Question

A sample of salary offers​ (in thousands of​ dollars) given to management majors is​ 48, 51,​ 46, 52,​ 47, 48,​ 47, 50,​ 51, and 59. Using this data to obtain a​ 95% confidence interval resulted in an interval from 47.19 to 52.61. Determine whether the statement is true or false. It is possible that the mean of the population is between 47.19 and 52.61.

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	48	3.61
	51	1.21
	46	15.21
	52	4.41
	47	8.41
	48	3.61
	47	8.41
	50	0.01
	51	1.21
	59	82.81
Total	499	128.9

Mean X̅ = Σ Xi / n
X̅ = 499 / 10 = 49.9
Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 128.9 / 10 -1 ) = 3.7845

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 10- 1 ) = 2.262
49.9 ± t(0.05/2, 10 -1) * 3.7845/√(10)
Lower Limit = 49.9 - t(0.05/2, 10 -1) 3.7845/√(10)
Lower Limit = 47.19
Upper Limit = 49.9 + t(0.05/2, 10 -1) 3.7845/√(10)
Upper Limit = 52.61
95% Confidence interval is ( 47.19 , 52.61 )

Yes, the statement is true.

Yes, it is possible that the mean of the population is between 47.19 and 52.61.