Question

Adult entrance fees to amusement parks in the United States are normally distributed with a population standard deviation of 2.5 dollars and an unknown population mean. A random sample of 22 entrance fees at different amusement parks is taken and results in a sample mean of 61 dollars. Find the margin of error for a 99% confidence interval for the population mean.

Answer 1

Answer:  1.509

Step-by-step explanation:

The formula of Margin of Error for (n<30):-

$E=t_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

Given : Sample size : n= 22

Level of confidence = 0.99

Significance level : $\alpha=1-0.99=0.01$

Using the t-distribution table ,

Critical value : $t_{n-1, \alpha/2}=t_{21,0.005}= 2.831$

Standard deviation: $\sigma=\text{ 2.5 dollars }$

Then, we have

$E=( 2.831)\dfrac{2.5}{\sqrt{22}}\approx1.509$

Hence, the margin of error for a 99% confidence interval for the population mean =1.509