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The problem in the image asks to perform a hypothesis test where the null hypothesis \( H_0: p = 0.5 \) and the alternative hypothesis \( H_a: p > 0.5 \), using a sample proportion \( \hat{p} = 0.55 \) for different sample sizes. The goal is to calculate the p-value and determine whether the results are statistically significant at the 5% significance level.

### Steps to solve:
We are performing a one-tailed z-test for proportions for each case. The formula for the z-score is:

\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}
\]

Where:
- \( \hat{p} \) is the sample proportion (0.55),
- \( p_0 \) is the hypothesized proportion under \( H_0 \) (0.5),
- \( n \) is the sample size,
- The p-value is found by looking up the z-score in a standard normal distribution table.

#### (a) Sample size = 100
- \( \hat{p} = 0.55 \), \( p_0 = 0.5 \), \( n = 100 \)
- Calculate the z-score and then find the p-value.

#### (b) Sample size = 500
- \( \hat{p} = 0.55 \), \( p_0 = 0.5 \), \( n = 500 \)
- Similarly, calculate the z-score and p-value.

#### (c) Sample size = 1000
- \( \hat{p} = 0.55 \), \( p_0 = 0.5 \), \( n = 1000 \)
- Again, calculate the z-score and p-value.

Finally, in part (d), compare the results to determine which sample size provides the strongest evidence for the alternative hypothesis. Generally, a larger sample size tends to yield more precise estimates, leading to stronger evidence for or against the null hypothesis.

Would you like me to perform these calculations for you?

In this exercise, test H₀: p = 0.5 vs Hₐ: p > 0.5, with p̂ = 0.55 for different sample sizes.

This exercise tests the hypothesis H₀: p = 0.5 vs Hₐ: p > 0.5 with a sample proportion of 0.55 across different sample sizes. It involves calculating the z-score and p-value for each sample size to determine statistical significance at the 5% level.

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