Question

Proportion problem. each year delta dental does a survey of parents concerning the tradition of the tooth fairy. a random sample of 1,058 parents were asked about how much money they leave for a tooth and questions concerning how their child responded to the concept of the tooth fairy. in one question, 508 parents indicated that their child saved the money they got for their tooth. you are asked to calculate a 99% confidence interval for this proportion. what is the boe for this confidence interval? i just want the answer. use 4 decimal places for your answer and use the proper rules of rounding.

Answer 1

The error bound for the 99% confidence interval of the proportion of parents who reported that their child saved the money from the tooth fairy is 0.0400, after calculating with a sample proportion of 0.4803 and a z-score of 2.576.

Step-by-step explanation:

To calculate a 99% confidence interval for the proportion of parents who reported that their child saved the money from the tooth fairy, we will use the sample proportion and the z-score for 99% confidence, which is approximately 2.576. The formula for a confidence interval is:

CI = p ± z*(sqrt(p(1-p)/n))

where p is the sample proportion (508/1058), n is the sample size (1058), and z is the z-score (2.576 for 99% confidence).

Firstly, calculate the sample proportion (p):

p = 508 / 1058 = 0.4803

Next, calculate the standard error (SE):

SE = sqrt(p(1-p)/n) = sqrt(0.4803(1-0.4803)/1058) = 0.0155

Then, calculate the error bound (EB):

EB = z * SE = 2.576 * 0.0155 = 0.0400

The 99% confidence interval is therefore:

CI = 0.4803 ± 0.0400

So, the error bound for this confidence interval is 0.0400, rounded to four decimal places.