In a recent yearout of 109857 arrests for Federal offences 2

In a recent year,out of 109,857 arrests for Federal offences, 29.1% of those (or about 31,968 offences) were for drug offences. Use a 1 % significance level to test the claim that the drug offence rate is 30%

Solution

Formulating the null and alternatuve hypotheses,          
          
Ho:   p   =   0.3
Ha:   p   =/=   0.3
As we see, the hypothesized po =   0.3      
Getting the point estimate of p, p^,          
          
p^ = x / n =    0.290996477      
          
Getting the standard error of p^, sp,          
          
sp = sqrt[po (1 - po)/n] =    0.001382598      
          
Getting the z statistic,          
          
z = (p^ - po)/sp =    -6.512034414      
          
As this is a    2   tailed test, then, getting the p value,  
          
p =    3.70699E-11      
significance level =    0.01      

Comparing p and the significance value, we   REJECT THE NULL HYPOTHESIS.      

Thus, there is a significant evidence that the population proportion is not 30%. [CONCLUSION]