175 188 152 168 173 193 138 151 143 208 Assume the sample is

1.75

1.88

1.52

1.68

1.73

1.93

1.38

1.51

1.43

2.08

Assume the sample is taken from a normally distributed population. Construct

90%

confidence intervals for

(a) the population variance

sigma squared 2

(b) the population standard deviation

(a) The confidence interval for the population variance is .(,).

(Round to three decimal places as needed.)

The number of hours of reserve capacity of 10
randomly selected automotive batteries is shown to the right.

1.75	1.88	1.52	1.68	1.73
1.93	1.38	1.51	1.43	2.08

Data : [ 1.75,1.88,1.52,1.68,1.73,1.93,1.38,1.51,1.43,2.08 ]

n =10

Mean = 1.689

Standard Deviation (s) = 0.23

Variance (s^2) = 0.0529

Thus for 90% confidence interval for ^2, we need

2 _0.950,9 = 3.3251

and

2 _0.050,9 = 16.9189

Thus, the 90% confidence interval for ^2 is :

= [ 16s^2 /16.9189 , 16s^2/ 3.3251]

= [ 16*0.0529 /16.9189 , 16*0.0529 / 3.3251]

= [0.050, 0.255] Answer

The 90% confidence interval for is :

= [ 0.050, 0.255]

= [ 0.224 , 0.505 ] Answer