Question

Consider the hypothesis test Ho : 41 – 42 = 0 against Hp : H1 - 12 € 0 samples below: 1 36 40 31 33 33 30 31 30 39 38 30 37 36 30 39 32 36 39 11 34 29 33 33 32 29 30 38 32 34 30 30 29 33 34 35 Variances: 01 = 3.8, 02 = 2.9. 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). P-value (b) Explain how this test could be conducted with a confidence interval. Round your answers to three decimal places (e.g. 98.765). We are 95% confident that value 1 exceeds value 2 by between (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 = Ish-4251

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
  
  
  n2 16
  
  
  620 α1 18 =1 1, 18 – = 34.44444444 18
  
  $\tiny \overline{x_2}=\frac{\sum_{i=1}^{16}x_{2_i}}{16}=\frac{515 }{16}=32.1875$
  
  Putting in the values, we get:
  
  34.44444444 32.1875 T= 1.958604799 = 1.9586 3.82 18 + 2.92 16
  
  
  
• p-value
  The p-value is calculated as:
  
  2 * min(P(Z > T), P(Z <-T))
  
  Now,
  $\tiny P(Z>1.9586)$
  $\tiny =1-P(Z<1.9586)$
  $\tiny =1-\Phi(1.9586)$
  $\tiny =1-0.97492018$
  $\tiny =0.02507982$
  
  Again,
  $\tiny P(Z<-1.9586)$
  $\tiny =\Phi(-1.9586)$
  $\tiny =0.02507982$
  
  Hence, the p-value is $\tiny =2*0.02507982=0.05015964 =0.0502$
  
  
• The 95% Confidence 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,
  $\tiny (34.44444444 -32.1875-\sqrt{\frac{3.8^2}{18}+\frac{2.9^2}{16}}*1.96,34.44444444 -32.1875+\sqrt{\frac{3.8^2}{18}+\frac{2.9^2}{16}}*1.96,)$ $\tiny =(-0.001607722 ,4.515496611 )$
  $\tiny =(-0.002 ,4.515)$
  
  
• We reject
  10
  at 5% level of significance iff
  
  T> 1.960
  
  $\tiny \Rightarrow \frac{|\overline{X_1}-\overline{X_2}|}{\sqrt{\frac{3.9^2}{18}+\frac{2.9^2}{16}}}>1.960$
  $\tiny \Rightarrow |\overline{X_1}-\overline{X_2}|>2.258552167$
  
  Now, power of the test is given by:
  $\tiny 1-P_{\mu=2}(|\overline{X_1}-\overline{X_2}|<2.258552167)$
  $\tiny =1-(P_{\mu=2}(\overline{X_1}-\overline{X_2}<2.258552167)-P_{\mu=2}(\overline{X_1}-\overline{X_2}<-2.258552167))$
  $\tiny =1-(P_{\mu=2}(\frac{\overline{X_1}-\overline{X_2}-2}{\sqrt{\frac{3.8^2}{18}+\frac{2.9^2}{16}}}<\frac{2.258552167 -2}{\sqrt{\frac{3.8^2}{18}+\frac{2.9^2}{16}}})-P_{\mu=2}(\frac{\overline{X_1}-\overline{X_2}-2}{\sqrt{\frac{3.8^2}{18}+\frac{2.9^2}{16}}}<\frac{-2.258552167 -2}{\sqrt{\frac{3.8^2}{18}+\frac{2.9^2}{16}}}))$ $\tiny =1-(P_{\mu=2}(Z<0.224374825 )-P_{\mu=2}(Z<-3.695625175 ))$
  $\tiny =1-(\Phi(0.224374825 )-\Phi(-3.695625175 ))$
  $\tiny =1-(0.58876529-0.00010968)$
  $\tiny =0.41134439$
  $\tiny =0.411$
  
  

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