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

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