Question

A sample containing years to maturity and yield for 40 corporate bonds is contained in the Excel Online file below. Construct a spreadsheet to answer the following questions.

Company Ticker	Years to Maturity	Yield
HSBC	24.75	0.865
GS	4.00	4.351
C	19.25	3.568
MS	27.75	5.084
C	22.50	1.641
TOTAL	17.50	2.257
MS	27.75	3.194
WFC	25.50	4.953
TOTAL	9.25	1.487
TOTAL	21.00	4.449
BAC	12.25	6.662
RABOBK	21.00	1.931
GS	29.00	3.648
AXP	12.25	1.623
MTNA	19.00	2.819
MTNA	4.25	3.390
JPM	10.50	6.067
GE	8.00	3.521
LNC	16.50	6.462
BAC	24.50	2.419
FCX	25.50	6.528
GS	29.50	3.202
RABOBK	29.00	7.138
GE	12.75	1.816
HCN	13.25	6.930
GE	21.75	3.658
VOD	21.50	4.763
NEM	6.50	5.419
GE	2.25	6.205
C	22.25	5.422
SHBASS	10.75	5.084
PAA	17.75	2.980
GS	13.75	3.396
TOTAL	3.00	6.604
MS	3.00	4.947
WFC	22.00	6.782
AIG	22.00	7.630
BAC	3.00	4.367
MS	6.00	7.136
T	18.00	2.452

a. What is the sample mean years to maturity for corporate bonds and what is the sample standard deviation?

Mean:?

Standard Deviation:?

b. Develop a 95% confidence interval for the population mean years to maturity. Please round the answer to four decimal places.

c. What is the sample mean yield on corporate bonds and what is the sample standard deviation?

Mean:?

Standard Deviation:?

d. Develop a 95% confidence interval for the population mean yield on corporate bonds. Please round the answer to four decimal places.

Answer 1

Answer(a): In this question, the variable (x) is years to maturity

Company Ticker	Years to Maturity(x)	x2
HSBC	24.75	612.5625
GS	4	16
C	19.25	370.5625
MS	27.75	770.0625
C	22.5	506.25
TOTAL	17.5	306.25
MS	27.75	770.0625
WFC	25.5	650.25
TOTAL	9.25	85.5625
TOTAL	21	441
BAC	12.25	150.0625
RABOBK	21	441
GS	29	841
AXP	12.25	150.0625
MTNA	19	361
MTNA	4.25	18.0625
JPM	10.5	110.25
GE	8	64
LNC	16.5	272.25
BAC	24.5	600.25
FCX	25.5	650.25
GS	29.5	870.25
RABOBK	29	841
GE	12.75	162.5625
HCN	13.25	175.5625
GE	21.75	473.0625
VOD	21.5	462.25
NEM	6.5	42.25
GE	2.25	5.0625
C	22.25	495.0625
SHBASS	10.75	115.5625
PAA	17.75	315.0625
GS	13.75	189.0625
TOTAL	3	9
MS	3	9
WFC	22	484
AIG	22	484
BAC	3	9
MS	6	36
T	18	324
Total	660	13688.5

n=40

Sample mean of years to maturity

  

The standard deviation of years to maturity is given by

Hence the Mean = 16.5

Standard deviation = 8.4709

Answer(b):

The 95% confidence interval of mean is given by

Lower limit= 13.7909

Upper limit = 19.2091

Answer(c): In this question, the variable (x) is yield

Company Ticker	Yield(x)	x2
HSBC	0.865	0.748225
GS	4.351	18.9312
C	3.568	12.73062
MS	5.084	25.84706
C	1.641	2.692881
TOTAL	2.257	5.094049
MS	3.194	10.20164
WFC	4.953	24.53221
TOTAL	1.487	2.211169
TOTAL	4.449	19.7936
BAC	6.662	44.38224
RABOBK	1.931	3.728761
GS	3.648	13.3079
AXP	1.623	2.634129
MTNA	2.819	7.946761
MTNA	3.39	11.4921
JPM	6.067	36.80849
GE	3.521	12.39744
LNC	6.462	41.75744
BAC	2.419	5.851561
FCX	6.528	42.61478
GS	3.202	10.2528
RABOBK	7.138	50.95104
GE	1.816	3.297856
HCN	6.93	48.0249
GE	3.658	13.38096
VOD	4.763	22.68617
NEM	5.419	29.36556
GE	6.205	38.50203
C	5.422	29.39808
SHBASS	5.084	25.84706
PAA	2.98	8.8804
GS	3.396	11.53282
TOTAL	6.604	43.61282
MS	4.947	24.47281
WFC	6.782	45.99552
AIG	7.63	58.2169
BAC	4.367	19.07069
MS	7.136	50.9225
T	2.452	6.012304
Total	172.85	886.1275

Sample mean of years to maturity

The standard deviation of yield is given by

Hence the Mean = 4.3213

Standard deviation = 1.8892

Answer(d):

The 95% confidence interval of mean is given by