*Identify*: Define \(p_1\) and \(p_2\) as follows:

\(p_1\text{:}\) the true proportion that would suffer a heart attack if given fish oil

\(p_2\text{:}\) the true proportion that would suffer a heart attack if given placebo

We will test the following hypotheses at the \(\alpha=0.10\) significance level.

\(H_0\text{:}\) \(p_1=p_2\) Fish oil and placebo are equally effective.

\(H_A\text{:}\) \(p_1 \lt p_2\) Fish oil is effective in reducing heart attacks.

*Choose*: Because we are testing whether two proportions equal each other, we choose the 2-proportion Z-test.

*Check*: We must verify that the difference of sample proportions can be modeled using a normal distribution. First we note that there are two randomly assigned treatments. Second, we calculate the pooled proportion as follows:

\begin{equation*}
\hat{p}_c = \frac{x_1+x _2}{n_1+n_2}=\frac{145 + 200}{12933 + 12938}=0.0133
\end{equation*}

We can now verify: \(12933(0.0133)\geq10\text{,}\) \(12933(1-0.0133)\geq10\text{,}\) \(12938(0.0133)\geq10\text{,}\) and \(12938(1-0.0133)\geq10\text{,}\) so both conditions are met.

*Calculate*: We will calculate the Z-statistic and the p-value.

\begin{gather*}
Z = \frac{\text{ point estimate } - \text{ null value } }{SE \text{ of estimate } }
\end{gather*}

The point estimate is the difference of sample proportions: \(\hat{p}_1-\hat{p}_2 = 0.0112 - 0.0155 = -0.0043\text{.}\)

The value hypothesized for the parameter in \(H_0\) is the null value: null value = 0.

The pooled proportion, calculated above, is: \(\hat{p}_c = 0.0133\text{.}\)

The \(SE\) of the difference of sample proportions, assuming \(H_0\) is true, is:

\(\sqrt{\hat{p}_c(1-\hat{p}_c)}\sqrt{\frac{1}{n_1} + \frac{1}{n_2}} = \sqrt{0.0133(1-0.0133)}\sqrt{\frac{1}{12933} + \frac{1}{12938}}=0.00142\text{.}\)

\begin{gather*}
Z = \frac{-0.0043 - 0}{0.00142} = -3.0
\end{gather*}

Because \(H_A\) uses a less than, meaning that it is a lower-tail test, the p-value is the area to the *left* of \(Z=-3.0\) under the standard normal curve. This area can be found using a normal table or a calculator. The area or p-value = \(0.0013\text{.}\)

*Conclude*: The p-value of 0.0013 is \(\lt 0.10\text{,}\) so we reject \(H_0\text{;}\) there is sufficient evidence that fish oil is effective in reducing heart attacks.