Question

A researcher believes that 8% 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 437 pet dogs would differ from the population proportion by greater than 3%? Round your answer to four decimal places.

Answer 1

Here, p = 0.08 , n = 437

we need to find,

z = (Pcap- p)/sqrt(p*(1-p)/n)

margin of error = 0.08

standard error = sqrt(p*(1-p)/n)
= sqrt(0.08 *(1-0.08)/437)
= 0.013

a = 0.08 - 0.03 = 0.05
b = 0.08 + 0.03 = 0.11

= P(0.05 < x < 0.11)
= P((0.05 - 0.08)/0.013 < z < (0.11 - 0.08)/0.013)
= P(-2.3077 < z < 2.3077)
= P(z < 2.3077) - P(z < -2.3077)
by using z standard normal table we get,
= 0.9895 - 0.0105
= 0.9790

P(x > 0.03 ) = 1 - P(x< 0.03)

= 1 - 0.9790

= 0.0210