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(MATH144)07F-Revision solution.pdf
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solution

Question 1
(a)
mean = 373/53 = 7.0371

= 2.7593
CV = 2.7593/7.0371 x 100% = 39.21%

(b)
53 * 0.2 = 10.6

2 + 5 + (x/2)*12 = 10.6 x = 0.6
The monthly sales is 10 C 0.6 = 9.4

(c)
P(X > 8.8) = [5+2+(1.2/2)*12]/53 = 0.2679


(d)
CV_xyz = 29.09%

ABC has greater variability



Question 2
(a)
(i) P(X<70)=0.017 = 14.1505

(ii) P(X > 115) P(Z>1.06)=0.1446

(b)
P(A) = 0.26, P(E) = 0.22 and P(A and E)=0.11


(i)
P(A or E) = 0.26 + 0.22 C 0.11 = 0.37


(ii)
P(E|A) = 0.11/0.26 = 0.423


(iii)
P(Ec | Ac) = (1 C 0.37)/(1 C 0.26) = 0.8514



(c)
P(C|D) = (0.2)(0.03)/[0.2x0.03 + 0.4x0.01 + 0.4x0.02] = 0.333




Question 3
(a)
(i) 90% CI :


(ii)

(iii) H0: p 60% vs H1: p < 60%



The null hypothesis is rejected. The manager cant claim it.



(b)
H0: = 15 vs H1: > 15



The null hypothesis is rejected. The mean age is above 15 years.



Question 4
(a) D = after-before The data is 0.9,0.6,-0.8,-0.8,0.3,0.8

H0: = 0 vs H1: > 0

The null hypothesis is not rejected. The campaign is not effective.

(b) H0: Equal preference vs H1:Unequal preference

2


The null hypothesis is not rejected. The preference is equal.

(c) H0: F = A vs H1: F A Critical region: t < -2.45 or t > 2.45

F = 18.5, A = 16.75
F = 5, F = 5.12




The null hypothesis is not rejected. There is no significant difference.

Question 5

Solution:









(a) (i)








(ii)





(iii) As

Therefore, 82.62% of the sum of squares of deviations of building permits reduced by using interest rates as a predicator rather than using the average annual building permits
as a predicator of y for these data.
.
(b) H0: Distribution is Binomial.
H1: Distribution is not Binomial.

\alpha = 0.05



\hat{p} = \frac{1.4}{4} = 0.35.$


Pr(X_k = k) = _4C_k \hat{p}^k (1-\hat{p})^{4-k}

we have

P(X=x)
0.1785
0.3845
0.3105
0.1265

E_i
21.42
46.14
37.21
15.18




Test Statistics = 1.78
Critical Region: 2 > 5.991 and v = 4-1-1 = 2
Conclusion: H0 is not rejected, the distribution is binomial.