Financial Analysis of Boeing Stock
VerifiedAdded on 2020/05/16
|13
|1870
|280
AI Summary
This assignment delves into a comprehensive financial analysis of Boeing's stock performance. It involves calculating various statistical measures such as mean, standard deviation, R-squared value, and confidence intervals to assess the stock's volatility, returns, and relationship with the overall market. The analysis utilizes regression statistics to determine the predictability of Boeing's stock price movements based on market trends. Furthermore, the assignment explores the normality of the error terms using the Jarque-Bera test and a normal probability plot.
Contribute Materials
Your contribution can guide someone’s learning journey. Share your
documents today.
Running Head: STATISTICS FOR FINANCE AND BUSINESS
Statistics for Finance and Business
Name of the Student
Name of the University
Author Note
Statistics for Finance and Business
Name of the Student
Name of the University
Author Note
Secure Best Marks with AI Grader
Need help grading? Try our AI Grader for instant feedback on your assignments.
1SFFBSTATISTICS FOR FINANCE AND BUSINESS
Table of Contents
Introduction......................................................................................................................................2
Answer 1..........................................................................................................................................2
Answer 2..........................................................................................................................................4
Part a............................................................................................................................................4
Part b............................................................................................................................................5
Part c............................................................................................................................................6
Answer 3..........................................................................................................................................6
Answer 4..........................................................................................................................................7
Answer 5..........................................................................................................................................8
Answer 6..........................................................................................................................................9
Answer 7..........................................................................................................................................9
Part a............................................................................................................................................9
Part b..........................................................................................................................................11
Part c..........................................................................................................................................11
Part d..........................................................................................................................................11
Answer 8........................................................................................................................................11
Answer 9........................................................................................................................................12
Table of Contents
Introduction......................................................................................................................................2
Answer 1..........................................................................................................................................2
Answer 2..........................................................................................................................................4
Part a............................................................................................................................................4
Part b............................................................................................................................................5
Part c............................................................................................................................................6
Answer 3..........................................................................................................................................6
Answer 4..........................................................................................................................................7
Answer 5..........................................................................................................................................8
Answer 6..........................................................................................................................................9
Answer 7..........................................................................................................................................9
Part a............................................................................................................................................9
Part b..........................................................................................................................................11
Part c..........................................................................................................................................11
Part d..........................................................................................................................................11
Answer 8........................................................................................................................................11
Answer 9........................................................................................................................................12
2SFFBSTATISTICS FOR FINANCE AND BUSINESS
Introduction
Analysis of performance of stocks is useful in taking a decision regarding investment in
the stocks. Stocks of an organization is evaluated on the basis of its returns and risk factors. In
the present assignment we have compared the stocks of Boeing and IBM. Based on my roll
number (19530373) the period of analysis of the stocks starts from 1/12/2010 and ends at
31/05/2016. The source of data is yahoo finance. Moreover, stock prices of S&P 500 and US TN
are also accessed for analysis of the stocks.
On the basis of analysis of the stocks an investment decision may be taken.
Answer 1
The time trend analysis of the stocks is being presented:
12/1/2010
6/1/2011
12/1/2011
6/1/2012
12/1/2012
6/1/2013
12/1/2013
6/1/2014
12/1/2014
6/1/2015
12/1/2015
0.00
500.00
1000.00
1500.00
2000.00
2500.00
Trend of S&P
Time Period
Price
Figure 1: Closing prices of S&P 500
Introduction
Analysis of performance of stocks is useful in taking a decision regarding investment in
the stocks. Stocks of an organization is evaluated on the basis of its returns and risk factors. In
the present assignment we have compared the stocks of Boeing and IBM. Based on my roll
number (19530373) the period of analysis of the stocks starts from 1/12/2010 and ends at
31/05/2016. The source of data is yahoo finance. Moreover, stock prices of S&P 500 and US TN
are also accessed for analysis of the stocks.
On the basis of analysis of the stocks an investment decision may be taken.
