Question

You must estimate the mean temperature (in degrees Fahrenheit) with the following sample temperatures:

23.5 55.9 33.5 45.5 7.4

Find the 90% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place)

. 90% C.I. =

Answer should be obtained without any preliminary rounding.

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	23.5	93.3156
	55.9	517.1076
	33.5	0.1156
	45.5	152.2756
	7.4	663.5776
Total	165.8	1426.392

Mean X̅ = Σ Xi / n
X̅ = 165.8 / 5 = 33.16
Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 1426.392 / 5 -1 ) = 18.8838

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.1 /2, 5- 1 ) = 2.132
33.16 ± t(0.1/2, 5 -1) * 18.8838/√(5)
Lower Limit = 33.16 - t(0.1/2, 5 -1) 18.8838/√(5)
Lower Limit = 15.1564 ≈ 15.16
Upper Limit = 33.16 + t(0.1/2, 5 -1) 18.8838/√(5)
Upper Limit = 51.1636 ≈ 51.16
90% Confidence interval is ( 15.16 , 51.16 )