Answer:
Confidence Interval = ( 5.6 < μ < 8.2 )
The interval shows that the mean attractiveness is Above average on a scale of 1 to 10
Explanation:
Data Given : 5, 8, 2, 8, 7, 6, 6, 8, 9, 9, 6, 9
a) construct a confidence interval at a 95% confidence level
confidence interval = x ± tcritical ( s / √ n ) ------ ( 1 )
n = 12
μ (mean) = ( 5+8+2+8+7+6+6+8+9+9+6+9 ) / 12 ≈ 6.92
std ( s ) = √ 1/11 *∑( xi- X )^2
= √ 1/11 * ( 5 - 6.92)^2 + ( 8 - 6.92)^2 + ( 2 - 6.92)^2 + ( 8 - 6.92)^2 + ( 7-6.92)^2 + ( 6 - 6.92)^2 + (6 - 6.92 )^2 + ( 8 - 6.92 )^2 + ( 9 - 6.92)^2 + ( 9 - 6.92)^2 + ( 6 - 6.92 )^2 + ( 9 - 6.92 )^2
= √ 1/11 * 46.92 = 2.07
tcritical at ∝ 0.05 , df = 11 = 2.20
Back to equation 1
6.92 ± 2.2 ( 2.07 / √ 12 )
( 6.92 ± 1.31 )
Confidence Interval = ( 5.6 < μ < 8.2 )