Question

Reserve Problems Chapter 10 Section 1 Problem 1 Your answer is partially correct. Try again. Consider the hypothesis test Ho: Mi - M2 = 0) against H1 : 41 - 42 7 0 samples below: I 35 38 31 32 33 30 31 29 39 38 30 38 36 31 39 32 36 39 II 32 29 33 32 31 29 30 36 31 34 30 30 30 32 34 34 Variances: 0 = 3.9, 02 = 0.6. Use a = 0.05. (a) Test the hypothesis and find the P-value. Find the test statistic. Round your answers to four decimal places (e.g. 98.7654). X 2.7818 ZO P-value = 0.0054 (b) Explain how this test could be conducted with a confidence interval. Round your answers to three decimal places (e.g. 98.765). 3.219 <M1-42 We are 95% confident that value 1 exceeds value 2 by between 1.963 (c) What is the power of the test for the true difference in means of 2? Round your answer to three decimal places (e.g. 98.765). The power of the test is 1-B =

Answer 1

• Here, we are to test:
  
  HO : μι - μ2 = 0 v/s H, :μι - μ2 40
  .
  
• The test statistic under known variance is given by:
  
  11 22 T = -2 + 05 112 ni
  
  Now,
  
  ni 18
  
  $n_2=16$
  $\overline{x_1}=\frac{\sum_{i=1}^{18}x_{1_i}}{18}=\frac{617 }{18}=34.27777778$
  $\overline{x_2}=\frac{\sum_{i=1}^{16}x_{2_i}}{16}=\frac{507 }{16}=31.6875$
  
  Now, putting in the values, we get,
  
  34.27777778 – 31.6875 T= 2.781067936 2.7811 0.62 3.92 18 + 16
  
  
  
• P-value
  The p-value is calculated as:
  
  Now,
  
  $=1-P(Z<2.7811)$
  $=1-\Phi(2.7811)$
  $=1-0.99795477$
  $=0.00204523$
  Also,
  
  $=\Phi(-2.7811)$
  $=0.00204523$
  
  Hence, the p-value is:
  
  
  
• The 95% Confidenc Interval is given by:
  
  ଏ o ଏ , os, T -4 (1 – 2 – S || + + * TO. 025, 11 - 12 + * T0.025, 1 2 1
  
  
  Now,
  
  T0.025 = 1.960
  
  
  Putting in the given values, we get,
  $\tiny (34.27777778 -31.6875 -\sqrt{\frac{3.9^2}{18}+\frac{0.6^2}{16}}*1.960,34.27777778 -31.6875 +\sqrt{\frac{3.9^2}{18}+\frac{0.6^2}{16}}*1.960)$
  $\tiny =(0.764740049 ,4.415815507 )$
  $\tiny =(0.765 ,4.416)$
  
  
• We reject $\tiny H_0$ at 5% level of significance iff
  $\tiny |T|>1.960$
  $\Rightarrow \frac{|\overline{X_1}-\overline{X_2}|}{\sqrt{\frac{3.9^2}{18}+\frac{0.6^2}{16}}}>1.960$
  $\Rightarrow |\overline{X_1}-\overline{X_2}|>1.825537729$
  
  Now, power of the test is given by:
  $1-P_{\mu=2}(|\overline{X_1}-\overline{X_2}|<1.825537729 )$
  $=1-(P_{\mu=2}(\overline{X_1}-\overline{X_2}<1.825537729 )-P_{\mu=2}(\overline{X_1}-\overline{X_2}<-1.825537729 ))$
  $\tiny =1-(P_{\mu=2}(\frac{\overline{X_1}-\overline{X_2}-2}{\sqrt{\frac{3.9^2}{18}+\frac{0.6^2}{16}}}<\frac{1.825537729-2}{\sqrt{\frac{3.9^2}{18}+\frac{0.6^2}{16}}} )-P_{\mu=2}(\frac{\overline{X_1}-\overline{X_2}-2}{\sqrt{\frac{3.9^2}{18}+\frac{0.6^2}{16}}}<\frac{-1.825537729-2}{\sqrt{\frac{3.9^2}{18}+\frac{0.6^2}{16}}} ))$
  $1-(P_{\mu=2}(Z<-0.187312508 )-P_{\mu=2}(Z<-4.107312508 ))$
  $1-(\Phi(-0.187312508 )-\Phi(-4.107312508 ))$
  $=1-(0.42571271-0.00002002)$
  $=0.57430731$
  $=0.574$
  
  

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