the answer is C hope this helps you
The 98% confidence interval for the mean of the population is (59, 68.2).
Step-by-step explanation:
Before building the confidence interval, we need to find the sample mean and the sample standard deviation.
Sample mean:
Sample standard deviation:
Confidence interval:
We have the standard deviation for the sample, and thus, we use the t-distribution.
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 = 10 - 1 = 9
98% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.821
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 63.6 - 4.6 = 59
The upper end of the interval is the sample mean added to M. So it is 63.6 + 4.6 = 68.2.
The 98% confidence interval for the mean of the population is (59, 68.2).