Step-by-step explanation:
To calculate the standardized test statistic and the p-value, we first need to compute the sample proportion (\( \hat{p} \)) and the standard error (\( SE \)):
\[ \hat{p} = \frac{x}{n} \]
\[ SE = \sqrt{\frac{p(1-p)}{n}} \]
Where:
- \( x \) is the number of students who typically walk to school in the sample (17)
- \( n \) is the sample size (100)
- \( p \) is the hypothesized proportion (0.13)
First, let's compute \( \hat{p} \):
\[ \hat{p} = \frac{17}{100} = 0.17 \]
Now, let's compute the standard error (\( SE \)):
\[ SE = \sqrt{\frac{0.13 \times (1-0.13)}{100}} \approx 0.0349 \]
Next, we calculate the z-score, which is the standardized test statistic:
\[ z = \frac{\hat{p} - p}{SE} = \frac{0.17 - 0.13}{0.0349} \approx 1.148 \]
Now, we need to find the p-value associated with this z-score. Since the alternative hypothesis is \( H_a: p > 0.13 \), we're interested in finding the probability that the proportion of students who walk to school is greater than 0.13. We'll use a standard normal distribution table or a calculator to find this probability.
Using a standard normal distribution table, the p-value corresponding to a z-score of 1.148 is approximately 0.1251.
Therefore, the correct answer is:
\[ z\text{-value} = 1.148; \text{p-value} \approx 0.1251 \]
None of the provided options exactly matches this calculation, but the closest one is:
\[ z\text{-value} = 1.19; \text{p-value} = 0.1170 \]
So, the closest option is:
\[ z\text{-value} = 1.19; \text{p-value} = 0.1170 \]
