Question

A package delivery service advertises that at least 90% of all packages brought in by 9:00 a.m. for delivery in the same city will be delivered by noon. A random sample of 80 packages had 65 that were delivered by noon.

a) Construct a 90% confidence interval for the proportion of packages delivered by noon.

b) Construct a 99% lower confidence interval for the true proportion.

c) What size sample is necessary to construct a 95% confidence interval with a margin of error of 0.05? Use an estimate of p of 0.90.

d) Can you conclude the service is falsely advertising? (Do a hypothesis test.)

Answer 1

(a) Here

sample proportion = p^ = 65/80 = 0.8125

90% confidence interval = p^ +- MOE

Margin of Error = MOE = Critical test statistic * standard error of proportion

critical test statistic = NORMSDIST(0.5 + 0.9/2) = 1.645

standard error of proportion = sqrt [0.8125 * (1 - 0.8125)/80] = 0.0436

Margin of error = 1.645 * 0.0436 = 0.0718

90% confidence interval = 0.8125 +- 0.0718 = (0.7407, 0.8843)

(b) Here 99% lower confidence interval is to be connected.

99% lower confidence interval = 0.8125 - NORMSINV(0.99) * 0.0436 = (0.7110, 1]

(c)  Margin of error = 0.05

Critical test statistic = NORMSINV(0.95) = 1.96

standard error = sqrt [0.9 * 0.1/n]

0.05 = 1.96 * sqrt [0.9 * 0.1/n]

n = 138.30 or 139

(d) For a hypothesis test.

H0 : p 0.90

Ha : p < 0.90

standard error = sqrt(0.9 * 0.1/80) = 0.0335

Test statistic

z = (0.8125 - 0.90)/0.0335 = -2.60875

p - value = P(z < -2.60875) = 0.0045 < 0.01

so here we reject the null hypothesis, so  we can conclude the service is falsely advertising.