Question

A forester studying the effects of fertilization on certain pine forests in the Southeast is interested in estimating the average diameters of pine trees. In studying diameters of similar trees for many years, he has discovered that these measurements ( in inches ) are normally distributed with mean 23 inches and standard deviation 4 inches.

a) IF the forester samples n=16 trees, find the probability that the sample mean will be less than 26 inches.

b. Suppose the population mean is unknown, and the mean diameters of the selected 16 trees is 21.5 inches, construct a 94% confidence interval for the population mean.

c. How many trees must be measured in order to obtain a 98% confidence interval with a width equal to .3 inches.

Answer 1

Solution :

a) Given that ,

mean = = 23

standard deviation = = 4

n = 16

= = 23

= / n = 4 / 16 = 1

P( < 26) = P(( - ) / < (26 - 23) /1 )

= P(z < 3 )

Using z table

= 0.9987

b) Given that,

Point estimate = sample mean = = 21.5

Population standard deviation =    = 4

Sample size = n = 16

At 94% confidence level

= 1 - 94%  

= 1 - 0.94 =0.06

/2 = 0.03

Z/2 = Z_{0.03 = 1.881}

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

= 1.881 * ( 4 /  16 )

= 1.88

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

  ± E

21.5 ± 1.88

( 19.62, 23.38 )

c) Given that,

Population standard deviation = = 4

Margin of error = E = width / 2 = 0.3 / 2 = 0.15

At 98% confidence level the z is,

= 1 - 98%

= 1 - 0.98 = 0.02

/2 = 0.01

Z_/2 = 2.326

sample size = n = [Z_/2* / E] ²

n = [2.326 * 4 / 0.15 ]²

n = 3847.30

Sample size = n = 3848 trees