Question

The developer of a line of key cutting machines would like to include as part of its advertising the number of keys a user can expect from each blade. A sample of 10 blades revealed the following number of keys cut.

   2,108     2,700     2,175     3,016     2,629     1,839     1,928     2,061     2,236     2,648

The Estimated Population Mean is 2334.  

1.) Determine the 95% confidence interval for the population mean. (Round answer to 2 decimal places.)

Confidence interval: ___________ and __________.

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	2108	51076
	2700	133956
	2175	25281
	3016	465124
	2629	87025
	1839	245025
	1928	164836
	2061	74529
	2236	9604
	2648	98596
Total	23340	1355052

Mean X̅ = Σ Xi / n
X̅ = 23340 / 10 = 2334
Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 1355052 / 10 -1 ) = 388.0223

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 10- 1 ) = 2.262
2334 ± t(0.05/2, 10 -1) * 388.0223/√(10)
Lower Limit = 2334 - t(0.05/2, 10 -1) 388.0223/√(10)
Lower Limit = 2056.44
Upper Limit = 2334 + t(0.05/2, 10 -1) 388.0223/√(10)
Upper Limit = 2611.56
95% Confidence interval is ( 2056.44 , 2611.56 )