Solutions for QMB Semester 2 2009 Final examination

Question 1

(a) (i) It is a symmetric distribution, therefore it is not skewed at all. [1mark] (ii) From the figure obviously, n=12 and p=0.5 [1mark] Then E[X] = np = 6 and Var(X) = np(1-p) = 3 [0.5 mark for each] (iii) Two assumptions Profitability of projects is independent of each other [0.5 mark] The probability of being profitable is the same for all projects [0.5 mark]

(b) X: number of crashes per day. Then X~Poisson(3)

(i) 0498.0!0

3)0(

03

e

XP (from the CDF table it is P(X≤0))[1 mark]

(ii) 084.0916.01)5(1)5( XPXP (from the table)[1 mark]

(iii) L: Revenue Loss, then L=6X E[L] = 6E[X] = 18, Var(L) = 62Var(X) = 108 STD(L) ≈ 10.39[1 mark each]

(c) (i) ̂ is more efficient relative to ~ means that ̂ has lower variance relative to ~ . [1 mark]

(ii) Let T1 and T2 be the random variables denoting the time it takes to complete tasks one and two. Then T1~N(35,36) and T2~N(22,9).

Let X be the total time it takes to produce one gadget. Then X= T1+T2 is also normal. [0.5 mark]

T1 and T2 are independent, therefore X~N(35+22,36+9). [0.5 mark]

60 57( 60) ( ) ( 0.447) 0.5 (0 0.447)

450.5 0.1725 0.6725

P X P Z P Z P Z

 

[1 mark]

(iii) n= 360, p=0.9, XB~Binomial(360,0.9)

Since n is large, we use normal approximation XN~N(360×0.9,360×0.9×0.1) or XN~N(324,32.4)

[0.5 mark]

7852.02852.05.0)791.00(5.0

)791.0(4.32

3245.319)5.0320(

ZP

ZPZpXP [0.5 mark]

Continuity correction is not very important here since np is very large. Give the full 0.5 mark if the student realizes this and explains.

Question 2.

(a)Bookwork. Expect them to say that, provided a r.v. X has a finite mean μ and variance

σ2 then, given a random sample of size n, then the sample mean X is such that

( ) / (0,1)dn X N . Of course, they will not know this limiting argument and

can be expected to answer via the asymptotic result that 2~ ( , / )a

X N n . A good student

may comment that the result is exact if sampling from a normal distribution, but then that it is not the CLT being used. Moreover, any statistic for which an average of independent

random variables is being used will satisfy the CLT (e.g. *2 1 2

1( )

n

iiS n X

)

provided the relevant moments exist. [2 marks]

(b) Assumption – normally distributed population of expenditures (plus random sampling).

90% CI requires 125.40 1.708*(27.30/5) (using t25 distribution). This is the interval

[116.07 134.73]. [3 marks]

(c) If mean expenditure was $116 and inflation is 3% then expenditure in terms of this year’s prices is 116.1.03, which is $119.48.

Now have a simple hypothesis test to conduct

0

1

: 119.46

: 119.46

0.05

H

H

Suppose

Test statistic is (125.40 119.48) /(27.3 / 5) 1.08t .

Therefore, at any reasonable significance level, H0 cannot be rejected (c.v. = 1.708).

[2 marks]

(d) Type I error – reject hypothesis that μ=119.46 erroneously in favour of H1. Type II error, failure to reject H0 in favour of H1 when H1 is, in fact, appropriate. [1 mark]

(e) Now expect a description of a test of difference between two means. (The students are NOT given the appropriate sample variance so they cannot attempt the computations.) They would again note that last year’s prices must be indexed to the present (or vice versa), so $116 in last year’s dollars would be $119.48 and the same would then need to

be done with the sample standard deviation. Then the desire is to test 0 0 1

1 1 0

:

:

H

H

at

some appropriate significance level. Note that the degrees of freedom of the t-test is now ‘large’ so the t-distn. is not really needed.

