Question

The following sample data are from a normal population: 13, 11, 15, 18, 16, 14, 9, 8.

(a)

What is the point estimate of the population mean?

(b)

What is the point estimate of the population standard deviation? (Round your answer to three decimal places.)

(c)

With 95% confidence, what is the margin of error for the estimation of the population mean? (Round your answer to one decimal place.)

(d)

What is the 95% confidence interval for the population mean? (Round your answer to one decimal place.)

to

Answer 1

Solution:

x	x2
13	169
11	121
15	225
18	324
16	256
14	196
9	81
8	64
∑x=104	∑x2=1436

a ) Mean ˉx=∑xn

=13+11+15+18+16+14+9+8/8

=104/8

=13

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

=√1436-(104)28/7

=√1436-1352/7

=√84/7

=√12

=3.4641

c ) Degrees of freedom = df = n - 1 = 25 - 1 = 24

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,24 =2.365

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

= 2.365 * (3.464 / 8)

= 2.9

Margin of error = 2.9

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

- E < < + E

13 - 2.9 < < 13 + 2.9

10.1 < < 15.9

(10.1, 15.9 )