Answer 1
The time trend analysis of the stocks is being presented:
12/1/2010
6/1/2011
12/1/2011
6/1/2012
12/1/2012
6/1/2013
12/1/2013
6/1/2014
12/1/2014
6/1/2015
12/1/2015
0.00
500.00
1000.00
1500.00
2000.00
2500.00
Trend of S&P
Time Period
Price
Figure 1: Closing prices of S&P 500
3SFFBSTATISTICS FOR FINANCE AND BUSINESS
12/1/2010
6/1/2011
12/1/2011
6/1/2012
12/1/2012
6/1/2013
12/1/2013
6/1/2014
12/1/2014
6/1/2015
12/1/2015
0.00
20.00
40.00
60.00
80.00
100.00
120.00
140.00
160.00
Trend of Boeing
Time Period
Period
Figure 2: Closing stock prices of Boeing
12/1/2010
6/1/2011
12/1/2011
6/1/2012
12/1/2012
6/1/2013
12/1/2013
6/1/2014
12/1/2014
6/1/2015
12/1/2015
0.00
50.00
100.00
150.00
200.00
250.00
IBM
Time Period
Price
Figure 3: Closing stock prices of IBM
The chart on the trend of S&P 500 and Boeing shows that there has been a growth during
the period. On the other hand, the chart on the trend of IBM shows that the stock prices rose
during the period 2010 to 2012 and post 2012 there has been a fall in the prices of the stocks.
12/1/2010
6/1/2011
12/1/2011
6/1/2012
12/1/2012
6/1/2013
12/1/2013
6/1/2014
12/1/2014
6/1/2015
12/1/2015
0.00
20.00
40.00
60.00
80.00
100.00
120.00
140.00
160.00
Trend of Boeing
Time Period
Period
Figure 2: Closing stock prices of Boeing
12/1/2010
6/1/2011
12/1/2011
6/1/2012
12/1/2012
6/1/2013
12/1/2013
6/1/2014
12/1/2014
6/1/2015
12/1/2015
0.00
50.00
100.00
150.00
200.00
250.00
IBM
Time Period
Price
Figure 3: Closing stock prices of IBM
The chart on the trend of S&P 500 and Boeing shows that there has been a growth during
the period. On the other hand, the chart on the trend of IBM shows that the stock prices rose
during the period 2010 to 2012 and post 2012 there has been a fall in the prices of the stocks.
Paraphrase This Document
Need a fresh take? Get an instant paraphrase of this document with our AI Paraphraser
4SFFBSTATISTICS FOR FINANCE AND BUSINESS
Answer 2
Part a
S&P 500 BA IBM
2.2393 6.2660 9.8798
3.1457 3.5766 -0.0741
-0.1048 2.6314 0.7324
2.8097 7.6154 4.5025
-1.3593 -2.2179 -0.9720
-1.8426 -5.3975 1.5390
-2.1708 -4.7932 5.8307
-5.8468 -5.2586 -5.6211
-7.4467 -9.9792 1.7072
10.2307 8.3659 5.4311
-0.5072 4.3136 1.8088
0.8497 6.5639 -2.2159
4.2660 1.1252 4.6332
3.9787 1.0327 2.1217
3.0851 -0.7769 5.8826
-0.7526 3.2152 -0.7553
-6.4699 -9.8296 -7.0933
3.8793 6.5203 1.3798
1.2519 -0.5263 0.2043
1.9571 -3.4550 -0.5783
2.3947 -2.5533 6.2660
-1.9988 1.1997 -6.4304
0.2843 5.3080 -2.3194
0.7043 1.4435 0.7756
4.9198 -1.9970 5.8402
1.1000 4.0199 -1.1092
3.5355 11.0096 6.0241
1.7924 6.2753 -5.1762
2.0550 7.9967 2.6697
-1.5113 3.3955 -8.4785
4.8278 2.5635 2.0355
-3.1798 -1.1291 -6.7716
2.9316 12.2817 1.5839
4.3630 10.4935 -3.2770
2.7663 2.8331 0.2619
2.3289 1.6548 4.2975
-3.6231 -8.5860 -5.9812
4.2213 2.8800 4.6934
Answer 2
Part a
S&P 500 BA IBM
2.2393 6.2660 9.8798
3.1457 3.5766 -0.0741
-0.1048 2.6314 0.7324
2.8097 7.6154 4.5025
-1.3593 -2.2179 -0.9720
-1.8426 -5.3975 1.5390
-2.1708 -4.7932 5.8307
