Solution
Abr Writing answered on
Apr 21 2020
Sheet1
Part 1 Ques no. Option
Year 2013 2014 2015 2016 2017 1 B
Return 5% -1% 2% 4% 7% 2 D
3 C
mean 3.40% 4 C
Mode NA 5 D
Median 4.00% 6 C
7 B
Part 2 8 B
Large cap stocks Small cap stocks 9 D
Year 1 16% 14% part 20 10 A
Year 2 -10% -8% Std e
or std dev/sqrt(n) 11 D
Year 3 12% 15% 0.25% 12 B
13 A
Std dev 14.0% 13.0% part 21 14 C
Std e
or std dev/sqrt(n) 15 B
Part 3 0.1875% 16 B
Year 1 2 3 Part 26 17 A
Ret 8% 10% 7% Std e
or 0.0277180765 18 C
UL 0.2127180765 19 D
Geometric mean 0.082425706 LL 0.1572819235 20 B
21 A
part11 P(s) Part 28 22 B
0.2 mean = n*p = 0.2*5 =1 Std e
or 0.0059950216 23 D
0.2 var = n*p*(1-p) = 0.8 UL 6.600% 24 A
0.2 LL 5.40% 25 C
0.2 26 D
0.2 Paart 30 27 A
alpha/2 0.005 28 B
Part 12 Z 2.5758293035 29 B
4C2*P^2*(1-p)^2 Size 200 30 A
0.5248 p 36% 31 D
SE 0.0339411255 32 A
Margin of e
or 0.0874265457 33 B
Part 13 UL 44.743% 34 C
S 58-49+1 LL 27.257% 35 B
T 100-0+ 36 C
Paart 31 37 C
P S/T 0.1 alpha/2 0.025 38 D
Z 1.9599639845 39 D
Part 14 Size 150 40 C
0.6914624613 p 22.00% 41 B
SE 0.0338230691 42 C
Part 15 Margin of e
or 0.0662919972 43 A
0.2511465536 UL 28.629% 44 B
LL 15.371% 45 B
Part 16 46 C
p(d) 0.2 Part 37 47 A
Mean n*p 0.8 x 350000 48 A
Variance n*p*(1-p) 0.64 x bar 300000 49 C
std dev 150000 50 A
Test stat 2.00 51 A
Part 17 Critic Val 2.326347874 52 D
10C1(0.04)^1*(0.96)^9+10c0*(0.96)^10 53 D
0.9418462343 Part 40 54 D
x 250 55 B
x bar 260 56 C
std dev 75 57 C
Test stat -0.73 58 B
Critic Val 2.326347874 59 B
60 D
Part 42
The z-statistic is computed as follows: 0.2436
z = \frac{\bar p - p_0}{\sqrt{p_0(1-p_0)/n}} = \frac{ 0.46 - 0.42 }{\sqrt{ 0.42(1- 0.42)/100}} = 0.81z=p0​(1−p0​)/n​p¯​−p0​​=0.42(1−0.42)/100​0.46−0.42​=0.81 0.81 0.002436
0.0493558507
0.8104408985
Part 45
 Sample 1  Sample 2 DifferenceÂ
Sample proportion 0.38 0.3 0.08
95% CI (asymptotic) 0.2736 - 0.4864 0.1996 - 0.4004 -0.2936
z-value 1.1
P-value 0.2855
Interpretation Not significant,
accept null hypothesis that
sample proportions are equal
n by pi n * pi >5, test ok
Part 52
Day SA SB
1 2.05% -0.08%
2 1.13% 2.23%
3 0.28% 0.45%
4 -0.15% -3.13%
5 2.75% 0.38%
First create a table of the variable scores, their ranks, differences in ranks (di) and the differences in ranks squared (di2).
Score 1 Score 2 Rank 1 Rank 2 di di2
2.05% -0.08% 4 1 3 9
1.13% 2.23% 3 5 2 4
0.28% 0.45% 2 4 2 4
-0.15% -3.13% 1 2 1 1
2.75% 0.38% 5 3 2 4
You now need to add up (sum) the differences in ranks squared (di2).
22
22
As you have no ties in your data use the following formula:
Substitute the sum of di2Â information from Step 2 and the number of scores in each variable (n) into the formula:
6 x 22
5[(5)2Â - 1]
132 0.08
120 7%
1.1 0.0056 5.3571428571
-0.1
Part 53
Co
(x,y) Covar(x.y)/stddev(x)*stddev(y)
Co
(x,y) 0.4441882553
Part 57
Co
el 0.0535714286
Wa 45%
Wb 55%
Std dev a 8%
Std dev b 7%
Var 0.29%
61-80
Part 61-65 Part Ans
Month Excess return (Xt) 61 D
Jan -0.02% 62 A
Feb -0.10% 63 A
Mar 0.10% 64 B
Apr 0.30% 65 C
May -0.50% 66 B
Jun 0.60% 67 A
Jul 0.20% 68 D
Aug 0.90% 69 C
Sep -0.20% 70 B
Oct 0.40% 71 D
Nov -0.10% 72 B
Dec 0.50% 73 C
Mean 0.17333% 74 C
Std dev 0.0038962297 75 B
SE 0.0011247446 76 A
The provided sample mean is \bar X = 0.001733X¯=0.001733 and the sample standard deviation is s = 0.00389623s=0.00389623, and the sample size is n = 12n=12. 77 C
(1)Â Null and Alternative Hypotheses 78 C
The following null and alternative hypotheses need to be tested: 79 B
Ho: \muμ = 00 80 B
Ha: \muμ > 00
This co
esponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.
