Question

In a survey on supernatural experiences, 712 of 4011 adult Americans surveyed reported that they had seen or been with a ghost.

(a) What assumption must be made in order for it to be appropriate to use the formula of this section to construct a confidence interval to estimate the proportion of all adult Americans who have seen or been with a ghost? We need to assume that there are only 712 adult Americans. We need to assume that the 4011 people formed a random sample of adult Americans. We need to assume that the 4011 people were surveyed at a supernatural convention. We need to assume that the 712 people are the only Americans who had seen or been with a ghost.

(b) Construct a 90% confidence interval for the proportion of all adult Americans who have seen or been with a ghost. (Round your answers to three decimal places.)

Answer 1

Solution :

Given that,

a)  We need to assume that the 712 people are the only Americans who had seen or been with a ghost.

b) n = 4011

x = 712

Point estimate = sample proportion = = x / n = 712 / 4011 = 0.178

1 - = 1 - 0.178 = 0.822

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

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

= 1.645 (((0.178 * 0.822) / 4011)

= 0.010

A 90% confidence interval for population proportion p is ,

± E

= 0.178 ± 0.010

= (0.168, 0.188 )