Given that [tex]\bar{x}=12.8\%[/tex] and [tex]\sigma=5.0\%[/tex], given that the population mean is [tex]\mu=11\%[/tex] and the sample size is n = 16, the test statistic is given by
[tex]z= \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} \\ \\ = \frac{12.8-11}{5/\sqrt{16}} \\ \\ = \frac{1.8}{5/4} = \frac{1.8}{1.25} \\ \\ =1.44 \\ \\ \Rightarrow p-value=p(z\ \textgreater \ 1.44) \\ \\ =1-P(z\ \textless \ 1.44)=1-0.92507 \\ \\ =0.07493[/tex]
Since the test is a two tailed test, the p-value is 2(0.07493) = 0.14986
Since the p-value is greater the significant level of α = 0.01, thus, we fail to reject the null hypothesis and conclude that there is no sufficient evidence that the percentage of wheat crop lost to hail in that county is different from the national mean of 11%.
