Skema Jawapan Final Bwm 11603/Bsm 1413 Semester I 2011/2012

SKEMA JAWAPAN FINAL BWM 11603/BSM 1413 SEMESTER I 2011/2012

QUESTION 1

(a) The sample mean,

xi / / M1 A1
6.1 / 37.21
5.7 / 32.49
5.8 / 33.64
6.0 / 36.00
5.8 / 33.64
6.3 / 39.69
A1 / A1

QUESTION 2

becomes .

Thus,

QUESTION 3

a)n1 = 50 and n2 = 55 (n1, n2 30), s1 = 38 and s2 = 12.

For 90%,  = 1 – 0.90 = 0.10, z/ 2 = z0.05 = 1.6449. A1

1 2 = () = 191 – 199 = –8.

z/ 212 + z/ 2

–8 – (1.6449)12 < –8 + (1.6449) M1

–8 – 9.231712 < –8 + 9.2317

–17.231712 < 1.2317 A2

Therefore, based on the selected samples, we are 90% confident that the difference between two means falls in 17.2317 acres and 1.2317 acres.

S = 5.3125 A1 s2 = 28.2227.  = 9.

 = 1 – 0.90 = 0.10.

== 16.919 A1

== 3.325 A1

2

2 M1

15.01292 < 76.3922 A2

Therefore, we are 90% confident that the true variance of the population is between RM15.01 and RM76.39

(i)s12 = 8 s22 = 7.

 = 1 – 0.95 = 0.05. 1 = 24 2 = 12.

A1 A1

12 is s12 = 82 22 is s22 = 72

< M1

0.4325 < < 3.3176 A2

Thus, we are 95% confident that the ratio of two variances falls in 0.4325 and 3.3176.

(ii)s12 = 8 and s22 = 7.

 = 1 – 0.98 = 0.02. 1 = 24 and 2 = 12.

12 is s12 = 82 22 is s22 = 72

< M1

0.3455 < < 3.9576

0.5878 < < 1.9894 A2

Thus, we are 98% confident that the ratio of two population standard deviation falls in 0.5878 and 1.9894.

QUESTION 4

(a)

Step 1: H0: 1 = 2 or H0: 12 = 0

H1: 12 (claim) H1: 12 < 0 (claim) M1

Step 2: -  = 0.05

- unknown but equal and n1, n2 < 30 (Case (d)), we use t-distribution for difference between two means.

 = n1 + n2 2 = 21 + 8  2 = 27. A1

-Since  = 0.05 and the test is a left-tailed test, the critical value is given by

t 0.05, 27 = 1.703. A1

Decision Rule: reject H0 if the test value falls in the rejection region,t1.703.

Step 3:

Tar (mg)
Filtered / Nonfiltered
n1 = 21 / n2 = 8
s1 = 3.7436 / s2 = 1.6903 / M2

= M1 A1

= 7.7328 A1

7.7328 1.703 0

Step 4: Since the test value, 7.7328 (< 1.703) falls in the rejection region, the decision is to reject the null hypothesis, as shown in figure. M1

Step 5: Based on the samples data, there is enough evidence to support the claim that the mean amount of tar in filtered cigarettes is less than the mean amount of tar in non-filtered cigarettes. M1

b) Step 1: H0:

H1: (claim) M1

Step 2: - = 0.05

-Since  = 0.05, the degrees of freedom, 1 = 24 and 2 = 12, and the test is a two-tailed test, the critical values for

. A1

Decision Rule: reject H0 if the test value falls in the rejection regions,f < 0.3937 and f > 3.02 M1

Step 3: M1 A1

0.3937 3.02 3.6

Step 4: Since the test value, 3.6 falls in the rejection regions, the decision is reject the null hypothesis, as shown in figure. M1

Step 5: Based on the samples data, it means that the variance of the heart rates of smokers are different from the nonsmokers. M1

QUESTION 5

a)By using the calculator, we have

, , , , , A2

Sum of squares:

Therefore,

, ,

The fitted linear regression line is,

This means thatif the number of order increased by one mark, the distribution cost would increase by 0.0139 thousands. A1

b)When ,

M1 A1

Step 2: , , this is a two-tailed test

Ttable =, A1

reject when T test is more than 2.306 or less than -2.306.

Step 3: Compute MSE, and Ttest.

SSE =  M1

Ttest = A1

Step 4 : Make decision

Reject since Ttest is larger than 2.306 M1

Step 5 : Make conclusion

We can conclude that the slope is not equal to zero. M1

(d)

M1 A1

70% of the total variation is explained by the regression line using the independent variable M1