-5.8468 -5.2586 -5.6211
-7.4467 -9.9792 1.7072
10.2307 8.3659 5.4311
-0.5072 4.3136 1.8088
0.8497 6.5639 -2.2159
4.2660 1.1252 4.6332
3.9787 1.0327 2.1217
3.0851 -0.7769 5.8826
-0.7526 3.2152 -0.7553
-6.4699 -9.8296 -7.0933
3.8793 6.5203 1.3798
1.2519 -0.5263 0.2043
1.9571 -3.4550 -0.5783
2.3947 -2.5533 6.2660
-1.9988 1.1997 -6.4304
0.2843 5.3080 -2.3194
0.7043 1.4435 0.7756
4.9198 -1.9970 5.8402
1.1000 4.0199 -1.1092
3.5355 11.0096 6.0241
1.7924 6.2753 -5.1762
2.0550 7.9967 2.6697
-1.5113 3.3955 -8.4785
4.8278 2.5635 2.0355
-3.1798 -1.1291 -6.7716
2.9316 12.2817 1.5839
4.3630 10.4935 -3.2770
2.7663 2.8331 0.2619
2.3289 1.6548 4.2975
-3.6231 -8.5860 -5.9812
4.2213 2.8800 4.6934
5SFFBSTATISTICS FOR FINANCE AND BUSINESS
0.6908 -2.6966 3.8770
0.6182 2.7741 2.0466
2.0812 4.7157 -6.3619
1.8879 -6.1128 -1.6903
-1.5195 -5.4513 5.5787
3.6964 5.1127 0.3282
-1.5635 0.4564 -1.2928
2.2936 -1.9581 -14.3826
2.4237 7.2929 -1.3657
-0.4197 -3.3142 -1.0725
-3.1533 11.1902 -4.5458
5.3439 3.7004 5.4764
-1.7549 -0.5118 -0.8932
0.8485 -4.5949 6.5064
1.0437 -1.9870 -0.9621
-2.1236 -1.2892 -4.2075
1.9550 3.8536 -0.4128
-6.4625 -9.8242 -9.1066
-2.6799 0.2064 -1.9942
7.9719 12.2870 -3.4313
0.0505 -1.7853 -0.4723
-1.7686 -0.5930 -1.2995
-5.2068 -18.5328 -9.7864
-0.4137 -1.6366 4.8794
6.3905 7.1506 14.4829
0.2696 6.0078 -3.7060
1.5208 -6.6321 5.2067
Returns during the period 1/12/2010 to 31/05/2016
Part b
Table 2: Summary Statistics for the returns on stock prices of Boeing and IBM
Statistics BA IBM
Mean 1.0140 0.0715
Standard Error 0.7427 0.6267
Median 1.1997 -0.0741
Mode #N/A #N/A
Standard Deviation 5.9877 5.0523
Sample Variance 35.8520 25.5255
Kurtosis 0.6987 0.6553
Skewness -0.4699 -0.1368
0.6908 -2.6966 3.8770
0.6182 2.7741 2.0466
2.0812 4.7157 -6.3619
1.8879 -6.1128 -1.6903
-1.5195 -5.4513 5.5787
3.6964 5.1127 0.3282
-1.5635 0.4564 -1.2928
2.2936 -1.9581 -14.3826
2.4237 7.2929 -1.3657
-0.4197 -3.3142 -1.0725
-3.1533 11.1902 -4.5458
5.3439 3.7004 5.4764
-1.7549 -0.5118 -0.8932
0.8485 -4.5949 6.5064
1.0437 -1.9870 -0.9621
-2.1236 -1.2892 -4.2075
1.9550 3.8536 -0.4128
-6.4625 -9.8242 -9.1066
-2.6799 0.2064 -1.9942
7.9719 12.2870 -3.4313
0.0505 -1.7853 -0.4723
-1.7686 -0.5930 -1.2995
-5.2068 -18.5328 -9.7864
-0.4137 -1.6366 4.8794
6.3905 7.1506 14.4829
0.2696 6.0078 -3.7060
1.5208 -6.6321 5.2067
Returns during the period 1/12/2010 to 31/05/2016
Part b
Table 2: Summary Statistics for the returns on stock prices of Boeing and IBM
Statistics BA IBM
Mean 1.0140 0.0715
Standard Error 0.7427 0.6267
Median 1.1997 -0.0741
Mode #N/A #N/A
Standard Deviation 5.9877 5.0523
Sample Variance 35.8520 25.5255
Kurtosis 0.6987 0.6553
Skewness -0.4699 -0.1368
6SFFBSTATISTICS FOR FINANCE AND BUSINESS
Range 30.8197 28.8656
Minimum -18.5328 -14.3826
Maximum 12.2870 14.4829
Sum 65.9092 4.6464
Count 65 65
The above table presents the summary statistics for the stocks of Boeing and IBM. From
the above summary it can be seen that the return on the prices of Boeing is higher than IBM’s.
