Answer:
a.
[tex]\bar p_1=0.05\\\bar p_2=0.067[/tex]
b-Check illustration below
c.(-0.0517,0.0177
Step-by-step explanation:
a.let [tex]p_1 \& p_2[/tex] denote processes 1 & 2.
For [tex]p_1[/tex]: T1=10,n1=200
For [tex]p_2[/tex]:T2=20,n2=300
Therefore
[tex]\bar p_1=\frac{t_1}{N_1}=\frac{10}{200}=0.05\\\bar p_2=\frac{t_2}{N_2}=\frac{20}{300}=0.067[/tex]
b. To test for hypothesis:-
i.
[tex]H_0:p_1=p_2\\H_A=p_1\neq p_2\\\alpha=0.05[/tex]
ii.For a two sample Proportion test
[tex]Z=\frac{\bar p_1-\bar p_2}{\sqrt(\bar p(1-\bar p)(\frac{1}{n_1}+\frac{1}{n_2})}\\[/tex]
iii. for [tex]\frac{\alpha}{2}=(-1.96,+1.96)[/tex] (0.5 alpha IS 0.025),
reject [tex]H_o[/tex] if[tex]|Z|>1.96[/tex]
iv. Do not reject [tex]H_o[/tex]. The noncomforting proportions are not significantly different as calculated below:
[tex]z=\frac{0.050-0.067}{\sqrt {(0.06\times0.94)\times \frac{1}{500}}}[/tex]
z=-0.78
c.[tex](1-\alpha).100\%[/tex] for the p1-p2 is given as:
[tex](\bar p_1-\bar p_2)\pm Z_0_._5_\alpha \times \sqrt \frac{ \bar p_1(1-\bar p_1)}{n_1}+\frac{\bar p_2(1-\bar p_2)}{n_2}\\\\=(0.05-0.067)\pm 1.645 \times \sqrt \ \frac{0.05+0.95}{200}+\frac{0.067+0.933}{300}\\[/tex]
=(-0.0517,+0.0177)
*CI contains o, which implies that proportions are NOT significantly different.
