Question

Thirty veterinary clinics in Oklahoma gave an average of 210 shots per week. Assume that σ is known to be 29 shots per week. For a 95% confidence interval, find:

a) Margin of Error

b) Confidence Interval

Answer 1

Solution :

Given that,

Point estimate = sample mean =     = 210

Population standard deviation =     =29

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   = Z_0.025 = 1.96   ( Using z table )

Margin of error = E =   Z / 2     * ( /n)
= 1.96* (29 / 30 )

= 10.3775
At 95% confidence interval estimate of the population mean
is,

- E <   < + E

210 - 10.3775 <   < 210+ 10.3775

199.6225 <   < 220.3775

( 199.6225,220.3775)