-(0.53)(6.91) = 4.67-3.6623 = 1.0077 = 1.01 (e) regression : Ŷ= bX + a = (0.53)(6) +1.01 =3.18+1.01 =3.21

(f)

Class interval

Frequency

Cum frequence

% frequence

90-99 80-89 70-79 60-69 50-59 40-49 30-39 20-29 0-19

6 14 35 44 65 45 25 12 9

255 249 235 200 156 91 46 21 9

2 5 14 17 25 18 10 5 4

g. Percentile Rank = 66 𝑠𝑐𝑜𝑟𝑒−𝑅𝐿𝐿

percentile rank = % below + 𝐶𝑙𝑎𝑠𝑠𝑖𝑛𝑡.𝑤𝑖𝑑𝑡ℎ (𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙%)

= 62 +

66−59.5 10

(17.3)

= 62 +11.25 =73.25 76.25% scores lie at or below a score of 68 h. 55th percentile score of p =

RLL +

= 49.5

+

𝑃𝑅−%𝑏𝑒𝑙𝑜𝑤 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 % 55−37 25

(𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑤𝑖𝑑𝑡ℎ)

(10)

= 49.5 + 7.2 = 56.7 The score at the 55th percentile is 56.7

Cum % frequence 100 98 93 79 62 37 19 9 4

QUESTION 2 (a) Proportion of students with raw score greater than 20? z= =

𝑋−µ 𝛿 20−35 6

= -2.5 Proportion = 0.99379 (b) Proportion of students with raw score greater than 40? z= =

𝑋−µ

𝛿 40−35 6

= 0.833 Proportion = 0.20327 X=20:Mean to Z score = 0.49379 X=40;Mean to Z score = 0.29673 Total

= 0.79052

0.79052 x 600 = 474,31 =474 students

QUESTION 3 (a) H0:µA = µB (b) H1:µA ≠ µB (c) t =

t=

𝑋1−𝑋2 2 √𝑆1 +𝑆2² 𝑁1 𝑁2

9.27−8.2 √

8.64 6.64 + 15 15

1.07

t = 1.01 t =1.06

(d) df = N1 + N2 -2 = 15 + 15 -2 = 28 (e) t 0.05 (28) = 2.0484 (f) 1.06 < 2.0484 Do not reject H0

(g) It can be concluded with 95% certainty that there is no significant different in productivity level of current employees and that of new employees.

QUESTION 4 SStotal = ∑ X² -

(∑𝑋)² 𝑁

= 1418- (182)²/24

df total = N -1 = 24-1 =23

= 1418-130.17 = 37.83 SSgroup = n ∑ (Xj-X…)²

df group= k-1

= 8[(7.58-7.13)² + (7.58 -7.38)² + (7.58-8.25)²]

= 3-1

= 8 [ (0.45)² + ( 0.2)² + (-0.67)²]

=2

= 8 (0.6914) = 5.53 SSerror = SStotal – Ssgroup = 37.83-5.53 =32.3 MSgroup = SSgroup / df group = 5.53/ 2 = 2.77

df error = k (n-1) =3 (8-1) = 21

MSerror = SSerror / df error =32.3/21 = 1.54 F = MSgroup / MSerror = 2.77/ 1.54 = 1.798 = 1.8 SOURCE GROUPS ERROR TOTAL

Df 2 21 23

SS 5.5 32.3 37.8

MS 2.8 1.5 4.3

F 1.8

F0.01 (2,21) = 5.78

(c) 1.8 < 5.78 Do not reject H0 (d) There is no significance difference in the career maturity of first, second and third year students. This can be said with 99% certainty.

QUESTION 5 (a) df = (R-1)(C-1) = (2-1)(2-1) = 1

𝑥 2 0.01 (1) = 6.6349 (b) 45.72 > 6.6349 Reject H0 (c) There is significance difference between opinions of students and opinions of recruiters. This can be said with 99% certainty.