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 8 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 = 8

Point estimate = sample proportion = = x / n = 8/35=0.2286

1 -   = 1- 0.2286 =0.7714

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= 1.645 *((0.2286*0.7714) / 35)

E = 0.1168

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.2286-0.1168 < p < 0.2286+0.1168

0.1118< p < 0.3454

The 90% confidence interval for the population proportion p is : lower limit=0.1118,upper limit=0.3454