A drug company claims that 75% of patients suffering from depression recover when treated with a new drug. A doctor believes that the claim is not true and thr percentage who will recover is lower than 75%.
1)Set up the null and alternative hypothesis. MY ANSWER:
Let $p$ be the percent of patients who will recover.
Let $\hat p$ be the sample percentage of patients who will recover
The hypotheses: $H_O$: $p=0.75$, $H_\alpha$: $p<0.75$
2) The doctor selects a random sample of 225 patients and uses the following rule:
If the sample shows that more than 162 patients recovered, conclude that the claim is acceptable. Otherwise conclude that the claim is not acceptable.
Find the probability of a Type 1 error.
MY ANSWER:
We conclude that the doctor's claim is true when we reject $H_o$.
Let $X$ be the number of patients in the sample who recover.
Reject $H_o$ if X>162 (reject $H_o$ if $X/n > 162/225$). Hence reject $H_o$ if $\hat p > 162/225 = 0.72$. The probability of a Type 1 error is:
$\begin{split}\alpha &= P(\text{reject $H_o$ when $H_o$ true}) \\ &= P(\hat p>0.72\text{ when }p=0.75) \\ &=P\left(\frac{\hat p-p}{\sqrt {pq/n}} > \frac{0.72 -0.75}{\sqrt{(0.75)(0.25)/225}}\right) \\ &=P(Z> - 1247.08) \\&=\ldots \end{split}$
I am trying to solve this question and that is my process work. Can someone please tell me where I am going wrong?
\begin{align} \alpha &= \mathbb{P}(\hat{p} < 0.72 |p = 0.75)\\ &= \Phi\left( \frac{0.72-0.75}{\sqrt{0.25(1-0.25)/225}} \right)\\ &= \Phi(-1.04)\\ &= 1- \Phi(1.04)\\ &=1 - 0.85\\ &= 0.15 \end{align}