On the other hand, the risks involved with Boeing is higher than IBM’s.
Part c
BA IBM
Jarque-Bera Statistics 16.7353 15.0915
p-value 0.0002 0.0005
Jarque- Bera test is used to test the normality of data.
Jarque−Bera Statistics=n∗[ skewnes s2
6 + ( kurtosis−3 ) 2
24 ]
Jarque-Bera statistics follows a Chi-square distribution with 2 degrees of Freedom.
From the above table it is found that the p-value for both the stocks is less than 0.05, level
of significance. Since the p-value < 0.05, hence it can be interpreted that the returns of both the
stocks are normally distributed.
The presence of normality is a basic requirement of statistical analysis. When a data is
normally distributed then further analysis on the data can be carried out.
Answer 3
One-sample t-test is used to check if the average return on the stock of Boeing is at least
3%. The one-sample t-test is used to evaluate if the average of a data is more than, equal to or
less than a particular value.
Null hypothesis: The average return of the stock prices of Boeing is less than 3%
H0 : μ<3 %
Alternate hypothesis: The average return of the stock prices of Boeing is at least 3%
H A : μ ≥3 %
Range 30.8197 28.8656
Minimum -18.5328 -14.3826
Maximum 12.2870 14.4829
Sum 65.9092 4.6464
Count 65 65
The above table presents the summary statistics for the stocks of Boeing and IBM. From
the above summary it can be seen that the return on the prices of Boeing is higher than IBM’s.
On the other hand, the risks involved with Boeing is higher than IBM’s.
Part c
BA IBM
Jarque-Bera Statistics 16.7353 15.0915
p-value 0.0002 0.0005
Jarque- Bera test is used to test the normality of data.
Jarque−Bera Statistics=n∗[ skewnes s2
6 + ( kurtosis−3 ) 2
24 ]
Jarque-Bera statistics follows a Chi-square distribution with 2 degrees of Freedom.
From the above table it is found that the p-value for both the stocks is less than 0.05, level
of significance. Since the p-value < 0.05, hence it can be interpreted that the returns of both the
stocks are normally distributed.
The presence of normality is a basic requirement of statistical analysis. When a data is
normally distributed then further analysis on the data can be carried out.
Answer 3
One-sample t-test is used to check if the average return on the stock of Boeing is at least
3%. The one-sample t-test is used to evaluate if the average of a data is more than, equal to or
less than a particular value.
Null hypothesis: The average return of the stock prices of Boeing is less than 3%
H0 : μ<3 %
Alternate hypothesis: The average return of the stock prices of Boeing is at least 3%
H A : μ ≥3 %
Secure Best Marks with AI Grader
Need help grading? Try our AI Grader for instant feedback on your assignments.
7SFFBSTATISTICS FOR FINANCE AND BUSINESS
Thus if the p-value for the test is less than 0.05 we reject Null hypothesis else accept Null
hypothesis.
Statistics Value
Mean 1.0140
Standard Deviation 5.9877
Count 65
Standard Error 0.7427
Hypothesized Mean 3% 0.05
Tails 1
Df 64
t stat 1.3249
p-value 0.0950
t crit 0.0630
Test summary no
The t-statistics is calculated as : 1.0140−0.03
0.7427 =1 . 3249
The p-value for the test statistics = 0.0950.
