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JANUARY 2018QUESTION 1a) Population – All customers who service their cars at the 40 service centers of the companyb) Variable – satisfaction level Type – qualitative Level of measurement – Ordinalc) Cluster sampling. All customers are grouped according to the service centers. Eight centers were sel...

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JANUARY 2018 QUESTION 1 a)

Population – All customers who service their cars at the 40 service centers of the company

b)

Variable – satisfaction level Type – qualitative Level of measurement – Ordinal

c)

Cluster sampling. All customers are grouped according to the service centers. Eight centers were selected from 40 available service centers. All customers from the eight service centers were chosen as samples.

QUESTION 2 a)

Skewness = (Mean – mode) / Std dev = (6.59-7.04)/0.74 = -0.61 The distribution of the life span of brand AA washing machine is skewed to the left.

b)

Most of the brand AA washing machines have a life span of 7.04 years.

c)

CVAA = 0.74 / 6.59 * 100 = 11% CVBB = sqrt(12.3) / 7.1 * 100 = 49% Brand AA washing machine has a more consistent life span compared to brand BB since it has lower coefficient of variation (CV).

QUESTION 3 a)

K = Standard error mean = 114.5695/ sqrt (20) = 25.6185

b)

90% CI for the mean monthly electricity bill: s  x  t  , n 1 2 n  705 .8  t 0.05,19 (25 .6185 )  705 .8  1 .729 (25 .6185 )   661 .5056 ,750 .0944 

c)

95% CI for the mean monthly electricity bill: s  x  t  , n 1 2 n  705 .8  t 0.025,19 (25 .6185 )  705 .8  2 .093 (25 .6185 )   652 .1804 ,759 .4196 

d)

Based on the interval in b), the manager’s claim is true since RM700 is in the interval.

e)

The higher the confidence level, the wider will be the confidence interval.

QUESTION 4 a)

t = mean / std error mean = -4.94 / 1.745 = -2.83

b)

H0: µd = 0 (No improvement in sales) H1: µd < 0 (There is an improvement in sales)

c)

Critical value = -t0.05,4 = -2.132 Reject H0 since |t| > |CV| There is sufficient evidence to indicate that the sales had improved.

QUESTION 5 a)

A = SSB  57 2 73 2 76 2  206 2 SSB      41.733  5 5  15  5

B = 164.933 – 41.733 = 123.2 C = 41.733 / 2 = 20.867 D = 123.2 / 12 = 10.267 b)

H0: µ1 = µ2= µ3 (There is no difference in the mean time among the three brands of medicine) H1: at least two means differ (There is a difference in the mean time among the three brands of medicine)

c)

p-value = 0.174 Decision: Do not reject H0 since p-value > 0.05 Conclusion: There is no difference in the mean time among the three brands of medicine

QUESTION 6 a)

P = 77 – 36.4 = 40.6 Q

 46  40.6 40.6

2

 36  29 .6 

2



29 .6



13  24 .8  24 .8

2



31  36 .4  36 .4

b)

H0: Student’s program and smoking habit are not related H1: Student’s program and smoking habit are related

c)

P-value = 0.000 Reject H0 since p-value < 0.01 Student’s program and smoking habit are related

2

20  26 .4 

2



26 .4

34  22 2. 

2



22 .2

 16 .34

QUESTION 7 a)

Dependent = monthly salary Independent – working experience

b)

R2 = 0.878 Interpretation: 87.8% of the total variation in monthly salary is explained by working experience.

c)

M = sqrt(0.878) = 0.937 Comment: There is a strong positive linear relationship between monthly salary and working experience.

d)

y = 23.470 + 1.194x Explanation of slope: When working experience increase by one year, monthly salary increase by RM1.194 (’00) @ RM119.40.

e)

y = 23.470 + 1.194(15) = 41.38 (’00) @ RM4138.00...