Question

Six measurements were made of mineral content (in percent) of spinach. It is reasonable to assume that the population is approximately normal.

19.3 20.5 20.7 21.2 20.8 19.2

construct confidence interval for the mean mineral content, round answer to two decimal place 98% confidence interval for the mean mineral content is ? < u < ?

Answer 1

Solution:

x	x2
19.3	372.49
20.5	420.25
20.7	428.49
21.2	449.44
20.8	432.64
19.2	368.64
∑x=121.7	∑x2=2471.95

Mean ˉx=∑xn

=19.3+20.5+20.7+21.2+20.8+19.2/6

=121.7/6

=20.2833


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

=√2471.95-(121.7)26/5

=√2471.95-2468.4817/5

=√3.4683/5

=√0.6937

=0.8329

Degrees of freedom = df = n - 1 = 6 - 1 = 5

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t /2,df = t0.005,5 =3.365

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

= 3.365* (0.83 / 6)

= 1.14

Margin of error = 1.14

The 98% confidence interval estimate of the population mean is,

- E < < + E

20.28- 1.14 < < 20.28 + 1.14

19.14 < < 21.42

(19.14, 21.42 )