Question

A researcher believes that 7% of pet dogs in Europe are Labradors.

If the researcher is right, what is the probability that the proportion of Labradors in a sample of 590 pet dogs would differ from the population proportion by more than 3%? Round your answer to four decimal places.

Answer 1

Solution:

Given ,

p = 7% = 0.07(population proportion)

1 - p = 0.93

n = 590 (sample size)

Let be the sample proportion.

The sampling distribution of is approximately normal with

mean = =  p = 0.07

SD =   =     

=  

=  0.01050423643

Now ,

P( will differ from p by more than 3%)

=  P( will differ from p by more than 0.03)

= 1 - P( will differ from μ by less than 0.03)

= 1 - P(p - 0.03 <   < p + 0.03 )

= 1 - P( 0.07 - 0.03 < < 0.07 + 0.03)

= 1 - P(0.04 < < 0.10)

= 1 - { P( < 0.10) - P( < 0.04) }

= 1 - { - }

= 1 - { P(Z <(0.10 - 0.07)/0.01050423643) - P(Z <(0.04 - 0.07)/0.01050423643) }

= 1 - { P(Z < 2.856) - P(Z < -2.856) }

= 1 - { 0.9979 - 0.0021} .. (use z table)

= 1 - 0.9958

= 0.0042