Answer:
The 99% (two-sided) confidence interval for the true average echo duration μ is between 0 sec and 1.99 sec.
Step-by-step explanation:
We have the sample standard deviation, so we use the student t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 101 - 1 = 100
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 100 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.99}{2} = 0.995. So we have T = 2.6259
The margin of error is:
M = T*s = 2.6259*0.45 = 1.18
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 0.81 - 1.18. Answer in seconds cannot be negative, so we use 0 sec.
The upper end of the interval is the sample mean added to M. So it is 0.81 + 1.18 = 1.99 sec
The 99% (two-sided) confidence interval for the true average echo duration μ is between 0 sec and 1.99 sec.
