Question

For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding.

In a survey of a random sample of 35 households in the Cherry Creek neighborhood of Denver, it was found that 7 households turned out the lights and pretended not to be home on Halloween.

(a) Compute a 90% confidence interval for p, the proportion of all households in Cherry Creek that pretend not to be home on Halloween. (Round your answers to four decimal places.)

lower limit:

upper limit:

Answer 1

Solution :

Given that,

n = 35

x = 7

Point estimate = sample proportion = = x / n = 7 / 35 = 0.20

1 - = 1 - 0.20 = 0.80

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05  = 1.645

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

= 1.645 (((0.20 * 0.80) / 35)

= 0.1112

A 90% confidence interval for population proportion p is ,

± E   

= 0.20 ± 0.1112

= ( 0.0888, 0.3112 )

lower limit = 0.0888

upper limit = 0.3112