Answer :
Answer:
Test statistic |Z| = 0.1040 < 1.96 at 0.05 level of significance
Step-by-step explanation:
Step(i):-
Given that a report on the nightly news broadcast stated that 10 out of 108 households with pet dogs were burglarized
First proportion
[tex]p_{1} = \frac{x_{1} }{n_{1} } = \frac{10}{108} = 0.0925[/tex]
Given that a report on the nightly news broadcast stated that 20 out of 208 households without pet dogs were burglarized
Second proportion
[tex]p_{2} = \frac{x_{2} }{n_{2} } = \frac{20}{208} = 0.0961[/tex]
where
[tex]P = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1} +n_{2} }[/tex]
P = 0.09486
Q = 1 - P = 1 - 0.09486 = 0.90514
Step(ii):-
Test statistic
[tex]Z = \frac{p_{1}-p_{2} }{\sqrt{PQ(\frac{1}{n_{1} } +\frac{1}{n_{2} } )} }[/tex]
[tex]Z = \frac{0.0925-0.0961 }{\sqrt{0.09486 X 0.90514(\frac{1}{108 } +\frac{1}{208 } )} }[/tex]
Z = -0.1040
|Z| = |-0.1040|
|Z| = 0.1040
Critical value Z = 1.96 at 0.05 level of significance