Question

Construct 90%, 95%, and 99% confidence intervals to estimate μ from the following data. State the point estimate. Assume the data come from a normally distributed population.

13.6	11.6	11.9	13.0	12.5	11.4	12.0
11.7	11.8	13.6

Answer 1

(a)

From the given data, the following statistics are calculated:

n = 10

= 123.1/10 = 12.31

Point Estimate = 12.31

s = 0.8212

SE = s/

= 0.8212/

= 0.2597

= 0.10

ndf = n - 1 = 10 - 1 = 9

From Table, critical values of t = 1.8331

90% Confidence Interval:

12.31 (1.8331 X 0.2597)

= 12.31 0.4760

= ( 11.8340 ,12.7860)

90% Confidence Interval:

11.8340 < <   12.7860

(b)

= 0.05

ndf = n - 1 = 10 - 1 = 9

From Table, critical values of t = 2.2622

90% Confidence Interval:

12.31 (2.2622 X 0.2597)

= 12.31 0.5875

= ( 11.7225 ,12.8975)

95% Confidence Interval:

11.7225 < <   12.8975

(c)

= 0.01

ndf = n - 1 = 10 - 1 = 9

From Table, critical values of t = 3.2498

90% Confidence Interval:

12.31 (3.2498 X 0.2597)

= 12.31 0.8440

= ( 11.4660 ,13.1540)

99% Confidence Interval:

11.4660 < <   13.1540

So,

Answers to questions asked:

Point Estimate = 12.31

90% Confidence Interval:

11.8340 < <   12.7860

95% Confidence Interval:

11.7225 < <   12.8975

99% Confidence Interval:

11.4660 < <   13.1540