Since p-value is more than level of significance (0.0950 > 0.05) hence we do not reject
Null hypothesis.
Hence, the average return of the stock prices of Boeing is less than 3%
Answer 4
F-test is used to compare the risks associated with the stocks.
Null Hypothesis: The risk on stock prices of both the stocks are similar
H0 :σ Boeing
2 =σIBM
2
Alternate Hypothesis: The risks on stock prices of both the stocks are similar
H0 :σ Boeing
2 ≠ σIBM
2
Thus if the p-value for the test is less than 0.05 we reject Null hypothesis else accept Null
hypothesis.
F-Test Two-Sample for Variances
BA IBM
Mean 1.0140 0.0715
Variance 35.8520 25.5255
Thus if the p-value for the test is less than 0.05 we reject Null hypothesis else accept Null
hypothesis.
Statistics Value
Mean 1.0140
Standard Deviation 5.9877
Count 65
Standard Error 0.7427
Hypothesized Mean 3% 0.05
Tails 1
Df 64
t stat 1.3249
p-value 0.0950
t crit 0.0630
Test summary no
The t-statistics is calculated as : 1.0140−0.03
0.7427 =1 . 3249
The p-value for the test statistics = 0.0950.
Since p-value is more than level of significance (0.0950 > 0.05) hence we do not reject
Null hypothesis.
Hence, the average return of the stock prices of Boeing is less than 3%
Answer 4
F-test is used to compare the risks associated with the stocks.
Null Hypothesis: The risk on stock prices of both the stocks are similar
H0 :σ Boeing
2 =σIBM
2
Alternate Hypothesis: The risks on stock prices of both the stocks are similar
H0 :σ Boeing
2 ≠ σIBM
2
Thus if the p-value for the test is less than 0.05 we reject Null hypothesis else accept Null
hypothesis.
F-Test Two-Sample for Variances
BA IBM
Mean 1.0140 0.0715
Variance 35.8520 25.5255
8SFFBSTATISTICS FOR FINANCE AND BUSINESS
Observations 65 65
Df 64 64
F 1.4046
P(F<=f) one-tail 0.0884
F Critical one-tail 1.5133
The p-value for the test statistics = 0.0884.
Since p-value is more than level of significance (0.0884 > 0.05) hence we do not reject
Null hypothesis.
Thus the risks associated on both the stocks is similar.
Answer 5
Independent sample t-test is used to compare the average return of stocks.
Null Hypothesis: The average return on the prices of Boeing is equal to the average
return on the stock prices of IBM.
H0 : μBoeing=μIBM
Alternate Hypothesis: The average return on the prices of Boeing is not equal to the
average return on the stock prices of IBM.
H A : μBoeing≠ μIBM
Thus if the p-value for the test is less than 0.05 we reject Null hypothesis else accept Null
hypothesis.
t-Test: Two-Sample Assuming Equal Variances
BA IBM
Mean 1.0140 0.0715
Variance 35.8520 25.5255
Observations 65 65
Pooled Variance 30.6887
Hypothesized Mean Difference 0
df 128
t Stat 0.9699
P(T<=t) one-tail 0.1670
t Critical one-tail 1.6568
P(T<=t) two-tail 0.3339
t Critical two-tail 1.9787
The p-value for the test statistics = 0.3339.
Observations 65 65
Df 64 64
F 1.4046
P(F<=f) one-tail 0.0884
F Critical one-tail 1.5133
The p-value for the test statistics = 0.0884.
Since p-value is more than level of significance (0.0884 > 0.05) hence we do not reject
Null hypothesis.
Thus the risks associated on both the stocks is similar.
Answer 5
Independent sample t-test is used to compare the average return of stocks.
Null Hypothesis: The average return on the prices of Boeing is equal to the average
return on the stock prices of IBM.
H0 : μBoeing=μIBM
Alternate Hypothesis: The average return on the prices of Boeing is not equal to the
average return on the stock prices of IBM.
