Question

A study is being conducted by a group of insurance brokers to analyze whether or not pet owners who do not have pet insurance would opt for medical treatment for their pet if the cost of the treatment exceeded $5,000. A random stratified sample of 140 pet owners without pet insurance was collected, and out of those sampled, 25 said they would opt for treatment when presented with the choice if the treatment would cost over $5,000 in total. Calculate the margin of error for this result at the 5% level of significance, and construct the corresponding confidence interval for the true proportion of pet owners that do not have pet insurance who would be willing to pay over $5,000 for medical treatment for their pet.

Standard Normal Distribution Table

E=E=

Round to four decimal places

Enter 0 if normal approximation cannot be used

  < p <  < p <  

Round to four decimal places

Enter 0 if normal approximation cannot be used

Please provide correct answers. thanks.

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 25 / 140 = 0.1786

1 - = 1 - 0.1786 = 0.8214

Z/2 = Z0.025 = 1.96

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.96 (((0.1786 * 0.8214) / 140)

= 0.0634

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.1786 - 0.0634 < p < 0.1786 + 0.0634

( 0.1152 < p < 0.2420 )