Question

A consulting firm wants to study communication deficiencies in the health care industry. A random sample of 73 health care clinicians reveals the​ following: Time wasted in a day due to outdated communication​ technologies: Upper X equals 37 ​minutes, S equals 8 minutes. Thirty dash eight health care clinicians cite inefficiency of pagers as the reason for the wasted time. Complete parts​ (a) and​ (b) below.

a. Construct a 90​% confidence interval estimate for the population mean time wasted in a day due to outdated communication technologies. ​(Round to two decimal places as​ needed.)

b. Construct a 95​% confidence interval estimate for the population proportion of health care clinicians who cite inefficiency of pagers as the reason for the wasted time.​(Round to four decimal places as​ needed.)

Answer 1

a)

sample mean, xbar = 37
sample standard deviation, s = 8
sample size, n = 73
degrees of freedom, df = n - 1 = 72

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.666

ME = tc * s/sqrt(n)
ME = 1.666 * 8/sqrt(73)
ME = 1.56

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (37 - 1.666 * 8/sqrt(73) , 37 + 1.666 * 8/sqrt(73))
CI = (35.44 , 38.56)

b)

sample proportion, = 0.5205
sample size, n = 73
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.5205 * (1 - 0.5205)/73) = 0.0585

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0585
ME = 0.1147

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.5205 - 1.96 * 0.0585 , 0.5205 + 1.96 * 0.0585)
CI = (0.4058 , 0.6352)