1.
+
+
+Suppose we have a coin that weβre told is not fair and has a probability of \(\theta = 0.6\) of coming up heads. We suspect the probability is even higher than this.
+
+Throughout the following parts, do not use a normal approximation.
+
+(a)
+We flip the coin 10 times and see 9 heads. Should we use a 1-tailed test or a 2-tailed test? Find the \(p\)-value. Do we have strong enough evidence to reject the null hypothesis at a 0.05 significance level?
+
+Solution.
+
+
+
+Using a 1-tailed test, the \(p\)-value is \(\Pr( \geq 9 \text{ heads})\text{:}\)
+
+
+\begin{align*}
+\Pr(9 \text{ heads}) \amp = {10 \choose 9} (0.6)^9 (1 - 0.6)^{10 - 9} = 10(0.6)^9(0.4)^1 \approx 0.04 \\
+\Pr(10 \text{ heads}) \amp = {10 \choose 10} (0.6)^{10} (1 - 0.6)^{10 - 10} = 1(0.6)^{10}(0.4)^{0} \approx 0.006
+\end{align*}
+
+The \(p\)-value is the sum: \(\Pr(\geq 9 \text{ heads}) \approx 0.046 \lt 0.05\text{,}\) so we can reject the null hypothesis.
+
+(b)
+What is the minimum number of heads we would need to see in 10 flips to reject the null hypothesis?
+
+Solution.
+
+
+
+We saw in part (a) that at least 9 heads is enough to reject. Then:
+
+\begin{gather*}
+\Pr(8 \text{ heads}) = {10 \choose 8} (0.6)^8 (1 - 0.6)^{10 - 8} = 45(0.6)^8(0.4)^2 \approx 0.121
+\end{gather*}
+
+Adding to the previous probabilities, \(\Pr(\geq 8 \text{ heads}) \approx 0.046 + 0.121 = 0.167 \gt 0.05\text{.}\) So 8 heads would not be enough to reject the null hypothesis, therefore the minimum number of heads to reject would be 9.
+
+(c)
+What is the power of the test if the true value of the parameter is \(\theta = 0.8\text{?}\)
+
+Solution.
+
+
+
+With \(\theta = 0.8\text{:}\)
+
+
+\begin{align*}
+\Pr(9 \text{ heads}) \amp = {10 \choose 9} (0.8)^9 (1 - 0.8)^{10 - 9} = 10(0.8)^9(0.2)^1 \approx 0.268 \\
+\Pr(10 \text{ heads}) \amp = {10 \choose 10} (0.8)^{10} (1 - 0.8)^{10 - 10} = 1(0.8)^{10}(0.2)^{0} \approx 0.107
+\end{align*}
+
+So the power of the test is \(0.268 + 0.107 = 0.375\text{.}\)
+
+
+
