MO S&P -5.7 -9 1.2 -5.5 4.1 -0.4 3.2 6.4 7.3 0.5 7.5 6.5 18.6 7.1 3.7 1.7 -1.8 0
ID: 3320367 • Letter: M
Question
MO S&P -5.7 -9 1.2 -5.5 4.1 -0.4 3.2 6.4 7.3 0.5 7.5 6.5 18.6 7.1 3.7 1.7 -1.8 0.9 2.4 4.3 -6.5 -5 6.7 5.1 9.4 2.3 -2 -2.1 -2.8 1.3 -3.4 -4 19.2 9.5 -4.8 -0.2 0.5 1.2 -0.6 -2.5 2.8 3.5 -0.5 0.5 -4.5 -2.1 8.7 4 2.7 -2.1 4.1 0.6 -10.3 0.3 4.8 3.4 -2.3 0.6 -3.1 1.5 -10.2 1.4 -3.7 1.5 -26.6 -1.8 7.2 2.7 -2.9 -0.3 -2.3 0.1 3.5 3.8 -4.6 -1.3 17.2 2.1 4.2 -1 0.5 0.2 8.3 4.4 -7.1 -2.7 -8.4 -5 7.7 2 -9.6 1.6 6 -2.9 6.8 3.8 10.9 4.1 1.6 -2.9 0.2 2.2 -2.4 -3.7 -2.4 0 3.9 4 1.7 3.9 9 2.5 3.6 3.4 7.6 4 3.2 1.9 -3.7 3.3 4.2 0.3 13.2 3.8 0.9 0 4.2 4.4 4 0.7 2.8 3.4 6.7 0.9 -10.4 0.5 2.7 1.5 10.3 2.5 5.7 0 0.6 -4.4 -14.2 2.1 1.3 5.2 2.9 2.8 11.8 7.6 10.6 -3.1 5.2 6.2 13.8 0.8 -14.7 -4.5 3.5 6 11.7 6.1 1.3 5.8 TASK 2: Using week 4 and week 5 inference techniques generate confidence intervals for both MO and S&P variables and perform hypothesis testing at the 95% confidence level to determine if the returns are statistically different from 0 (zero) for the covered period. Use the Student's t distribution as we do not know the true population standard deviation. Compute a 95% confidence interval for the difference in returns between the two equities. Assume MO - S&P for the confidence interval. Perform a hypothesis test for MO - S&P to determine the p-value (observed level of significance) and compare it to the desired level of significance (alpha) of 0.05 at the 95% confidence level. Compare the conclusions based on the confidence interval to that of the hypothesis test. (Hint: is 0 inside the confidence interval or not?)
Explanation / Answer
Answer
1. Confidence Interval :
Xbar -t/2 *(s/n) <µ< Xbar +t/2 *(s/n)
MO
S&P
MO-S&P
xbar
1.878313
1.54759
xbar
0.330723
std devn(s)
7.553937
3.274559
std devn(s)
6.659052
n
83
83
n
83
df
82
82
df
82
t0.025
1.96
1.96
t0.025
1.96
s/sqrt(n)
0.829152
0.35943
s/sqrt(n)
0.730926
t*(s/sqrt(n)
1.625138
0.704482
t*(s/sqrt(n)
1.432615
CI 1
0.253175
0.843108
CI 1
-1.10189
CI2
3.503452
2.252072
CI2
1.763338
Hence Confidence Interval of Mean
for MO is 1.878313 +/- 1.625 => 0.25317 - 3.50
for S&M 1.54759 +/- 0.704482 => 0.84311 - 2.25207
Hence for both the cases 0 is not included in confidence interval
2. For MO -S&P
MO – S&P
So confidence interval is 0.330723 +/- 1.43 => -1.10189 -- 1.763338
So 0 is included within confidence interval .
Also for test of Hypothesis tcal =( xbar-0)/( sdiff/n) = 0.330723/0.730926=0.45247 which is less than t0.025
So Ho is accepted , I,e difference of two equities is not significantly different from zero
calculation summary
MO
S&P
MO-S&P
xbar
1.878313
1.54759
xbar
0.330723
std devn(s)
7.553937
3.274559
std devn(s)
6.659052
n
83
83
n
83
df
82
82
df
82
t0.025
1.96
1.96
t0.025
1.96
s/sqrt(n)
0.829152
0.35943
s/sqrt(n)
0.730926
t*(s/sqrt(n)
1.625138
0.704482
t*(s/sqrt(n)
1.432615
CI 1
0.253175
0.843108
CI 1
-1.10189
CI2
3.503452
2.252072
CI2
1.763338
Note : Data is not given in a clear format. Hence there is likely comutation error due to data copy error. Please recalculate if there is mistake using correct data and following same method explained in answer
MO
S&P
MO-S&P
xbar
1.878313
1.54759
xbar
0.330723
std devn(s)
7.553937
3.274559
std devn(s)
6.659052
n
83
83
n
83
df
82
82
df
82
t0.025
1.96
1.96
t0.025
1.96
s/sqrt(n)
0.829152
0.35943
s/sqrt(n)
0.730926
t*(s/sqrt(n)
1.625138
0.704482
t*(s/sqrt(n)
1.432615
CI 1
0.253175
0.843108
CI 1
-1.10189
CI2
3.503452
2.252072
CI2
1.763338
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