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We have that $H_0:p=0.35$ and H_A:$p>0.35$ \[z-score=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.42-0.35}{\sqrt{\frac{0.35*0.65}{100}}}=1.47\]  and p-value\[Prob(\hat{p} > 0.42 \vert p=0.35=prob(z>1.47)=1-0.9292=0.0708\]  \subsection{Decision based on the p-value}  \begin{itemize}  \item When the data are consistent with the model from the null hypothesis, the p-value is high and we are unable to reject the null hypothesis.  \begin{itemize}  \item In that case, we have to“retain” the null hypothesis.  \item We can't claim to have proved it; instead we fail to reject the null hypothesis and conclude that the difference we observed between the null hypothesis (p = 0.35) and the observed outcome ($\hat{p}$ = 0.42) is due to natural sampling variability (or chance).  \end{itemize}  \end{itemize}  \begin{itemize}  \item If the p-value is low enough, we reject the null hypothesis since what we observed would be very unlikely if in fact the null model was true.  \begin{itemize}  \item We call such results statistically significant  \end{itemize}  \end{itemize}