H A : μBoeing≠ μIBM
Thus if the p-value for the test is less than 0.05 we reject Null hypothesis else accept Null
hypothesis.
t-Test: Two-Sample Assuming Equal Variances
BA IBM
Mean 1.0140 0.0715
Variance 35.8520 25.5255
Observations 65 65
Pooled Variance 30.6887
Hypothesized Mean Difference 0
df 128
t Stat 0.9699
P(T<=t) one-tail 0.1670
t Critical one-tail 1.6568
P(T<=t) two-tail 0.3339
t Critical two-tail 1.9787
The p-value for the test statistics = 0.3339.
9SFFBSTATISTICS FOR FINANCE AND BUSINESS
Since p-value is more than level of significance (0.3339 > 0.05) hence we do not reject
Null hypothesis.
Thus the average return on the stocks of Boeing and IBM are equal.
However, since the average return of Boeing is higher than IBM’s hence further analysis
of the stocks of Boeing are done.
Answer 6
The excess return and excess return for Boeing is calculated.
Answer 7
Part a
Excess Return Excess Market Return CAPM
yt xt
2.8880 -1.1387 3.6343
0.1626 -0.2683 0.5838
-0.8226 -3.5588 -0.5190
4.3194 -0.4863 5.2366
-5.2679 -4.4093 -5.4947
-8.5555 -5.0006 -9.1745
-7.5982 -4.9758 -8.1030
-7.4766 -8.0648 -7.9669
-11.9032 -9.3707 -12.9217
6.1909 8.0557 7.3314
2.2456 -2.5752 2.9153
4.6929 -1.0213 5.6546
-0.6738 2.4670 -0.3524
-0.9443 2.0017 -0.6553
-2.9929 0.8691 -2.9482
1.3002 -2.6676 1.8571
-11.4106 -8.0509 -12.3704
4.8613 2.2203 5.8431
-2.0183 -0.2401 -1.8573
-5.0170 0.3951 -5.2139
-4.1903 0.7577 -4.2886
-0.4863 -3.6848 -0.1426
3.7020 -1.3217 4.5455
-0.3125 -1.0517 0.0520
-3.9820 2.9348 -4.0553
Since p-value is more than level of significance (0.3339 > 0.05) hence we do not reject
Null hypothesis.
Thus the average return on the stocks of Boeing and IBM are equal.
However, since the average return of Boeing is higher than IBM’s hence further analysis
of the stocks of Boeing are done.
Answer 6
The excess return and excess return for Boeing is calculated.
Answer 7
Part a
Excess Return Excess Market Return CAPM
yt xt
2.8880 -1.1387 3.6343
0.1626 -0.2683 0.5838
-0.8226 -3.5588 -0.5190
4.3194 -0.4863 5.2366
-5.2679 -4.4093 -5.4947
-8.5555 -5.0006 -9.1745
-7.5982 -4.9758 -8.1030
-7.4766 -8.0648 -7.9669
-11.9032 -9.3707 -12.9217
6.1909 8.0557 7.3314
2.2456 -2.5752 2.9153
4.6929 -1.0213 5.6546
-0.6738 2.4670 -0.3524
-0.9443 2.0017 -0.6553
-2.9929 0.8691 -2.9482
1.3002 -2.6676 1.8571
-11.4106 -8.0509 -12.3704
4.8613 2.2203 5.8431
-2.0183 -0.2401 -1.8573
-5.0170 0.3951 -5.2139
-4.1903 0.7577 -4.2886
-0.4863 -3.6848 -0.1426
3.7020 -1.3217 4.5455
-0.3125 -1.0517 0.0520
-3.9820 2.9348 -4.0553
Paraphrase This Document
Need a fresh take? Get an instant paraphrase of this document with our AI Paraphraser
10SFFBSTATISTICS FOR FINANCE AND BUSINESS
2.1319 -0.7880 2.7880
