A horticulturist wishes to estimate the mean growth of seedlings in a large timber plot last year. A random sample of n = 100 seedlings is selected and the one-year growth for each is measured. The sample yields = 5.62 cm. The known standard deviation for this popualtiopn is 2.50 cm. The 95 percent confidence interval for the mean growth is:

Question

Question options:

(3.12, 8.12)

(4.98, 6.26)

(5.13, 6.11)

(5.37, 5.87)

(5.57, 5.67)

Answer 1

Formula:

CI95 = mean ± 1.96(sd/√n)

Your data:

mean = 5.62
sd = 2.50
n = 100

Substitute the values into the formula and calculate the interval. You should then be able to select the correct answer.

Answer 2

Uuh

Answer 3

(5.37, 5.87)