Question

In a recent poll, 160 people were asked if they liked dogs, and 55% said they did. Find the margin of error of this poll, at the 99% confidence level.

Give your answer to three decimals

If n=28, ¯xx¯(x-bar)=33, and s=7, find the margin of error at a 80% confidence level

Give your answer to two decimal places.

Answer 1

Solution :

Given that,

n =  160

Point estimate = sample proportion = =0.55

1 -   = 1-0.55 =0.45

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

  Margin of error = E = Z/2   * (((( * (1 - )) / n)

= 2.576* (((0.55*0.45) / 160 )

E = 0.1013

  Margin of error = E =0.10

(B)

Solution :

Given that,

s =7

n = 28Degrees of freedom = df = n - 1 =28 - 1 = 27

At 80% confidence level the t is ,

= 1 - 80% = 1 - 0.80 = 0.20

  / 2= 0.20 / 2 = 0.10

t /2,df = t0.10,27=1.314 ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 1.314 * (7 / 28)

= 1.74

Margin of error = E =1.74