Respuesta :
Answer:
The 80% confidence interval for the mean number of seeds for the species is of 50.848 seeds.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
T interval
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 = 18 - 1 = 17
80% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 17 degrees of freedom(y-axis) and a confidence level(as we want the one-sided upper-bound confidence interval) of [tex]1 - \frac{1 - 0.8} = 0.8[/tex]. So we have T = 0.863
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}} = 0.863\frac{14}{\sqrt{18}} = 2.848[/tex]
In which s is the standard deviation of the sample and n is the size of the sample.
The upper end of the interval is the sample mean added to M. So it is 48 + 2.848 = 50.848 seeds.
The 80% confidence interval for the mean number of seeds for the species is of 50.848 seeds.
