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

Explanation:

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

`E=t_(\alpha/2)(\sigma)/(√(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)(2.5)/(√(22))\approx1.509`

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