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# HWK 4 - If Ho is true mu = 238 sampling distribution is t...

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Homework #4 #27 Part 1: Business Issue:  Unemployment insurance benefits of Virginia residents versus rest of US Business Quest: Is the mean unemployment insurance benefits of Virigina Residents less than \$238? Data: 100 residents of Virginia will be examined, with and x-bar of 231 and Sigmus of x-bar of 80 Source: World Almanac 2003 Analysis: Normal Test For Mean Hypothesis 1) Test Statistic 2) x-bar the mean of 100 sample members Sig. Level 3) Reject Rule  4) Norminv (.05, 238, 80/10) = 224.84 Calculation 5) x-bar =  \$231  p-value =  0.19 Normdist (231,238, 8,1) = 0.191 Conclusion 6) Reject Ho  Business Ans: The sample mean of \$231 is insufficent evidence to be able to conclude that the mean unemployment insurance benefits of Virginia Residents is less than \$238 Part 2: Analysis: t-Test For Mean Hypothesis 1) Test Statistic 2) Calculate t = (x-bar -238 ) / (s / 100^.5)
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Unformatted text preview: If Ho is true mu = 238, sampling distribution is t distributed with Df = n-1 = 99 Sig. Level 3) Reject Rule 4) Tinv ( .1, 99 ) = 1.66 Calculations 5) x-bar = 231 s = 80 t = -0.88 (231-238)/(80/10) = -0.88 P-value = 0.1905 Tdist: ( .88, 99,1) = 0.19 Conclusion 6) Do not reject Ho Based on a sample of 100, with a sample mean of 231 and a sample standard deviation of 80, there is insufficent evidence to be able to conclude that the mean insurance benefits are less than \$238 in Virginia Ho: μ ≥ 238 Ha: μ < 238 If Ho is true (μ=\$238), x-bar is normal with (mean=\$238 and std error = 80/100^.5) Alpha = .05 Reject Ho if x-bar < 224.84 Ho: mu ≥ 238 Ha: mu < 238 Alpha = .05 Reject Ho if t < 1.66...
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