Of 575 broiler chickens purchased from various kinds of food stores in different regions of a...

Of 575 broiler chickens purchased from various kinds of food stores in different regions of a country and tested for types of bacteria that cause food-borne illnesses, 66% were infected with a particular bacterium. Construct a 95% confidence interval and explain what your confidence interval says about chicken sold in the country.

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Expert Solution

Solution :

Given that,

n = 575

Point estimate = sample proportion = = 0.66

1 - = 1 - 0.66 = 0.34

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_{/ 2} * $\sqrt$ (( * (1 - )) / n)

= 1.96 * ( $\sqrt$ ((0.66 * 0.34) / 575)

= 0.039

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.66 - 0.039 < p < 0.66 + 0.039

0.621 < p < 0.699

(0.621 , 0.699)

A 95% confidence interval of chicken sold in the country is : (0.621 , 0.699)

orchestra answered 2 years ago

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