Analysis for the data collected by survey regarding purchase decision and customer satisfaction (for cars)

9. Microsoft Office
Excel Worksheet

11. In our sample, the following pattern of male and
female candidates is observed. We want to test if
there is randomness in our data or not.
F M F F F F F F M F M F F M F M M F M F F F M F F
F F F M F F M F F M F M M F M M M F F F F M M M
M F F F M F
Significance level = 20%

12. n1 (Males) = 21 , n2 (Females) = 34
R (no. of runs) = 29
Null = randomness is there.
µ of r = (2n1n2/n1+n2)+1
= [(2*21*34)/55]+1
= (1428/55)+1
= 26.96
Std. dev. Of r = √{2n1n2(2n1n2-n1-n2)}/(n1+n2)^2(n1+n2-1)
= √1960644/163350
= 3.46

13. Z value (from table) =1.28
Z = (r - µ of r)/std. dev. Of r
= (29-26.96)/3.46
= 0.59
Accept null hypotheses.
Therefore, randomness is there in our sample.

14. As per our sample, the following information is given:
Probability of male is 38%, Probability of females is
62%
, Probability that a Hyundai car is purchased by male
is 0.53 and by female is 0.18.
Find the probability that:
i. The car purchased by male is Hyundai.
ii. The car purchased by female is Hyundai.

15. Elementary events = Male , Female
A = Hyundai
Event P(E) P(A/E) P(A&E)
MALE 0.38 0.53 0.20
FEMALE 0.62 0.18 0.11
P(A) = 0.31

16. i. P(M/A) = P(M & A)/P(A)
= 0.20/0.31
= 0.645
ii. P(F/A) = P(F & A)/P(A)
= 0.11/0.31
= 0.354

17. From a sample of 55, 34 people are using
loan/EMI to finance their car. What is the
probability that more than 50% of sample used
loan/EMI as finance?

18. p = probability of success = 34/55 = 0.62
q = 1-p = 1-0.62 = 0.38
µ p⁻ = p = 0.62
P (p>0.5) = ?
Std. dev. Of p = √pq/n⁻
= √(0.62)(0.38)/55
= 0.065

19. Z = (p - µp )/std. dev. Of p⁻ ⁻ ⁻
Z = (0.5 – 0.62)/0.065
Z = -1.85
Area at 1.85 = 0.4678
Required area = 0.5+0.4678 = 0.9678
Therefore, 0.9678 is the probability that more than
50% of the sample use loan as finance.

20. A random sample of 55 people are examined;
sample mean of the brands available is found to be
12. If sample std. deviation is 4.60, estimate the
average people who chose the particular brand
from the given list. Using a 95% confidence
interval.

21. n = 55 , = 12 , s(sample std. dev.) = 4.60ẍ
Estimated standard error of = s/√n = 4.60/√55ẍ
= 4.60/7.42 = 0.62
Confidence interval = ± (z * estimated std. error of )ẍ ẍ
z = 0.95/2 = 0.475 (area)
Value of z under area 0.475 = 1.96
C.I. = 12 ± 1.96 (0.62) = 12 ± 1.2152
= [ 10.7848 , 13.2152]
So, average will lie in between 10.7848 and 13.2152.

22. According to the data, conclude the following:
i. What is the proportion of petrol, diesel and CNG?
ii. Find the probability that diesel will be preferred by
female.
iii. Find the probability that a male will prefer CNG.
iv. Determine the probability for preference of petrol or
diesel.
v. Find the probability that a female will prefer petrol.

23. Information collected from the data:
P(MALE) = 0.38 , P(FEMALE) = 0.62 , P(PETROL) = 0.62
P(DIESEL) = 0.25 , P(CNG) = 0.13
JOINT PROBABILTIES:
P(AB) = P(A) * P(B)
P(MP) = 0.24 , P(MD) = 0.09 , P(MC) = 0.05
P(FP) = 0.38 , P(FD) = 0.16 , P(FC) = 0.08

24. PETROL DIESEL CNG TOTAL
MALE 0.24 0.09 0.05 0.38
FEMALE 0.38 0.16 0.08 0.62
TOTAL 0.62 0.25 0.13

25. i. Proportion of petrol = 0.62 or 62%
Proportion of diesel = 0.25 or 25%
Proportion of CNG = 0.13 or 13%
ii. P(F/D) = P(FD)/P(D)
= 0.16/0.25 = 0.64
iii. P(C/M) = P(CM)/P(M)
= 0.05/0.38 = 0.13
iv. P(P U D) = P(P) + P(D)
= 0.62 + 0.25 = 0.87
v. P(P/F) = P(PF)/P(F)
= 0.38/0.62 = 0.61

26. Amongst the different brands available in the
market, it is assumed that the clientage of Honda
and Hyundai is 45%. In our sample of 55, we found
that proportion of people using Honda and Hyundai
is 43.6%. At 5% level of significance, can we
conclude that proportion of Honda and Hyundai
has decreased in the market?

27. pho (hypothesised proportion) = 0.45
qho= 1 - pho = 0.55
n = 55
p = 0.436⁻
Ho : p ≥ 0.45
Ha : p < 0.45
Std. error of proportions = √(pho*qho)/n
= √(0.45*0.55)/55
= √0.0045 = 0.067

28. Level of significance = 0.05
Z (area) = 0.5 – 0.05 = 0.45
Value of z at 0.45 area = -1.64
Critical value = pho – z (std. error of proportions)
= 0.45 – 1.64 (0.067)
= 0.45 – 0.109 = 0.339
Therefore, 43.6 lies in the acceptance region. So,
we accept the null and conclude that the
proportion of Honda and Hyundai is not decreased.