Homework SourseWILL GIVE THUMBS UP THANK YOU D The following 8 questions ar
WILL GIVE THUMBS UP!!! THANK YOU :D
The following 8 questions are based on this information.
The table below shows the annual return data for 10 firms in the gold industry and 10 firms in the oil industry. Here we assume that these two samples are from tow normal populations.
Annual returns (in percent)
Gold
Oil
6
-3
15
15
19
28
26
18
2
32
16
31
31
15
14
12
15
10
16
15
Let (1) be the mean return for the gold industry and (2) be the mean return for the oil industry. We wish to test if the average returns in the two industries are different? Here we assume that these two samples are from two independent normal populations.
1.
The competing hypotheses should be formulated as
Select one:
a. H(0): (1)-(2)0 versus H(a):(1)-(2)=0
b. H(0): (1)-(2)=0 versus H(a):(1)-(2)0
c. H(0): (1)-(2)<0 versus H(a):(1)-(2)0
2.
The test statistics value is
Select one:
a. -0.3023
b. 0
c. 16
d. 17.3
3.
The p-value of the test is
Select one:
a. 0.3830
b. 0.7661
c. 1.7396
d. 2.1098
4.
At =0.1,
Select one:
a. We can reject H(0) in favor of H(a)
b. We cannot reject H(0)
c. We can conclude that average return for gold industry is less than the average return for the oil industry
5.
What is the standard error of (x(1)bar- x(2)bar?
Select one:
a. 4.30
b. 3.32
c. 16
d. 17.3
6.
Suppose you would like to construct the 90% confidence interval for (1)-(2). The critical value, t(/2) is
Select one:
a. 2.11
b. 1.96
c. 1.64
d. 1.74
7. The 90% confidence interval for (1)-(2) is
Select one:
a. -1.3 ± 6.44
b. -1.3 ±5.78
c. -1.3 ±7.48
d. -1.3±9.07
8.Using the 90% confidence interval estimate, we do not reject H_0 because
Select one:
a. 1 is included in the interval
b. -1 is included in the interval
c. 2 is included in the interval
d. 0 is included in the interval
| Gold | Oil |
| 6 | -3 |
| 15 | 15 |
| 19 | 28 |
| 26 | 18 |
| 2 | 32 |
| 16 | 31 |
| 31 | 15 |
| 14 | 12 |
| 15 | 10 |
| 16 | 15 |
Solution
Set Up Hypothesis
Null Hypothesis , There Is No-Significance between them Ho: u1 = u2
Alternate Hypothesis, There Is Significance between them - H1: u1 != u2
Test Statistic
X(Mean)=16
Standard Deviation(s.d1)=8.406 ; Number(n1)=10
Y(Mean)=17.3
Standard Deviation(s.d2)=10.688; Number(n2)=10
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =16-17.3/Sqrt((70.66084/10)+(114.23334/10))
to =-0.3023
| to | =0.3023
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 9 d.f is 2.262
We got |to| = 0.30233 & | t | = 2.262
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -0.3023 ) = 0.769
Hence Value of P0.05 < 0.769,Here We Do not Reject Ho
ANS:
b. H(0): (1)-(2)=0 versus H(a):(1)-(2)0
a. -0.3023
Ha : ( P != -0.3023 ) = 0.7661
b. We cannot reject H(0)
