Question

1.The marketing department for a cereal company would like to know what "proportion" of households that receive free samples of the cereal with their newspapers later purchased the cereal. A random sample of 100 showed 55 purchased the cereal.Find a 95% confidence interval for the true proportion.

Answer 1

Solution :

Given that,

n = 100

x = 55

Point estimate = sample proportion = = x / n = 55/100=0.55

1 -   = 1-0.55 =0.45

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 (((0.55*0.45) /100 )

E = 0.098

A 95% confidence interval for proportion p is ,

- E < p < + E

0.55-0.098 < p < 0.55+0.098

0.452< p < 0.648

(0.452 , 0.468)