Question

Consider the hypothesis test Ho: M1 - H2 = 0 against H : M -12 € 0 samples below: I 36 38 32 33 33 30 31 29 38 38 31 37 37 32 39 30 35 39 11 32 29 33 33 32 29 30 37 31 34 30 29 30 32 34 35 Variances: 0 = 3.7,02 -1.9. 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). 20 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 3? Round your answer to three decimal places (e.g. 98.765). The power of the test is 1 - B = ISH-MS

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
  
  
  618 α1 18 =1 1, 18 – Ξ 34.33333333 18
  
  
  Σ121 12 510 t2 = – = 31.875. 16 16
  
  
  Putting in the values, we get:
  
  34.33333333 – 31.875 T= 2.475497843 = 2.4755 3.72 18 + 1.92 16
  
  
• p-value
  The p-value is calculated as:
  
  2 * min(P(Z > T), P(Z <-T))
  
  Now,
  
  P(Z > 2.4755
  
  
  =1- P(Z < 2.4755)
  
  
  =1 - Φ(2.4755)
  
  $\tiny =1-0.99334751$
  $\tiny =0.00665249$
  
  Also,
  $\tiny =P(Z<-2.4755 )$
  $\tiny =\Phi(-2.4755)$
  
  0.00665267
  
  
  Hence, the p-value is:
  $\tiny =2*0.00665249$
  $\tiny =0.01330534$
  $\tiny =0.0133$
  
  
• 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,
  $\tiny (34.33333333 -31.875 -\sqrt{\frac{3.7^2}{18}+\frac{1.9^2}{16}}*1.960,34.33333333 -31.875 +\sqrt{\frac{3.7^2}{18}+\frac{1.9^2}{16}}*1.960)$ $\tiny =(0.511923504, 4.404743162 )$
  $\tiny =(0.512, 4.405 )$
  
• We reject
  10
  at 5% level of significance iff
  
  T> 1.960
  
  $\tiny \Rightarrow \frac{|\overline{X_1}-\overline{X_2}|}{\sqrt{\frac{3.7^2}{18}+\frac{1.9^2}{16}}}>1.96$
  $\tiny \Rightarrow |\overline{X_1}-\overline{X_2}|>1.946409829$
  
  Now, power of the test is given by:
  $\tiny 1-P_{\mu=3}(|\overline{X_1}-\overline{X_2}|<1.946409829 )$
  $\tiny =1-(P_{\mu=3}(\overline{X_1}-\overline{X_2}<1.946409829)-P_{\mu=3}(\overline{X_1}-\overline{X_2}<-1.946409829 ))$ $\tiny =1-(P_{\mu=3}(\frac{\overline{X_1}-\overline{X_2}-3}{\sqrt{\frac{3.9^2}{18}+\frac{1.9^2}{16}}}<\frac{1.946409829-3}{\sqrt{\frac{3.9^2}{18}+\frac{1.9^2}{16}}})-P_{\mu=3}(\frac{\overline{X_1}-\overline{X_2}-3}{\sqrt{\frac{3.9^2}{18}+\frac{1.9^2}{16}}}<\frac{-1.946409829-3}{\sqrt{\frac{3.9^2}{18}+\frac{1.9^2}{16}}}))$
  $\tiny =1-(P_{\mu=3}(Z<-1.06094652 )-P_{\mu=3}(Z<-4.98094652 ))$
  $\tiny =1-(\Phi(-1.06094652 )-\Phi(-4.98094652 ))$
  $\tiny =1-(0.14436767-0.00000032)$
  $\tiny =0.85563265$
  $\tiny =0.856$
  

As per HOMEWORKLIB POLICY, we are to answer the first four subparts only.

I hope this clarifies your doubt. If you're satisfied with the solution, hit the Like button. For further clarification, comment below. Thank You. :)