Question

A survey given to consumers leaving a supermarket three days before Thanksgiving asked whether turkey would be part of the Thanksgiving meal. Of the 182 people surveyed, 26 said "no." Give a 95% confidence interval for the population proportion who claimed not to be eating turkey.

Answer 1

Answer:

(0.1392,0.1466)

Step-by-step explanation:

Given that a survey given to consumers leaving a supermarket three days before Thanksgiving asked whether turkey would be part of the Thanksgiving meal

People surveyed = 182

claimed not eating turkey = 26

Sample proportion = $\frac{26}{182} \\=0.1429$

Std error =$\sqrt{\frac{pq}{n} } \\=0.0260$

Margin of error = 1.96*SE=$0.00378$

Confidence interval = $(0.1429-0.0037,0.1429+0.0037)\\=(0.1392,0.1466)$