Let’s analyze and solve the problems in the image. The task requires calculating the P-values for given $z$-scores based on the hypothesis tests.

Problem 1:

Hypotheses:

$H_0: p = 0.23 \quad \text{vs.} \quad H_1: p > 0.23$ The test statistic is $z = 2.05$.

Solution:

Since this is a right-tailed test, the $P$-value is the area to the right of $z = 2.05$ under the standard normal curve. Use the standard normal distribution table or a calculator to find:

$P(Z > 2.05) = 1 - P(Z \leq 2.05)$

From the standard normal table: $P(Z \leq 2.05) \approx 0.9798$

So: $P(Z > 2.05) = 1 - 0.9798 = 0.0202$

Thus, the $P$-value is 0.0202.

Problem 2:

Hypotheses:

$H_0: p = 0.56 \quad \text{vs.} \quad H_1: p < 0.56$ The test statistic is $z = -1.77$.

Solution:

Since this is a left-tailed test, the $P$-value is the area to the left of $z = -1.77$. From the standard normal table:

$P(Z < -1.77) \approx 0.0384$

Thus, the $P$-value is 0.0384.

Problem 3:

Hypotheses:

$H_0: p = 0.36 \quad \text{vs.} \quad H_1: p \neq 0.36$ The test statistic is $z = -2.13$.

Solution:

Since this is a two-tailed test, we calculate the area in both tails. First, find the area to the left of $z = -2.13$:

$P(Z < -2.13) \approx 0.0166$

For a two-tailed test, double this value: $P \text{-value} = 2 \cdot P(Z < -2.13) = 2 \cdot 0.0166 = 0.0332$

Thus, the $P$-value is 0.0332.

Summary of Results:

1. Problem 1: $P$-value = 0.0202
2. Problem 2: $P$-value = 0.0384
3. Problem 3: $P$-value = 0.0332

Would you like a detailed explanation for any part of these calculations or the reasoning behind them?

Here are 5 related questions for further exploration:

1. What is the significance level ($\alpha$) for rejecting a null hypothesis?
2. How does the $z$-score relate to the critical values in hypothesis testing?
3. Why is the $P$-value doubled in a two-tailed test?
4. How do you compute $P$-values for very large or small $z$-scores not found in standard tables?
5. What are the differences between one-tailed and two-tailed hypothesis tests?

Tip: Always compare the $P$-value with the significance level ($\alpha$) to determine whether to reject $H_0$.