Question

Thirty-two small communities in Connecticut (population near 10,000 each) gave an average of x = 138.5 reported cases of larceny per year. Assume that o is known to be 42.1 cases per year. (a) Find a 90% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place.) lower limit 11.8 upper limit margin of error (b) Find a 95% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place.) lower limit upper limit margin of error (c) Find a 99% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place.) lower limit upper limit margin of error

Answer 1

Date Page Solutions Given that, X= 138.5 6= 42.1 n = 32 U AT 90 % confidence level, x=1- 90 olo dol-0.00 = 0.10 lg = 0.10 = 0.05_ 2212= Zooos - 1.645 Margin of error=E= Z612x6 ví E=1.645 X 22.1 132 E = 12.2 his At go % confidence interval for X- E <M exte | 138. 5-12.2 cu< 138.5+ 12.2 126.3 CM 150.7 (126:3, 150. 7) Lower limit = [126.3 upper limit = [150.71 Margin of errora [ 12.2

Date Page At as lo confidence level, x=1-95 olo d=1-0.95 = 0.05 ♡ = 0.05 = 0.025 Z 412= 20.025 = 1.96 un E- 1.96% 42 - 1 V32 E_1406 At 95 % confidence interval, - X-E cua xtE 138.5-14.6 cuc 13805 +14.6 123.9 cu< 153.1 (123.9, 153:1) Lower limit = [123.9 ] upper limit = 153:) Margin of error = [ 14.6 02

05 At aq % confidence level, 2- 1-99 % del-0.99 = 0.01 Q = 0.01 = 0.005 2212 - Zooous = 2.576 Margin of error = E= 2x12 X6 vñ E=2.576 X 42.1 V 32 E=19.2 At aq % confidence interval, X- E <l<t te 138.5 -19.02 c44 138.5 +19.2 119.3 cu < 157.7 (119.3, 157.7 lower limit = [119.31 Margin of error = [19. 2]