Question

The National Association for Gardening says that the average vegetable garden size has a standard deviation of 247 sq ft. A random sample of 210 households shows an average size of 642 sq ft. Find a 90% confidence interval for the average size vegetable garden.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 642

Population standard deviation =    = 247

Sample size = n =210

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= 1.645 * (247 /  210 )

= 28.0384
At 90% confidence interval estimate of the population mean
is,

- E < < + E

642 - 28.0384 <   < 642 + 28.0384

613.9616<   < 670.0384

(  613.9616 ,670.0384 )