To test the given hypothesis, we will calculate the test statistic using the sample data and then find the corresponding p-value.
Given:
- Null Hypothesis \( H_0: p = 0.12 \)
- Alternative Hypothesis \( H_a: p < 0.12 \)
- Sample size \( n = 703 \)
- Number of successes \( X = 71 \)
- Significance level \( \alpha = 0.005 \)
Step 1: Calculate the sample proportion
The sample proportion \( \hat{p} \) is calculated as:
\[ \hat{p} = \frac{X}{n} = \frac{71}{703} \approx 0.101 \]
Step 2: Calculate the test statistic (Z-score)
The test statistic for a proportion is calculated using the formula:
\[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]
Where:
- \( p_0 = 0.12 \) (the value under the null hypothesis)
First, we calculate the standard error:
\[ \text{Standard Error} = \sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.12(1 - 0.12)}{703}} = \sqrt{\frac{0.12 \cdot 0.88}{703}} = \sqrt{\frac{0.1056}{703}} \approx \sqrt{0.000150} \approx 0.012247 \]
Now, substituting in the values:
\[ Z = \frac{0.101 - 0.12}{0.012247} \approx \frac{-0.019}{0.012247} \approx -1.5504 \]
Rounding to three decimal places, we have:
\[ \text{test statistic } Z \approx -1.550 \]
Step 3: Calculate the p-value
To find the p-value, we will look up the Z-score in the standard normal distribution table. The p-value for a one-tailed test (lower tail) is found using the cumulative distribution function (CDF).
Using standard normal distribution tables or a calculator for \( Z = -1.550 \):
The p-value is approximately \( 0.0606 \).
Final Answers:
- Test statistic: \( \text{test statistic} = -1.550 \)
- p-value: \( \text{p-value} = 0.0606 \)
Thus:
- Test statistic = -1.550
- p-value = 0.0606