A good answer would talk about the samples being taken to be independent and how an appropriate standard error would be computed (and, indeed, explicitly how the test statistic would be computed) but a satisfactory answer is obtained without this. (This last is worth the final 0.5 marks.) [2 marks]

(f) At large sample sizes, there is no need to consider sampling from Normal populations and the t-distribution is not necessary (though, of course, random sampling is still assumed). Reliance can be placed in the CLT to validate the use of asymptotic normality. Since it is quite likely that expenditures will have a distribution that is skew to the right, this may be advantageous. [2 marks]

Question 3.

(a) In certain circumstances, telephone polling is likely to induce some bias over the ideal situation of random sampling (e.g. sampling individual members of society when the homeless are likely to be under-represented). However, it seems unlikely that any member of the ACCI does not have a telephone associated with the business, so this is likely to be a minimal consideration in this case. Since sampling was done across states, industries and size of business, a serious attempt has been made to eliminate sources of bias. Having said that, there could be remaining ones lurking (e.g. urban/rural business split might be inappropriate). [2 marks]

(b) Sample proportion is 296 /1711 0.173p . Since sample is large, there is no need to

consider anything other than a suitable normal distribution to construct the interval estimate,

which, with a confidence coefficient of 0.99, is 0.173 2.58 0.173*0.827 /1711 . This

interval is 0.173 2.58.(0.0094) 0.173 0.024 , i.e. [0.149 0.197]. [3 marks]

(c) The width of the current interval is 0.048, so to get a width of 0.07, they need fewer firms than were in fact interviewed. If the desired target is to be met, we must solve the equation

0.035 2.58 0.173*0.827 / n for the value of n. Simple calculations show n=777.4, or

n=778 to choose a suitable integer will suffice. Hence the firm has actually interviewed just over twice as many firms as they need have done to achieve their aim. An alternative (perfectly satisfactory answer) would be to argue that, before sampling started, sampling error would be maximized if the sample proportion were to be 0.5. Under these circumstances, the

equation to be solved for required sample size becomes 0.035 2.58 0.5*0.5 / n . This

yields n=1359 as being sufficient for purpose, which is still a saving of some 25% or so.

If the confidence coefficient is changed to 0.8, the above methods go through, except that the number 2.58 should be changed to 1.28 everywhere used. To all intents and purposes the required sample sizes are divided by 4 (so, dependent on the solution proposed, sample sizes are now 191, or 335. Hence, a desire to halve the precision can save 75% of the sampling costs. [3 marks]

(d) If the specified null is to be tested against a 2-sided alternative, the rejection region yielding a

P-value of 0.05 is 1.96z . Given the information, the test statistic is computed from

( 1400) /(275 / 50)z X . Solving this equation for X shows that the sample mean must

be $1476.23 (or, of course, a similar amount less than $1400, viz. $1323.77). [2 marks]

(e) Strictly speaking, we now need to assume sampling from a normal population, but the sample

size is fairly large, so this won’t be too important. The relevant reference distribution is 249 ,

but they will not have tables for this and should, therefore, use 250 . Thus the standard means

of constructing the CI is the probability expression 2 2 2 250 50( (0.025) ( 1).275 / (0.975)) 0.95P n . Manipulation of this provides the

required 95% CI for σ2 as [49*275*275/71.42 4*.275*275/32.36] or [51884.98 114512.52]. The value of s2=275.275=75625 and it is clear that this is not in the middle of the CI. This is obviously because of asymmetry in the Chi-squared distribution (although with 50 observations this is not pronounced) and because of the inversions used to construct the interval. [2 marks]

Question 4.

(a) Bookwork. They should define both of sample covariance 1

1( 1) ( )( )

n

xy i iis n X X Y Y

and population covariance ( )( )xy x yE X Y

in the first case and their standardized counterparts /( )xy xy x yr s s s and /( )xy xy x y .

