In a comparison between two. major brands of broiler chickens, 70 of 160 Perdue chickens in a SRS tested positive for food-borne illness bacteria. In

Question

In a comparison between two. major brands of broiler chickens, 70 of 160 Perdue chickens in a SRS tested positive for food-borne illness bacteria. In a SRS of 160 Tyson chickens, 132 tested positive. Wed

Answer 1

Solution:
Given: - For Perdue chikens, $n_{1} = 160, x_{1} = 70$
$p_{1} = s a \mp \leq \propto^{n} = \frac{x_{1}}{n_{1}} = \frac{70}{160} = 0.4375$
For Tyson chickens, $n_{2} = \frac{x_{2}}{n_{2}} = \frac{132}{160} = 0.825$
The problem is to test
$H_{0} : p_{1} = p_{2}$
Vs $H_{1} : p_{1} \neq q p_{2}$
where $p_{1} and p_{2}$ are population props of posotove results in Perdue and Tyson chikens respectively.
Hence it is the problem of generalition inference method and can be solved using two $\propto^{n}$ z method.
Because sample size of 160 is large.

In a comparison between two. major brands of broiler chickens, 70 of 160 Perdue chickens in a SRS tested positive for food-borne illness bacteria. In

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