In: Statistics and Probability
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.
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 = Z0.025 = 1.96
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.96 * (((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)