Question

Problem 10-28 Some citizens complained to city council members that there should be equal protection under the law against the occurrence of crimes. The citizens argued that this equal protection should be interpreted as indicating that high-crime areas should have more police protection than low-crime areas. Therefore, police patrols and other methods for preventing crime (such as street lighting or cleaning up abandoned areas and buildings) should be used proportionately to crime occurrence. The city has been broken down into 20 geographic areas, each containing 4,950 residences. The police recognize that not all crimes and offenses are reported: people do not want to become involved, consider the offenses too small to report, are too embarrassed to make a police report, or do not take the time, among other reasons. Every month, because of this, the police are contacting by phone a random sample of 990 of the 4,950 residences for data on crime. (Respondents are guaranteed anonymity.) The 990 sampled from each area showed the following incidence of crime during the past month: AREA 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 NUMBER OF CRIMES 15 5 20 19 15 30 11 18 13 4 22 16 -13 15 11 33 5 22 7 33 327 SAMPLE SIZE CRIME RATE 990 0.015 990 0.005 990 0.020 990 0.019 990 0.015 990 0.030 990 0.011 990 0.018 990 0.013 990 0.004 990 0.022 990 0.016 990 0.013 990 0.015 990 0.011 990 0.033 990 0.005 990 0.022 990 0.007 990 0.033 a. Determine P.S.UCL and LCL for a p chart of 95 percent confidence (at Z - 1.96). (Round your answers to 4 decimal places.) Se UCL LCL

Answer 1

a) Now, Using given data:- Sample size (n) = 800 Fraction defective (p] = {(krimes) no. of Somples xn > È 327/620x990) - » Standard deviation of sampling distribution (Sp): Spi ē ci-e 0.0165( 1-0.0165) 0.0165 u 00040 990 Control limite are :-(2=1.96) → UCL p= 8 + 2Sp 0.0165+1.96(0.004) = 0.0243 LCL p = 8 - 2Sp 0.0165 -1.96(0.004) 0.0087