A builder claims that heat pumps are installed in $70\%$ of all homes being constructed today in Chennai. Would you agree with this claim if a random survey of new homes in this city showed that $8$ out of $15$ had heat pumps installed? Use a $10\%$ level of significance.
Here the null hypothesis(H$_0$) is $p = 0.7$ and alternative hypothesis(H$_A$) : $p \neq 0.7$. So, it is a both-ended test.
I want to compute the value of the type-1 error using the definition P ($A^c$ | $H_0$ is true), where A is the acceptance set, but I cannot formulate it correctly. How do we define the acceptance set?
I have completed the following two steps.
$\bullet$ Reject $H_0$ if $\big|T-10.5\big| > c$.
$\bullet$ P($A^c$ | $H_0$ is true ) = $\text{P} (T > c+10.5 \,\, \text{or} \,\, T < 10.5-c \,\, | \,\, p = 0.7 \,)= \text{P} (T > c+10.5 \,\,| \,\, p =0.7) + P(\,\, T < 10.5-c \,\, | \,\, p = 0.7 \,)$.
The above step is creating confusion. Any help would be appreciated.
For a two-sided test that controls the Type I error, the critical values for a sample size of $n = 15$ are integers $L$ and $U$ such that $$\Pr[X \le L \mid H_0] \le \alpha/2, \quad \Pr[X \ge U \mid H_0] \le \alpha/2,$$ where $\alpha = 0.10$ is the overall significance level of the test, and $$X \mid H_0 \sim \operatorname{Binomial}(n = 15, p = 0.7)$$ is the random number of homes with heat pumps installed in a sample of $15$ homes surveyed.
Thus we want to find the largest such $L$ and smallest such $U$. To this end, we calculate the cumulative distribution of $X$ under the null hypothesis:
$$\begin{align} \Pr[X = 0 \mid H_0] &\approx 1.43489 \times 10^{-8} \\ \Pr[X \le 1 \mid H_0] &\approx 5.16561 \times 10^{-7} \\ \Pr[X \le 2 \mid H_0] &\approx 8.71935 \times 10^{-6} \\ \Pr[X \le 3 \mid H_0] &\approx 9.16587 \times 10^{-5} \\ \Pr[X \le 4 \mid H_0] &\approx 0.000672234 \\ \Pr[X \le 5 \mid H_0] &\approx 0.00365252 \\ \Pr[X \le 6 \mid H_0] &\approx 0.0152425 \\ \Pr[X \le 7 \mid H_0] &\approx 0.0500125. \end{align}$$ We stop at $X \le 7$ because this is the first value that exceeds $\alpha/2 = 0.05$. So the largest $L$ we can choose is $L = 6$, even though $L = 7$ is just barely above $\alpha/2$.
We perform the same process to locate $U$: $$\begin{align} \Pr[X = 15 \mid H_0] &\approx 0.00474756 \\ \Pr[X \ge 14 \mid H_0] &\approx 0.0352676 \\ \Pr[X \ge 13 \mid H_0] &\approx 0.126828. \end{align}$$ Here we stop much sooner and obtain $U = 14$. Thus the rejection region is $$X \in \{0, 1, \ldots, 6\} \cup \{14, 15\}.$$ This region has a cumulative Type I error probability of $0.0152425 + 0.0352676 \approx 0.0505101$, which is not the full $0.10$ that we were allowed to "spend." This is due to the fact that the sample size is small and the number of homes must be a nonnegative integer, so the test statistic has a discrete distribution. Even if we allowed some of the overall significance level to be slightly unevenly distributed between the two tails of the test and included $X = 7$ in our rejection region, the Type I error would still fall short of what we could allocate.