The professional organization for private colleges and universities professors reported that more than 17% of professors attended a national convention in the past year. To test this claim, a researcher surveyed 200 professors and found that 45 has attended a national convention in the past year. At 𝛼 = 0.05, test the claim that this figure is correct using p -value method.

$H_0:p=p_0=0.17$

$H_1:p>0.17$

Test statistics

$T={\frac {{\frac m n}-p_0} {p_0*(1-p_0)}}*\sqrt{n}$ , where m - number of satisfiyung observations, n - sample size

In the given case we have

$T={\frac {{\frac {45} {200}}-0.17} {0.17*0.83}}*\sqrt{200}\approx 5.51$

Since the sample size is big(>30), then it is appropriate to use z-score as critical value C, so, the p-value would be

$p=P(Z>T)=P(Z>5.51)=1.8*10^{-8}$

Since $p<{\frac {\alpha} 2}\implies p<0.05$ , then it is appropriate to reject the null hypothesis and conclude that p is indeed greater than 17%