In: Statistics and Probability
A sample containing years to maturity and yield for corporate bonds are contained in the data given below.
Years to Maturity | Yield | Years to Maturity | Yield | |||
---|---|---|---|---|---|---|
5.75 | 7.728 | 3.25 | 2.820 | |||
25.25 | 2.415 | 5.25 | 4.835 | |||
1.75 | 5.382 | 15.75 | 6.905 | |||
23.50 | 5.106 | 1.50 | 4.849 | |||
26.25 | 1.576 | 23.75 | 5.054 | |||
18.50 | 7.145 | 2.50 | 3.966 | |||
20.50 | 0.811 | 14.25 | 5.300 | |||
29.75 | 4.275 | 16.50 | 3.114 | |||
1.50 | 7.320 | 24.75 | 5.949 | |||
5.75 | 8.140 | 11.50 | 2.934 | |||
13.50 | 5.170 | 21.50 | 6.921 | |||
14.50 | 7.286 | 27.75 | 7.810 | |||
13.25 | 3.505 | 14.25 | 6.998 | |||
6.25 | 7.009 | 12.00 | 7.759 | |||
13.75 | 7.062 | 14.00 | 3.196 | |||
10.75 | 7.764 | 27.50 | 1.789 | |||
12.75 | 7.687 | 17.75 | 8.066 | |||
29.50 | 1.145 | 22.75 | 3.984 | |||
6.75 | 1.055 | 13.25 | 3.992 | |||
28.25 | 5.902 | 17.00 | 0.794 |
a. What is the sample mean years to maturity for corporate bonds and what is the sample standard deviation?
Mean | (to 4 decimals) |
Standard deviation | (to 4 decimals) |
b. Develop a confidence interval for the population mean years to maturity. Round the answer to four decimal places.
, years
c. What is the sample mean yield on corporate bonds and what is the sample standard deviation?
Mean | (to 4 decimals) |
Standard deviation | (to 4 decimals) |
d. Develop a confidence interval for the population mean yield on corporate bonds. Round the answer to four decimal places.
, percent
Years to Maturity | Yield | ||
Mean | 15.3625 | Mean | 5.01295 |
Standard Error | 1.352465 | Standard Error | 0.366903 |
Median | 14.25 | Median | 5.138 |
Mode | 5.75 | Mode | #N/A |
Standard Deviation | 8.553742 | Standard Deviation | 2.320498 |
Sample Variance | 73.16651 | Sample Variance | 5.384711 |
Kurtosis | -1.01337 | Kurtosis | -1.09878 |
Skewness | 0.019778 | Skewness | -0.35711 |
Range | 28.25 | Range | 7.346 |
Minimum | 1.5 | Minimum | 0.794 |
Maximum | 29.75 | Maximum | 8.14 |
Sum | 614.5 | Sum | 200.518 |
Count | 40 | Count | 40 |
a) Sample mean years to maturity
Mean=15.3625
Sample standard deviation years to maturity
Std DEv= 8.553742
b) : Population mean years to maturity.
Since population standard deviation is not known. We use t distribution to find confidence interval for population mean.
The 95% confidence interval for population mean years is to maturity is
Alpha: level of significance = 0.05
From t-table
Hence 95% confidence interval is
=(15.3625 -8.553742*2.0227/sqrt(40) ,15.3625 -8.553742*2.0227/sqrt(40))
=(12.6269,18.0981)
95% confidence interval for population mean years to maturity is (12.6269,18.0981) years.
c) Sample mean yield on corporate bond
5.01295
Sample standard deviation yield on corporate bond
2.320498
d) : Population mean yield on corporate bond
Since population standard deviation is not known. We use t distribution to find confidence interval for population mean.
The 95% confidence interval for population mean yield on corporate bond is
Alpha: level of significance = 0.05
From t-table
Hence 95% confidence interval is
=(5.01295 -2.320498*2.0227/sqrt(40) ,5.01295 +2.320498*2.0227/sqrt(40))
=(4.2708,5.7551)
95% confidence interval for population mean yield on corporate bond is (4.2708,5.7551) percent.