9.1576 1.6835 10.6520
4.6003 0.1174 5.5510
5.8327 -0.1090 6.9304
0.9175 -3.9893 1.4288
-0.0295 2.2348 0.3687
-3.8781 -5.9288 -3.9391
9.6667 0.3166 11.2219
7.9515 1.8210 9.3020
0.0921 0.0253 0.5048
-1.3712 -0.6971 -1.1331
-11.2540 -6.2911 -12.1950
0.2220 1.5633 0.6503
-5.4196 -2.0322 -5.6645
0.1261 -2.0298 0.5430
2.2587 -0.3758 2.9300
-8.6288 -0.6281 -9.2566
-8.0073 -4.0755 -8.5609
2.7697 1.3534 3.5020
-2.0516 -4.0715 -1.8947
-4.2931 -0.0414 -4.4036
5.0989 0.2297 6.1091
-5.4842 -2.5897 -5.7368
9.5152 -4.8283 11.0522
1.6984 3.3419 2.3028
-2.4458 -3.6889 -2.3358
-6.6409 -1.1975 -7.0315
-4.0820 -1.0513 -4.1672
-3.6242 -4.4586 -3.6549
1.6486 -0.2500 2.2470
-12.0242 -8.6625 -13.0571
-1.8536 -4.7399 -1.6730
10.1360 5.8209 11.7471
-4.0033 -2.1675 -4.0792
-2.8620 -4.0376 -2.8018
-20.4638 -7.1378 -22.5037
-3.3766 -2.1537 -3.3777
5.3646 4.6045 6.4064
4.1888 -1.5494 5.0903
-8.4661 -0.3132 -9.0744
CAPM for Boeing is calculated from: rt −r f ,t =β0 + β1 (r M ,t −rf ,t )
2.1319 -0.7880 2.7880
9.1576 1.6835 10.6520
4.6003 0.1174 5.5510
5.8327 -0.1090 6.9304
0.9175 -3.9893 1.4288
-0.0295 2.2348 0.3687
-3.8781 -5.9288 -3.9391
9.6667 0.3166 11.2219
7.9515 1.8210 9.3020
0.0921 0.0253 0.5048
-1.3712 -0.6971 -1.1331
-11.2540 -6.2911 -12.1950
0.2220 1.5633 0.6503
-5.4196 -2.0322 -5.6645
0.1261 -2.0298 0.5430
2.2587 -0.3758 2.9300
-8.6288 -0.6281 -9.2566
-8.0073 -4.0755 -8.5609
2.7697 1.3534 3.5020
-2.0516 -4.0715 -1.8947
-4.2931 -0.0414 -4.4036
5.0989 0.2297 6.1091
-5.4842 -2.5897 -5.7368
9.5152 -4.8283 11.0522
1.6984 3.3419 2.3028
-2.4458 -3.6889 -2.3358
-6.6409 -1.1975 -7.0315
-4.0820 -1.0513 -4.1672
-3.6242 -4.4586 -3.6549
1.6486 -0.2500 2.2470
-12.0242 -8.6625 -13.0571
-1.8536 -4.7399 -1.6730
10.1360 5.8209 11.7471
-4.0033 -2.1675 -4.0792
-2.8620 -4.0376 -2.8018
-20.4638 -7.1378 -22.5037
-3.3766 -2.1537 -3.3777
5.3646 4.6045 6.4064
4.1888 -1.5494 5.0903
-8.4661 -0.3132 -9.0744
CAPM for Boeing is calculated from: rt −r f ,t =β0 + β1 (r M ,t −rf ,t )
11SFFBSTATISTICS FOR FINANCE AND BUSINESS
Part b
Coefficients Standard Error t Stat P-value
Intercept 0.4018 0.6294 0.6383 0.5256
xt 1.1193 0.1710 6.5461 0.0000
The coefficient of the stocks of Boeing is 1.1193. Thus the volatility of the prices of the
stock is 111.93%. Since it is more volatile hence it is riskier and thus less profitable.
Part c
Regression Statistics
Multiple R 0.6363
R Square 0.4048
Adjusted R Square 0.3954
Standard Error 4.6577
Observations 65
The R2 for Boeing is 0.4048. R2 is the predictability between Boeing and the market.
Thus 40.48% of the market is predictable from the prices of the stocks.
Part d
At 0.05 level of significance and 63 degrees of freedom the critical value of is 1.9983.
