Question

Use Excel to obtain a 95% confidence level for the mean satisfaction rating of the new sandwich

Satisfaction
7
5
5
6
8
7
6
7
10
7
9
5
5
8
8
6
7
8
7
5
5
5
5
5
2
5
8
7
6
6
4
6
6
5
6
7
8
9
5
4

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	7	0.5625
	5	1.5625
	5	1.5625
	6	0.0625
	8	3.0625
	7	0.5625
	6	0.0625
	7	0.5625
	10	14.0625
	7	0.5625
	9	7.5625
	5	1.5625
	5.0	1.5625
	8	3.0625
	8	3.0625
	6	0.0625
	7	0.5625
	8	3.0625
	7	0.5625
	5	1.5625
	5	1.5625
	5	1.5625
	5	1.5625
	5	1.5625
	2	18.0625
	5	1.5625
	8	3.0625
	7	0.5625
	6	0.0625
	6	0.0625
	4	5.0625
	6	0.0625
	6	0.0625
	5	1.5625
	6	0.0625
	7	0.5625
	8	3.0625
	9	7.5625
	5	1.5625
	4	5.0625
Total	250	99.5

Mean X̅ = Σ Xi / n
X̅ = 250 / 40 = 6.25

Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 99.5 / 40 -1 ) = 1.5973

CONFIDENCE.T(0.05,STDEV.S(F3:F42),COUNT(F3:F42)) = 0.5108

CONFIDENCE.T(alpha,standard_dev,size)

alpha = 0.05

standard_dev = STDEV.S(F3:F42) = 1.5973

size = COUNT(F3:F42) = 40

X̅ = AVERAGE(F3:F42) = 6.25

Where, Range(F3:F42) is the range of data values in excel

Confidence interval X̅ ± CI = 6.25 ± 1.5973

95% Confidence interval is ( 5.7391 , 6.7609 )