ATT would like to test the hypothesis that the average revenue per retail user

# Att would like to test the hypothesis that the

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170)AT&T would like to test the hypothesis that the average revenue per retail user for Verizon Wireless customersequals \$50. A random sample of 32 Verizon Wireless customers provided an average revenue of \$54.70. Assumethat the actual population mean for revenue per retail user is \$44.30 and the population standard deviation forthe revenue per retail user is \$11.00. AT&T would like to set ΅= 0.05. Calculate the probability of a Type II error.
171) Hotels.com would like to test the hypothesis that the proportion of American travelers in Europe that prefer anAmerican-branded hotel equals 0.42. A random sample of 90 Americans found that 49 preferredAmerican-branded hotels. Assume that the actual proportion of American travelers in Europe who preferAmerican-branded hotels is 0.368. Hotels.com would like to set ΅= 0.05. Calculate the probability of a Type IIerror. Answer: 172)Historically, voter turnout for political elections in Texas have been reported to be 54%. You have been assignedby a polling company to test the hypothesis that voter turnout during the most recent election was higher than54%. You have collected a random sample of 90 registered voters from this elections and found that 54 actuallyvoted. Use the critical value approach to test this hypothesis with ΅=0.02. 50
173) Hotels.com would like to test the hypothesis that the proportion of American travelers in Europe that prefer an American-branded hotel equals 0.42. A random sample of 90 Americans found that 49 preferred American-branded hotels. Hotels.com would like to set ΅ = 0.05. Use the critical value approach to test this hypothesis. Answer: Because z p = 2.38 is greater than z ΅ /2 = 1.96, we reject the null hypothesis. Therefore, Hotels.com can conclude that the proportion of American travelers in Europe that prefer an American-branded hotel does not equal 0.42. 174) The Department of Labor would like to test the hypothesis that the average hourly wage for recent college graduates is less than \$20. A random sample of 24 recent college graduates averaged \$19.30 per hour with a standard deviation of \$3.20 per hour. The Department of Labor would like to set ΅ = 0.10. Use the p -value approach to test this hypothesis. Answer: The p -value is between 0.10 and 0.20. Because the p -value L ΅ , we fail to reject the null hypothesis. Therefore, the Department of Labor cannot conclude that the average hourly wage for recent college graduates is less than \$20. 51
Answer Key Testname: C9 1) D 2) C 3) C 4) B 5) C 6) A 7) A 8) C 9) B 10) D 11) B 12) A 13) A 14) C 15) D 16) A 17) B 18) B 19) C 20) C 21) A 22) A 23) C 24) B 25) C 26) A 27) D 28) C 29) A 30) D 31) B 32) B 33) D 34) A 35) A 36) B 37) B 38) A 39) D 40) C 41) D 42) A 43) A 44) A 45) A 46) A 47) B 48) A 49) A 50) A 52
Answer Key Testname: C9 51) B 52) C 53) B 54) A 55) C 56) B 57) C 58) C 59) C 60) A 61) B 62) B 63) B 64) C 65) D 66) C 67) D 68) A 69) D 70) A 71) B 72) C 73) D 74) A 75) A 76) D 77) B 78) D 79) C 80) B 81) C 82) A 83) D 84) C 85) B 86) B 87) C 88) B 89) B 90) A 91) C 92) A 93) B 94) A 95) D 96) B 97) C 98) A 99) A 100) C 53
Answer Key Testname: C9 101) A 102) C 103) B 104) C 105) TRUE 106) TRUE 107) FALSE 108) TRUE 109) FALSE 110) FALSE 111) TRUE 112) FALSE 113) FALSE 114) TRUE 115) FALSE 116) FALSE 117) TRUE 118) FALSE 119) TRUE 120) TRUE 121) TRUE 122) FALSE 123) TRUE 124) TRUE 125) FALSE 126) TRUE 127) TRUE 128) TRUE 129) FALSE 130) TRUE 131) TRUE 132) TRUE 133) FALSE 134) FALSE 135) TRUE 136) FALSE 137) FALSE
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