Respuesta :
Answer:
The 90% for the average weights of men is between 137.24 lb and 185.76 lb.
Step-by-step explanation:
We have the standard deviation for the sample, so we use the 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 = 14 - 1 = 14
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 14 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 1.7709
The margin of error is:
M = T*s = 1.7709*13.7 = 24.26
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 161.5 - 24.26 = 137.24 lb
The upper end of the interval is the sample mean added to M. So it is 161.5 + 24.26 = 185.76 lb
The 90% for the average weights of men is between 137.24 lb and 185.76 lb.
