Question

Question 10

Recorded here are the germination times (in days) for fifteen randomly chosen seeds of a new type of bean:

11, 13, 11, 10, 15, 15, 10, 20, 10, 20, 13, 21, 16, 14, 12

Assuming that germination times are normally distributed, find a 95% confidence interval for the mean germination time for all beans of this type. Then complete the table below.

Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)

What is the lower limit of the confidence interval? What is the upper limit of the confidence interval?

Answer 1

Solution:

x	x2
11	121
13	169
11	121
10	100
15	225
15	225
10	100
20	400
10	100
20	400
13	169
21	441
16	256
14	196
12	144
∑x=211	∑x2=3167

Mean ˉx=∑xn

=11+13+11+10+15+15+10+20+10+20+13+21+16+14+12/15

=211/15

=14.0667

Sample Standard deviation S=√∑x2-(∑x)2nn-1

=√3167-(211)215/14

=√3167-2968.0667/14

=√198.9333/14

=√14.2095

=3.7696

Degrees of freedom = df = n - 1 =15 - 1 = 14

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,14 =2.145

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

= 2.145 * (3.8 / 15)

= 2.1

Margin of error = 2.1

The 95% confidence interval estimate of the population mean is,

- E < < + E

14.1 - 2.1 < < 14.1 + 2.1

12.0 < < 16.2

The lower limit = 12.0

The upper limit = 16.2