mont4e_sm_ch09_sec03 - 9-16 Section 9-39-44 a) α=0.01,...

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Unformatted text preview: 9-16 Section 9-39-44 a) α=0.01, n=20, the critical values are 861.2±b) α=0.05, n=12, the critical values are 201.2±c) α=0.1, n=15, the critical values are 761.1±9-45 a) α=0.01, n=20, the critical value = 2.539 b) α=0.05, n=12, the critical value = 1.796 c) α=0.1, n=15, the critical value = 1.345 9-46 a) α=0.01, n=20, the critical value = -2.539 b) α=0.05, n=12, the critical value = -1.796 c) α=0.1, n=15, the critical value = -1.345 9-17 9-47 a) 05.*2025.*2≤≤pthen 1.05.≤≤pb) 05.*2025.*2≤≤pthen 1.05.≤≤pc) 4.*225.*2≤≤pthen 8.5.≤≤p9-48 a) 05.025.≤≤pb) 025.105.1−≤≤−pthen 975.95.≤≤pc) 4.25.≤≤p9-49 a) 025.105.1−≤≤−pthen 975.95.≤≤pb) 05.025.≤≤pc) 25.14.1−≤≤−pthen 75.6.≤≤p9-50 a. 1) The parameter of interest is the true mean interior temperature life, μ. 2) H: μ= 22.5 3) H1: μ≠22.5 4) α= 0.05 5)nsxt/μ−=6) Reject Hif |t| > tα/2,n-1where tα/2,n-1= 2.776 7) 22.496=x, s= 0.378 n=5 00237.5/378.5.22496.22−=−=t8) Since –0.00237 >- 2.776, we cannot reject the null hypothesis. There is not sufficient evidence to conclude that the true mean interior temperature is not equal to 22.5 °C at α= 0.05. 2*0.4 <P-value< 2* 0.5 ; 0.8 < P-value <1.0 b.) The points on the normal probability plot fall along the line. Therefore, there is no evidence to conclude that the interior temperature data is not normally distributed. 21.522.523.5151020304050607080909599DataPercentNormal Probability Plot for tempML Estimates - 95% CIMeanStDev22.4960.338384ML Estimates9-18 c.) d= 66.378.|5.2275.22|||=−=−=σμμσδUsing the OC curve, Chart VII e) for α= 0.05, d= 0.66, and n = 5, we get β≅0.8 and power of 1−0.8 = 0.2. d) d= 66.378.|5.2275.22|||=−=−=σμμσδUsing the OC curve, Chart VII e) for α= 0.05, d= 0.66, and β≅0.1 (Power=0.9), 40=n. e) 95% two sided confidence interval ⎟⎟⎠⎞⎜⎜⎝⎛+≤≤⎟⎟⎠⎞⎜⎜⎝⎛−nstxnstx4,025.4,025.μ965.22027.225378.776.2496.225378.776.2496.22≤≤⎟⎟⎠⎞⎜⎜⎝⎛+≤≤⎟⎟⎠⎞⎜⎜⎝⎛−μμWe cannot conclude that the mean interior temperature is not equal to 22.5 since the value is included inside the confidence interval. 9-51 a. 1) The parameter of interest is the true mean female body temperature, μ. 2) H: μ= 98.6 3) H1: μ≠98.6 4) α= 0.05 5)nsxt/μ−=6) Reject Hif |t| > tα/2,n-1where tα/2,n-1= 2.064 7) 264.98=x, s = 0.4821 n=25 48.325/4821.6.98264.98−=−=t8) Since 3.48 > 2.064, reject the null hypothesis and conclude that the there is sufficient evidence to conclude that the true mean female body temperature is not equal to 98.6 °F at α= 0.05. P-value= 2* 0.001 = 0.002 b)9-19 Data appear to be normally distributed....
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This note was uploaded on 10/12/2009 for the course IND E 315 taught by Professor Kailashkapur during the Spring '09 term at University of Washington.

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mont4e_sm_ch09_sec03 - 9-16 Section 9-39-44 a) α=0.01,...

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