In a study of speed​ dating, male subjects were asked to rate the attractiveness of their female​ dates, and a sample of the results is listed below ​(1 = not ​attractive; 10 = extremely ​attractive). Construct a confidence interval using a ​95% confidence level. What do the results tell about the mean attractiveness ratings of the population of all adult​ females? 5, 8, 2, 8, 7, 6, 6, 8, 9, 9, 6, 9 What is the confidence interval for the population mean ​μ? <μ

Confidence Interval =  ( 5.6 < μ < 8.2 )

The interval shows that the mean attractiveness is Above average on a scale of 1 to 10

Step-by-step 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 )    

tickets for the harlem globetrotters

show at michigan state university in

2010 cost $16, $23, or, for vip seats, $40.

total revenue = 16x + 23y + 40z

if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, sales of all three kinds of tickets would

total $46,575.

16x + 23y + 40z = 46575, how many of each kind of ticket would

have been sold?

solve for x, y and z

x − 9z = 0

x − y − z = 55

16x + 23y + 40z = 46575

x = 1170

y = 985

z = 130

welcome!