H_0:p=p_0=0.2H 0 ​ :p=p 0 ​ =0.2 H_1:p\not=0.2H 1 ​ :p  =0.2 Test statistics T={\frac {{\frac m n}-p_0} {p_0*(1-p_0)}}*\sqrt{n}T= p 0 ​ ∗(1−p 0 ​ ) n m ​ −p 0 ​ ​ ∗ n ​ , where m - number of satisfiyung observations, n - sample size In the given case we have T={\frac {{\frac {45} {250}}-0.2} {0.2*0.8}}*\sqrt{250}\approx -1.98T= 0.2∗0.8 250 45 ​ −0.2 ​ ∗ 250 ​ ≈−1.98 Since the sample size is big(>30), then it is appropriate to use z-score as critical value C, so, since we run two-tailed test P(Z>C)={\frac {0.01} 2}=0.005\implies C=2.58P(Z>C)= 2 0.01 ​ =0.005⟹C=2.58 Since T\in(-2.58,2.58)T∈(−2.58,2.58) , then we should accept the null hypothesis and conclude that, based on the data, there is no statisticsal evidence at \alpha=0.01α=0.01 to conclude that p\not=0.2p  =0.2