Question

A sample of 18 observations taken from a normally distributed population produced the following data: 28.1 27.2 25.1 25.2 31.2 23.5 26.4 24.3 28.2 37.2 23.6 28.9 27.9 25.5 27.3 25.3 22.4 22.7 Round your answers to three decimal places. a. What is the point estimate of ? b. Make a 90% confidence interval for u. c. What is the margin of error of estimate for p in part b? E =

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	28.1	2.0543
	27.2	0.2844
	25.1	2.4545
	25.2	2.1512
	31.2	20.5508
	23.5	10.028
	26.4	0.0711
	24.3	5.6013
	28.2	2.351
	37.2	110.9504
	23.6	9.4046
	28.9	4.9876
	27.9	1.521
	25.5	1.3612
	27.3	0.4011
	25.3	1.8679
	22.4	18.2047
	22.7	15.7347
Total	480	209.9798

Mean X̅ = Σ Xi / n
X̅ = 480 / 18 = 26.6667
Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 209.9798 / 18 -1 ) = 3.5145

Part a)

Point estimate of   $\mu$

X̅ = 26.667

Part b)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.1 /2, 18- 1 ) = 1.74
26.6667 ± t(0.1/2, 18 -1) * 3.5145/√(18)
Lower Limit = 26.6667 - t(0.1/2, 18 -1) 3.5145/√(18)
Lower Limit = 25.225
Upper Limit = 26.6667 + t(0.1/2, 18 -1) 3.5145/√(18)
Upper Limit = 28.108
90% Confidence interval is ( 25.225 , 28.108 )

Part c)

Margin of Error = t(α/2, n-1) S/√(n) = 1.441