To find the $P$-value for this hypothesis test, let's go through the steps systematically:

1. Identify the parameters:

   • Sample size ($n$) = 1023
   • Number of successes ($X$) = 537
   • Sample proportion ($\hat{p}$) = $\frac{X}{n} = \frac{537}{1023} \approx 0.5249$

2. Set up the hypotheses:

   • Null hypothesis $H_0: p = 0.5$
   • Alternative hypothesis $H_1: p > 0.5$

3. Calculate the test statistic (z-score): Since we are conducting a one-sample proportion test, we can use the following formula for the test statistic: $z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}}$ where $p_0 = 0.5$ (the hypothesized proportion).

   Substituting the values: $z = \frac{0.5249 - 0.5}{\sqrt{\frac{0.5 \times (1 - 0.5)}{1023}}}$ $z \approx \frac{0.0249}{\sqrt{\frac{0.25}{1023}}}$ $z \approx \frac{0.0249}{0.0156} \approx 1.5962$

4. Find the P-value: Since this is a right-tailed test (as per $H_1: p > 0.5$), we will look up the probability corresponding to $z = 1.5962$ in the standard normal distribution.

   Using standard normal tables or a calculator, we find that the area to the right of $z = 1.5962$ is approximately $P \approx 0.0554$.

Thus, the $P$-value for this test is approximately 0.0554 (rounded to four decimal places).

Would you like further clarification on any of these steps, or do you have any additional questions?

Here are some follow-up questions to expand your understanding:

1. How does the choice of significance level $\alpha$ impact the decision to reject or fail to reject the null hypothesis?
2. What are the consequences of rejecting $H_0$ when it is actually true?
3. How would a change in the sample size impact the $z$-score and $P$-value?
4. What is the interpretation of the $P$-value in the context of this problem?
5. How would the test differ if we were testing $H_1: p < 0.5$ instead of $H_1: p > 0.5$?

Tip: Remember that the $P$-value is the probability of observing a result as extreme as, or more extreme than, the sample result, assuming the null hypothesis is true.