Question

Consider the hypothesis test Ho : H1 – uz = 0 against H1 : Hi – M2 0 samples below: I 36 39 32 33 33 30 32 29 39 38 31 37 37 30 39 32 35 38 II 3330 34 33 32 29 30 36 32 34 30 30 31 33 33 34 Variances: j = 3.8, 02 = 2.4. 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). 2.1515 ZO P-value = 0.0314 (b) Explain how this test could be conducted with a confidence interval. Round your answers to three decimal places (e.g. 98.765). <H1-42 [4.4324 We are 95% confident that value 1 exceeds value 2 by between 0.2064 (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). -2.113 The power of the test is 1-B =

please answer it correctly because this is my last attempt thankyou!

Answer 1

Answer:

Given

H="1 - V2 0

$H_{a}=\mu_{1}-\mu_{2}\neq 0$

$\bar{x_{1}}=\frac{\sum x_{1i}}{n_{1}}=\frac{620}{18}=34.444$

Γ2 ΣΤ2 no 514 16 - 32.125

$\sigma_{1}=3.8\ \ ,\sigma_{2}=2.4\ \ \alpha=0.05$

(a)

$Test\ statistics : Z_{0}=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}$

  $Z_{0}=\frac{34.444-32.125}{\sqrt{\frac{3.8^{2}}{18}+\frac{2.4^{2}}{16}}}$

${\color{Green} Z_{0}=2.1511}$

$P-value = 2(1-\phi (z_{0}))$

${\color{Green} P-value= 0.0314}$

(c)

$\delta =3$

$\beta=\phi \left ( Z_{\frac{\alpha}{2}}-\frac{\delta }{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}} \right )-\phi \left ( -Z_{\frac{\alpha}{2}}-\frac{\delta }{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}} \right )$

$\beta=\phi \left ( 1.96-\frac{3 }{\sqrt{\frac{3.8^{2}}{18}+\frac{2.4^{2}}{16}}} \right )-\phi \left ( -1.96-\frac{3 }{\sqrt{\frac{3.8^{2}}{18}+\frac{2.4^{2}}{16}}} \right )$

$\beta=\phi (-0.8227)-\phi(-4.7428)$

$\beta=(1-0.79466)-0$

$\beta=0.20534$

$Power\ of\ test = 1-\beta=1-0.20534$

  $=0.79466$

  $\mathbf{{\color{Green} 1-\beta=0.795}}$