Question

A simple random sample of germination times (in days) for 7 seeds of a new strain of snap beans has a sample mean of 15.57 and standard deviation of 3.21.

(a) Determine a 99% confidence interval for the true mean germination time for this strain.

(b) Choose a correct interpretation to your result in part (a):

i. We can be 99% sure the sample average of 7 germination times is in the interval calculated in part (a).

ii. We can be 99% sure the population average of germination times is in the interval calculated in part (a).

iii. We can be 99% sure all germination times lie in the interval calculated in part (a).

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 15.57

sample standard deviation = s = 3.21

sample size = n = 7

Degrees of freedom = df = n - 1 = 7 - 1 = 6

a)

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,6 = 3.707

Margin of error = E = t_/2,df * (s /n)

= 3.707* (3.21 / 7)

= 4.50

The 99% confidence interval estimate of the population mean is,

- E < < + E

15.57 - 4.50 < < 15.57 + 4.50

11.07 < < 20.07

(11.07 , 20.07)

b)

2. We can be 99% sure the population average of germination times is in the interval calculated in part (a).