(2)Â Rejection Region
Based on the information provided, the significance level is \alpha = 0.05α=0.05, and the critical value for a right-tailed test is t_c = 1.796tc​=1.796.
The rejection region for this right-tailed test is R = {t: t > 1.796}R=t:t>1.796
(3)Â Test Statistics
The t-statistic is computed as follows:
t = =1.541
(4)Â Decision about the null hypothesis
Since it is observed that t = 1.541 \le t_c = 1.796t=1.541≤tc​=1.796, it is then concluded that the null hypothesis is not rejected.
Using the P-value approach: The p-value is p = 0.0758p=0.0758, and since p = 0.0758 \ge 0.05p=0.0758≥0.05, it is concluded that the null hypothesis is not rejected.
(5)Â Conclusion
It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean \muμ is greater than 0, at the 0.05 significance level.
Confidence Interval
The 95% confidence interval is -0.001 < \mu < 0.004−0.001<μ<0.004.
Part 66-70
year Quarter Manager Fund Excess
2015 1 15% 14% 1.00%
2015 2 6.10% 5.10% 1.00%
2015 3 12% 11.05% 0.95%
2015 4 16.20% 14.10% 2.10%
2016 1 7.35% 6.35% 1.00%
2016 2 6.80% 5.70% 1.10%
2016 3 11.25% 10.05% 1.20%
2016 4 11.25% 10.25% 1.00%
2017 1 0.50% 0% 0.50%
2017 2 3% 2.05% 0.95%
2017 3 14% 11.65% 2.35%
2017 4 9% 7% 1.80%
Mean 9.35% 8.11% 1.2458%
Std dev 4.81% 4.49% 0.5437%
Std e
or 1.39% 1.30% 0.1570%
The provided sample mean is \bar X = 0.012458X¯=0.012458 and the sample standard deviation is s = 0.005437s=0.005437, and the sample size is n = 12n=12.
(1)Â Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
Ho: \muμ = 00
Ha: \muμ > 00
This co
esponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.
(2)Â Rejection Region
Based on the information provided, the significance level is \alpha = 0.01α=0.01, and the critical value for a right-tailed test is t_c = 3.106 tc​=3.106
The rejection region for this right-tailed test is R = {t: t > 2.718}R=t:t>2.718
(3)Â Test Statistics
The t-statistic is computed as follows:
t =7.937
(4)Â Decision about the null hypothesis
Since it is observed that t = 7.937 > t_c = 2.718t=7.937>tc​=2.718, it is then concluded that the null hypothesis is rejected.
Using the P-value approach: The p-value is p = 0p=0, and since p = 0 < 0.01p=0<0.01, it is concluded that the null hypothesis is rejected.
(5)Â Conclusion
It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean \muμ is greater than 0, at the 0.01 significance level.
Confidence Interval
The 99% confidence interval is 0.008 < \mu < 0.0170.008<μ<0.017.
Part 71-75
Month Stock 1 Stock2
1 2.75% -0.03%
2 0.79% 2.30%
3 1.78% 2.73%
4 1.67% -1.73%
5 -2.09% -2.65%
6 3.53% 1.57%
7 1.55% -3.68%
8 3.67% -0.27%
9 3.72% 3.11%
10 2.96% 5.91%
11 2.35% -3.76%
12 -1.91% 4.97%
Mean
t-Test: Paired Two Sample for Means
Variable 1 Variable 2
Mean 0.0173083333 0.0070583333
Variance 0.0003878681 0.0010572699
Std dev 0.0196943674 0.0325156874
Observations 12 12
Pearson Co
elation 0.0591192783
Co var 0.0038%
df 11
t Stat 0.9595022801
P(T<=t) one-tail 0.1789607378
t Critical one-tail 1.7958848187
P(T<=t) two-tail 0.3579214756
t Critical two-tail 2.2009851601
Part 76-80
Std dev A 0.0358887169
Std dev B 0.0532916504
Covar 0.01%
Co
el 0.0726771183
A B
Expected ret 1% 1.26%
x*1% + (1-x)*1.26% = 1.05%
X 0.8076923077
WtA WtB
65% 35%
Portfolio std dev 0.0309083322