Question

The Bureau of Meteorology of the Australian Government provided the mean annual rainfall (in millimeters) in Australia 1983–2002 as follows (http://www.bom.gov.au/ climate/change/rain03.txt) 499.2, 555.2, 398.8, 391.9, 453.4, 459.8, 483.7, 417.6, 469.2, 452.4, 499.3, 340.6, 522.8, 469.9, 527.2, 565.5, 584.1, 727.3, 558.6, 338.6 Construct a 99% two-sided confidence interval for the mean annual rainfall. Assume population is approximately normally distributed. Round your answers to 2 decimal places. less-than-or-equal-to mu less-than-or-equal-to

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	499.2	32.49
	555.2	3806.89
	398.8	8968.09
	391.9	10322.56
	453.4	1608.01
	459.8	1135.69
	483.7	96.04
	417.6	5760.81
	469.2	590.49
	452.4	1689.21
	499.3	33.64
	340.6	23378.41
	522.8	858.49
	469.9	556.96
	527.2	1135.69
	565.5	5184
	584.1	8208.36
	727.3	54662.44
	558.6	4238.01
	338.6	23994.01
Total	9376.5	156260.29

Mean X̅ = Σ Xi / n
X̅ = 9376.5 / 19 = 493.5
Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 156260.29 / 19 -1 ) = 93.1726

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01 /2, 19- 1 ) = 2.878 ( Critical value from t table )
493.5 ± t(0.01/2, 19 -1) * 93.1726/√(19)
Lower Limit = 493.5 - t(0.01/2, 19 -1) 93.1726/√(19)
Lower Limit = 431.98
Upper Limit = 493.5 + t(0.01/2, 19 -1) 93.1726/√(19)
Upper Limit = 555.02
99% Confidence interval is ( 431.98 , 555.02 )