Question

An animal shelter wants to estimate the mean number of animals housed daily. A sample of 30 days resulted in a standard deviation of 4.2 animals. If they want to find a 95 percent confidence interval, what is the margin of error for the number of animals housed daily?

1.661

1.503

1.568

0.829

Answer 1

Solution :

Given that,

Point estimate = sample mean =     =

Population standard deviation =    = 4.2

Sample size n =30

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E =   Z/2    * ( /n)
= 1.96 * ( 4.2 / 30 )

= 1.503