Question

In a random sample of 67 professional actors, it was found that 42 were extroverts.

(a) Find a 95% confidence interval for p. (Round your answers to two decimal places.)

lower limit
upper limit

In a survey of a random sample of 35 households in the Cherry Creek neighborhood of Denver, it was found that 12 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

a)

p̂ = X / n = 42/67 = 0.627
p̂ ± Z(α/2) √( (p * q) / n)
0.627 ± Z(0.05/2) √( (0.627 * 0.373) / 67)
Z(α/2) = Z(0.05/2) = 1.96
Lower Limit = 0.627 - Z(0.05) √( (0.627 * 0.373 / 67) = 0.51
upper Limit = 0.627 + Z(0.05) √( (0.627 * 0.373) / 67) = 0.74

b)

p̂ = X / n = 12/35 = 0.343
p̂ ± Z(α/2) √( (p * q) / n)
0.343 ± Z(0.1/2) √( (0.343 * 0.657) / 35)
Z(α/2) = Z(0.1/2) = 1.645
Lower Limit = 0.343 - Z(0.1) √( (0.343 * 0.657) / 35) = 0.2110
upper Limit = 0.343 + Z(0.1) √( (0.343 * 0.657) / 35) = 0.4750