Question

2. A software engineering firm began stocking flavored sparkling water in addition to soda in the break room. A survey found that out of 180 engineers, 45 of them switching to the sparking water.

a. Determine a 95% CI.

b. Determine the margin of error for the CI in part (a). Round to three decimal places.

Answer 1

Solution :

Given that,

n = 180

x = 45

Point estimate = sample proportion = = x / n = 45/180=0.25

1 -   = 1- 0.25 =0.75

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 (((0.25*0.75) /180 )

E = 0.0633

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.25-0.0633 < p <0.25+ 0.0633

0.1867< p < 0.3133

The 95% confidence interval for the population proportion p is : 0.1867, 0.3133