They measure the strength of bivariate linear association between the two variables. Comment should also be made that the covariance(s) are measured in the products of the

units of X and Y, whereas the latter two are unit free (and [ 1 1] ). [2 marks]

(b) Assumptions made about the error term: ( ) 0, 1,i iE X i n ; 2 2( ) , 1,iE i n ;

and ( ) 0,i jE i j . These are the important ones; some comment about them being

needed for efficient estimation when the model is correct is sufficient. They might add others

such as: 2~ (0, )i NID for inference purposes; not all { }iX take the same value; and the

model is correctly specified. If they put ( ) 0iE for the first of these, without conditioning

on X, this is ‘standard’ in texts (including Keller 8th ed.) and they should not be heavily penalized for this. (The assumption given here implies X and ε are uncorrelated.) [2 marks]

(c) (i) A=26.37; B=0.0002. [2 marks]

(ii) Intercept – price is always positive whenever capacity is, but no real interpretation. Price predicted to increase by $55 for each 1Gb increase in capacity. This corresponds to prior expectation. [1 mark]

(iii) Using t(11): [55.467-2.201*9.632 55.467+2.201*9.632] or [34.27, 76.66]. [ [1 mark]

(iv) 0 1

1 1

: 65

: 65

H

H

and the appropriate test statistic is (55.467 65) / 9.632 0.990t .

Thus the P-value is approx. 0.1611 (using N(0,1) table. Should be t(11) – reward method and any comment about this.). It is to be interpreted as the probability of observing a more extreme value in a (hypothetical) resampling if the null is, in fact true; it is the marginal significance level, i.e 0.99 would be the critical value if the test were conducted at a 16.11% significance level with a 1 tail test. [1 mark]

(v) R2=0.768 tells us that 76.8% of the variability in price is accounted for by variations in capacity. The SE of the regression tells us the estimated standard deviation of the residuals and is instrumental in setting up confidence intervals for predictions (among many other things, of course). [2 marks]

(vi) The point prediction is ˆ 6.667 55.467*4 228.54Y (A$). Though we are not told the range of values of X in the results given, this prediction is about 15% greater than the largest price of the 12 MP3 players on which the results are based. They should comment on the inadvisability of predicting far out of the range of the original data. [1 mark]

Question 5

(a) In a multiple regression model, adding more regressors to the right hand side of the model always increases R2 towards 1[1 mark], while the added variables might not be necessarily meaningful. The adjusted R2 penalizes the number of regressors [1 mark], therefore it increases as variables with great explanatory power are added and decreases as irrelevant variables are added.

(b) (i) Two-sided, t-distribution with 45 degrees of freedom: critical value for 95% confidence interval is 2.021 (from 40 degrees of freedom), then [1 mark] CI = 1.588±2.674×2.021 ≈ [-3.816,6.992] [1 mark] (ii) Let b1 be the coefficient on MBA. Then H0: b1 = 0 versus H1: b1 > 0 [0.5 mark] From the table of results t = 5.6869 One-sided test, 5% significance, t-distribution with 45 degrees of freedom: critical value = 1.684 The critical region is therefore (1.684, ∞) [1 mark] Conclusion: t-statistics is out of the critical region, therefore can reject null. Holding an MBA has positive effect on the pay of CEO.[0.5 mark] (iii) Let b3 be the coefficient on Stock Chg., then H0: b3 = 0 versus H1: b3 ≠ 0 [0.5

mark] Two-sided, t-distribution with 45 degrees of freedom (10% significance): critical value = 1.684. t-statistics is 0.77 and falls into the non-critical region. [1 mark]

Cannot reject null, the effect of stock prices on CEO pay is insignificant. [0.5 mark]

(iv) Y ≈ -1.593+190.59×0+1.588×(-5)+1.012×11+304.7×5 = 1525.1(×$1000) [1

mark]

(v) Fit to the data is good as R2=0.75. It means that 75% of variation in CEO pay is explained by the factors included.[1 mark]

(vi) Gender of CEO, women might be paid less than men. OR the industry of the company, In some industries CEO might get paid more. OR

the size of company (number of employees), CEO of larger companies might get paid more. [2 marks]