Asurvey of households in a small town showed that in 500 of 1,200 sampled households, at least one member attended a town meeting during the year. using the 95% level of confidence, what is the confidence interval for the proportion of households represented at a town meeting?

Divide the number of homes by the sample size:

500 out of 1200 houses = 0.4166

Now find the sample proportion:

p(1-p)/n = 0.4166((1-0.4166)/1200)= 0.0002

Take the square root: √0.0002 = 0.014

Multiply that by the Z value for 95% which is 1.96:

1.96 x 0.014 = 0.028 x 100 = 2.8%

The confidence interval is the percentage of homes from above ( 0.4166 plus and minus the sample proportion:

0.4166 - 0.028 = 0.3886

0.4166 + 0.028 = 0.4446

[0.3888, 0.4446]

$30.32

step-by-step explanation:

65.00*.12=2.80

65.00-2.80=57.20

57.20*1.06=60.632

60.632/2=30.316

b

step-by-step explanation:

u take $75 minus $60.the anwer/original($75)multiply by 100