Question

(2 points) Over half of all American teens (ages 12 to 17 years) have an online profile, mainly on Facebook. A random sample of 549 teens with profiles found that 402 included photos of themselves. (a) Give the 99% large-sample confidence interval for the proportion p of all teens with profiles who include photos of themselves. Large-sample Interval: (b) Give the plus four 99% confidence interval for p. (The plus-four interval always pulls the results away from 0% or 100%, whichever is closer. Even though the condition for using the large-sample interval is met, the plus four interval is more trustworthy.) Plus four Interval:

Answer 1

Answer Date: 07/12/2019 A 99% large-sample confidence interval for the proportion of all teens with profiles who include photos of themselves is, First, compute proportion, the z-critical value then find confidence interval. The proportionis, P _ 402 549 = 0.7322 The z-critical value is, From the standard normal table, the z-critical value at 99% confidence is 2.576.

D) 99%C.I. = pzcx = 0.7322 +2.576x, 0.7322(1-0.7322) 549 = 0.6835 to 0.7809 A 99% large-sample confidence interval for the proportion of all teens with profiles who include photos of themselves is 0.6835 to 0.7809. b) The plus four 99% confidence interval for the proportion of all teens with profiles who include photos of themselves is, First, compute proportion for plus four method, the z-critical value then find confidence interval.

The proportion for plus four method is, PL+2 n+4 402 +2 549+4 = 0.7306 The Z-critical value is, From the standard normal table, the z-critical value at 99% confidence is 2.576. P(1-P) 99%C.I. = pzcx Vn+4 0.7306(1 – 0.7306) = 0.7306+ 2.5 549+4 = 0.6820 to 0.7792 the episodes of the engine is 0920 for the groportion of all teens with The plus four 99% confidence interval for the proportion of all teens with profiles who include photos of themselves is 0.6820 to 0.7792