Question

A botanist wishes to estimate the mean number of seeds for a certain fruit. She samples 18 specimens and finds the average number of seeds is 48 with a standard deviation of 14. Construct a one-sided upper-bound 80% confidence interval for the mean number of seeds for the species using the appropriate critical value to three decimal places.

Answer 1

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 $1 - \frac{1 - 0.8} = 0.8$. So we have T = 0.863

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 0.863\frac{14}{\sqrt{18}} = 2.848$

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.