Answer
Since $P(ρ ̂\geq0.208)\lt0.05$, it is not unusual.
Work Step by Step
18% = 0.18
$μ_{ρ ̂}=0.18$
$σ_{ρ ̂}=\sqrt {\frac{p(1-p)}{n}}=\sqrt {\frac{0.18(1-0.18)}{250}}=0.0243$
The proportion in the sample: $ρ ̂=\frac{52}{250}=0.208$
Let's find the z-score for 0.208:
$z=\frac{ρ ̂-μ_{ρ ̂}}{σ_{ρ ̂}}=\frac{0.208-0.18}{0.0243}=1.15$
According to Table V, the area of the standard normal curve to the left of z-score equal to 1.15 is 0.8749.
But, we want the area of the standard normal curve to the right of z-score equal to 1.15:
$P(ρ ̂\geq0.208)=1-0.8749=0.1251$
