To calculate the test statistic and p-value based on the sample data provided, we first need to define the hypotheses and the proportions involved.
Given Data
- Sample size (\(n\)) = 300
- Proportion of cat owners in the sample (\(\hat{p}\)) = 0.93 (93%)
Null and Alternative Hypotheses
- \(H_0\): The proportion of cat owners in the population is \(p_0\).
- \(H_a\): The proportion of cat owners in the population is not \(p_0\).
For this calculation, we'll need an expected population proportion (\(p_0\)). If not given, let's assume \(p_0 = 0.90\) as a hypothetical value (you may adjust this based on your specific context).
Step 1: Compute the Test Statistic
The test statistic for a proportion is calculated using the formula:
\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \]
Let's fill in the values:
- \(\hat{p} = 0.93\)
- \(p_0 = 0.90\)
- \(n = 300\)
Calculate the standard error:
\[ \text{Standard Error} = \sqrt{\frac{p_0 (1 - p_0)}{n}} = \sqrt{\frac{0.90 \times (1 - 0.90)}{300}} = \sqrt{\frac{0.90 \times 0.10}{300}} = \sqrt{\frac{0.09}{300}} \approx 0.1732 \]
Now plug it into the formula for \(z\):
\[ z = \frac{0.93 - 0.90}{0.1732} \approx \frac{0.03}{0.1732} \approx 0.173 \]
Rounding to two decimals:
\[ z \approx 0.17 \]
Step 2: Calculate the p-value
Since we are conducting a two-tailed test, we will find the p-value associated with the calculated \(z\) score. We can use standard normal distribution tables or a calculator.
Using a \(z\)-table or normal distribution calculator:
- For \(z = 0.17\): The one-tailed p-value is approximately 0.4325 (from tables or calculators).
Since this is a two-tailed test: \[ \text{p-value} = 2 \times \text{(p-value from the z-table for } |z| \text{)} \]
So:
\[ \text{p-value} = 2 \times 0.4325 \approx 0.865 \]
Final Answer
- The test statistic is: 0.17
- The p-value is: 0.87
(Note: Make sure to adjust \(p_0\) according to your context if needed.)