Coefficient = 1.1193
Standard Error = 0.1710
Thus the confidence interval: 1.1193 ±1.9983∗0.1 710 0.7776, 1.4610
Thus, 95% confidence interval for the coefficient of Boeing is 0.7776, 1.4610.
Answer 8
Statistics Values
Mean 1.0140
Standard Deviation 5.9877
Count 65
Standard Error 0.7427
Significance Level 0.05
z-value 1.9600
Margin of Error 1.4556
Lower Limit -0.4416
Part b
Coefficients Standard Error t Stat P-value
Intercept 0.4018 0.6294 0.6383 0.5256
xt 1.1193 0.1710 6.5461 0.0000
The coefficient of the stocks of Boeing is 1.1193. Thus the volatility of the prices of the
stock is 111.93%. Since it is more volatile hence it is riskier and thus less profitable.
Part c
Regression Statistics
Multiple R 0.6363
R Square 0.4048
Adjusted R Square 0.3954
Standard Error 4.6577
Observations 65
The R2 for Boeing is 0.4048. R2 is the predictability between Boeing and the market.
Thus 40.48% of the market is predictable from the prices of the stocks.
Part d
At 0.05 level of significance and 63 degrees of freedom the critical value of is 1.9983.
Coefficient = 1.1193
Standard Error = 0.1710
Thus the confidence interval: 1.1193 ±1.9983∗0.1 710 0.7776, 1.4610
Thus, 95% confidence interval for the coefficient of Boeing is 0.7776, 1.4610.
Answer 8
Statistics Values
Mean 1.0140
Standard Deviation 5.9877
Count 65
Standard Error 0.7427
Significance Level 0.05
z-value 1.9600
Margin of Error 1.4556
Lower Limit -0.4416
12SFFBSTATISTICS FOR FINANCE AND BUSINESS
Upper Limit 2.4696
The lower and upper 95% confidence interval for the return on the stock prices of Boeing
is -0.4416, 2.4696. Thus, taking a risk of 5% it can be said that the returns on the prices of
Boeing vary from -0.4416, 2.4696. This can be interpreted as, that if the stock prices of Boeing
are taken for any other period then 95% of the return values would lie within the interval of -
0.4416, 2.4696.
Answer 9
Statistics Value
Count 65
Skewness 0.0927
Kurtosis 0.8889
Jarque-Bera 12.1631
p-value 0.0023
The Jarque-Bera test for is used to check the normality of the error values.
From the above table it is found that the p-value for the error is less than 0.05, level of
significance. Since the p-value < 0.05, hence it can be interpreted that the error (residuals) of the
stocks of Boeing are normally distributed.
The normal probability plot also shows that the error (residuals) are normally distributed.
0 20 40 60 80 100 120
-30
-20
-10
0
10
20
Normal Probability Plot
Sample Percentile
yt
Figure 4: Normal Probability plot for Boeing
Upper Limit 2.4696
The lower and upper 95% confidence interval for the return on the stock prices of Boeing
is -0.4416, 2.4696. Thus, taking a risk of 5% it can be said that the returns on the prices of
Boeing vary from -0.4416, 2.4696. This can be interpreted as, that if the stock prices of Boeing
are taken for any other period then 95% of the return values would lie within the interval of -
0.4416, 2.4696.
Answer 9
Statistics Value
Count 65
Skewness 0.0927
Kurtosis 0.8889
Jarque-Bera 12.1631
p-value 0.0023
The Jarque-Bera test for is used to check the normality of the error values.
From the above table it is found that the p-value for the error is less than 0.05, level of
significance. Since the p-value < 0.05, hence it can be interpreted that the error (residuals) of the
stocks of Boeing are normally distributed.
The normal probability plot also shows that the error (residuals) are normally distributed.
0 20 40 60 80 100 120
-30
-20
-10
0
10
20
Normal Probability Plot
Sample Percentile
yt
Figure 4: Normal Probability plot for Boeing
1 out of 13
Related Documents
Your All-in-One AI-Powered Toolkit for Academic Success.
+13062052269
info@desklib.com
Available 24*7 on WhatsApp / Email
Unlock your academic potential
© 2024 | Zucol Services PVT LTD | All rights reserved.